Ted Kaczynski
The Mathematical Work of Ted Kaczynski
Introduction by Jørgen Veisdal
University of Michigan (1962–67)
University of California, Berkeley (1967–69)
An Advanced Explanation of His Breakthrough by Lara Pudwell
A family of 4 and 5digit solutions
Ted's Work as a Michigan PhD Student
1. June 1964  Another Proof of Wedderburn’s Theorem
A NOTE ON PRODUCT SYSTEMS OF SETS OF NATURAL NUMBERS
2. 1964  Distributivity and (−1)x = −x (Advanced Problem 5210)
3. 1964  Distributivity and (−1)x = −x (Advanced Problem 5210, with Solution by Bilyeu, R.G.)
4. 1965  Boundary Functions for Functions Defined in a Disk
Explanation by John D. Bullough
2. Boundary functions for homeomorphisms.
3. Boundary functions for continuous functions.
4. Boundary functions for Baire functions.
5. Boundary functions for measurable functions.
5. 1966  On a Boundary Property of Continuous Functions
Explanation by John D. Bullough
6. 1967  PhD thesis at University of Michigan  Boundary Functions
CHAPTER I. BOUNDARY FUNCTIONS FOR CONTINUOUS FUNCTIONS
4. Domain of the Boundary Function
5. Boundary Functions for Continuous Functions
CHAPTER II. BOUNDARY FUNCTIONS FOR DISCONTINUOUS FUNCTIONS
6. Boundary Functions for Baire Functions
7. Boundary Functions for LebesgueMeasurable Functions
Ted's Work as an Assistant Professor of Mathematics at the Uni. of California
7. 1968  Note on a Problem of Alan Sutcliffe
8. March 1969  Boundary Functions for Bounded Harmonic Functions
Explanation by John D. Bullough
9. July 1969  Boundary functions and sets of curvilinear convergence for continuous functions
Explanation by John D. Bullough
10. Nov 1969  The Set of Curvilinear Convergence…
Explanation by John D. Bullough
Ted's Work from his Parents Home in Illinois
Problem 786. [January, 1971] Proposed by T. J. Kaczynski, Lombard, Illinois.
I. Solution by Richard A. Gibbs, Hiram Scott College, Nebraska.
II. Solution by Richard L. Breisch, Pennsylvania State University.
Ted's Work as a Montana Hermit
13. FourDigit Numbers that Reverse Their Digits When Multiplied
Original PDFs

2. 1964 Distributivity and (−1)x = −x (Advanced Problem 5210).pdf

3. Distributivity and (−1)x = −x (Advanced Problem 5210, with Solution by Bilyeu, R.G.).pdf

4. 1965 Boundary Functions for Functions Defined in a Disk.pdf

6. 1967  PhD thesis at University of Michigan  Boundary Functions.pdf

8. March 1969  Boundary Functions for Bounded Harmonic Functions.pdf

9. July 1969  Boundary functions and sets of curvilinear convergence for continuous functions.pdf

11. Problem 787

13. Unknown Date  FourDigit Numbers that Reverse Their Digits When Multiplied.pdf

14. Unknown date  Handwritten Math equations and procedures.pdf
Introduction by Jørgen Veisdal
The Mathematics of Ted Kaczynski
Disclaimer: As should be fairly evident, this essay is in no way meant to glorify Ted Kaczynski. Rather, it was written with two goals in mind: 1. To orient to reality some of the myths of Kaczynski's "genius" and 2. To illustrate yet another example of a mathematician whose abstract endeavours ultimately defeated him.
Before terrorist Theodore John Kaczynski (1942) began sending mailbombs to faculty members at various American universities, he had a promising career in mathematics. In particular, between 1964–69, he published a total of six singleauthored research papers in renowned mathematical journals, including The American Mathematical Monthly and Proceedings of the American Mathematical Society.
The young Kaczynski did work in analysis, specifically geometric function theory in the narrow subfield of boundary values of continuous functions. The purpose of this article is to give an introduction to this work.
Education (1958–67)
Kaczynski grew up in Illinois, where he attended Sherman Elementary School and Evergreen Park Central Junior High school. At the age of 10 years old, his IQ was evaluated to be 167, and so he skipped the sixth grade (Chicago Tribune, 2017), an event later described as pivotal to his development (Chase, 2004):
“Previously he had socialized with his peers and was even a leader, but after skipping ahead he felt he did not fit in with the older children and was bullied.”
Harvard University (1958–62)
Kaczynski entered Harvard University in 1958 at the age of 16 years old. A mathematical prodigy since he was a child, he was described by other undergraduates as “shy”, “quiet” and “a loner” who “never talked to anyone“ (Song, 2012):
“He would just rush through the suite, go into his room, and slam the door […] When we would go into his room there would be piles of books and uneaten sandwiches that would make the place smell”
His personality notwithstanding, Kaczynski’s talent was however still recognized among his Harvard peers, one of which in 2012 stated:
“It’s just an opinion — but Ted was brilliant [...]. He could have become one of the greatest mathematicians in the country”
Kaczynski graduated Harvard with a B.A. in mathematics in 1962. When he graduated, his GPA was 3.12, scoring B’s in the History of Science, Humanities and Math, C in History and A’s in Anthropology and Scandinavian (Stampfl, 2006).
University of Michigan (1962–67)
With an IQ of 167, Kaczynski had been expected to perform better at Harvard. After graduating, he applied to the University of California at Berkeley, The University of Chicago and the University of Michigan. Although accepted at all three, he ended up choosing Michigan because the university offered him an annual grant of $2,310 and a teaching post. The “darling of the math department”, he would graduate from the University of Michigan in 1964 with a M.Sc. in mathematics and markedly improved grades — 12 A’s and five B’s, which he himself later attributed to the standing of the university:
“[My] memories of the University of Michigan are not pleasant […] The fact that I not only passed my courses (except one physics course) but got quite a few A’s shows how wretchedly low the standards were at Michigan”
Nonetheless, as the story goes, while there once a professor named George Piranian told his students — including Kaczynski — about an unsolved problem in boundary functions. Weeks later, Kaczynski came to his office with a 100page correct, handwritten proof. Kaczynski graduated with a Ph.D. in mathematics in 1967. His dissertation, entitled simply “Boundary Functions” regarded the same topic as his proof of Piranian’s problem. His doctoral committee consisted of professors Allen L. Shields, Peter L. Duren, Donald J. Livingstone, Maxwell O. Reade, ChiaShun Yin. Every professor approved it. His supervisor Shields later called his dissertation
“The best I have ever directed”
An additional testament to its quality was it being awarded the Sumner Myers Prize for the best mathematics thesis of the university, accompanying a prize of $100 and a plaque in the East Quad Residence Hall entrance listing his accomplishment. Of the complexity (or perhaps narrow implications) of his dissertation, one of the members of his dissertation committee, Maxwell Reade, said
“I would guess that maybe 10 or 12 men in the country understood or appreciated it”
Another, Peter Duren, stated
“He was really an unusual student”
Kaczynski at UCB in 1967 (Photo: Wikimedia Commons)
University of California, Berkeley (1967–69)
In late 1967, at 25 years old Kaczynski was hired as the youngestever assistant professor of mathematics at the University of California at Berkeley. There, he taught undergraduate courses in geometry and calculus, although with mediocre success. His student evaluations suggest that he was not particularly wellliked because he taught “straight from the textbook and refused to answer questions”.
He resigned on June 30th, 1969 without explanation.
Work (1964–69)
Wedderburn’s Theorem
Kaczynski’s only published paper relating to topics other than boundary functions was his first journal paper, written before he started his Ph.D. It is entitled:

Kaczynski, T.J. (1964). “Another proof of Wedderburn’s theorem”. The American Mathematical Monthly 71(6), pp. 652–653.
The paper concerned a 1905 result of Joseph H. M. Wedderburn that every finite skew field is commutative. His paper provided a grouptheoretic proof of the theorem, which had previously been proved at least seven times.
Boundary Functions
Kaczynski’s Ph.D. dissertation concerned boundary values of continuous functions and was entitled, simply

Kaczynski, T.J. (1967). Boundary Functions. Ann Arbor: University of Michigan.
Let H denote the set of all points in the Euclidean plane having positive ycoordinate, and let X denote the xaxis. If p is a point of X, then by an arc at p we mean a simple arc γ, having one endpoint at p, such that γ = {p} ⊆ H. Let f be a function mapping H into the Riemann sphere.
Boundary Functions By a boundary function for f we mean a function φ defined on a set E ⊆ X such that for each p ∈ E there exists an arc γ at p for whichlim (s p, s ∈ γ) f(z) = φ(p)
Kaczynski’s dissertation begins by reproving a theorem of J. E. McMillan which states that if f(H) is a a continuous function mapping H into the Riemann sphere, the the set of curvilinear convergence of F (the largest set on which a boundary function for f can be defined) is of a certain type. This proof also shows that if A is a set of the same type in X, then there exists a bounded continuous complexvalued function in H having A as its set of curvilinear convergence. The dissertation contains two additional new proofs related to boundary functions, and a list of problems for future research. Of the results, Professor Donald Rung later stated:
What Kaczynski did, greatly simplified, was determine the general rules for the properties of sets of points of curvilinear convergence. Some of those rules were not the sort of thing even a mathematician would expect.
Kaczynski would publish five journal papers related to the work from his dissertation between 1965–69:

Kaczynski, T.J. (1965). “Boundary functions for functions defined in a disk”. Journal of Mathematics and Mechanics. 14(4), pp. 589–612.

Kaczynski, T.J. (1966). “On a boundary property of continuous functions”. Michigan Math. J. 13, pp. 313–320.

Kaczynski, T.J. (1969). “The set of curvilinear convergence of a continuous function defined in the interior of a cube”. Proceedings of the American Mathematical Society 23(2), pp. 323–327.

Kaczynski, T.J. (1969). “Boundary functions and sets of curvilinear convergence for continuous functions”. Transactions of the American Mathematical Society. 141, pp. 107–125.

Kaczynski, T.J. (1969). “Boundary functions for bounded harmonic functions”. Transactions of the American Mathematical Society. 137, pp. 203–209.
The Distributivity Problem
The only other trace of Kaczynski in a mathematical journal is two notes in the American Monthly in 1964 and 65:

Kaczynski, T.J. (1964). “Distributivity and (−1)x = −x (Advanced Problem 5210)”. The American Mathematical Monthly. 71(6), pp. 689.

Kaczynski, T.J. (1965). “Distributivity and (−1)x = −x (Advanced Problem 5210, with Solution by Bilyeu, R.G.)”. The American Mathematical Monthly 72(6), pp. 677–678.
In the first note, Kaczynski proposes the following problem, concerning group theory:
Let K be an algebraic system with two binary operations (one written additively, the other multiplicatively), satisfying:1. K is an abelian group under addition, 2. K  {0} is a group under multiplication, and 3. x(y+z) = xy + xz for all x,y,z ∈ K.Suppose that for some n, 0=1+1+1....+1 (n times). Prove that, for all x ∈ K, (1)x = x.
In the second note, the solution to the problem is — somewhat dismissively — provided by R. G. Bilyeu:
The last part of the hypothesis is unnecessary. If z denotes 1, then z+z+zz = z(1+1+z) = z, so zz = 1. Now z(x+zx) = zx+x = x+zx, so either x+zx = 0 or z = 1. In either case zx = x.
Conclusion
Theodore J. Kaczynski was a very promising young undergraduate, graduate and postgraduate student in the 1960s. His work — although pertaining to vary narrow topics — was undoubtedly, technically, first rate.
As is the case however, elegance or complexity do not themselves raise the importance of problems, achievements or for that matter, mathematicians. As expressed by his fellow graduate student Professor Peter Rosenthal in a 1996 Toronto Star article (after Kaczynski was charged):
[The] topic was only of interest to a very small group of mathematicians and does not appear to have broader implications; thus, his work had little impact. Kaczynski might have quit mathematics because he was discouraged by the resultant lack of recognition.
In another 1996 article, in the Los Angeles Times article, Professor Donald Rung similarly expressed:
“The field that Kaczynski worked in doesn’t really exist today […]. He probably would have gone on to some other area if he were to stay in mathematics,” Rung said. “As you can imagine, there are not a thousand theorems to be proved about this stuff.”
An Advanced Explanation of His Breakthrough by Lara Pudwell
Original PDF: Digit Reversal Without Apology.pdf
Digit Reversal Without Apology
Lara Pudwell Rutgers University Piscataway, NJ 08854 lpudwell@math.rutgers.edu
In A Mathematician’s Apology [1] G. H. Hardy states, “8712 and 9801 are the only fourfigure numbers which are integral multiples of their reversals”; and, he further comments that “this is not a serious theorem, as it is not capable of any significant generalization.”
However, Hardy’s comment may have been shortsighted. In 1966, A. Sutcliffe [2] expanded this obscure fact about reversals. Instead of restricting his study to base 10 integers and their reversals, Sutcliffe generalized the problem to study all integer solutions of
k(ahn^{h} + a_{h1}n^{h}1 + • • • + a1 n + a_{0}) = a_{0}n^{h} + a1n^{h1} + • • • + a_{h1}n + ah
with n >2, 1 < k < n, 0 < ai < n — 1 for all i, a_{0} = 0, a_{h} = 0. We shall refer to such an integer a0...ah as an (h + 1)digit solution for n and write k(a_{h}, a_{h1}, ..., a_{1}, a_{0})_{n} = (a_{0}, a_{1}, ..., a_{h1}, a_{h})_{n}. For example, 8712 and 9801 are 4digit solutions in base n = 10 for k = 4 and k = 9 respectively. After characterizing all 2digit solutions for fixed n and generating parametric solutions for higher digit solutions, Sutcliffe left the following open question: Is there any base n for which there is a 3digit solution but no 2digit solution?
Two years later T. J. Kaczynski{1}[3] answered Sutcliffe’s question in the negative. His elegant proof showed that if there exists a 3digit solution for n, then deleting the middle digit gives a 2digit solution for n. Together with Sutcliffe’s work, this proved that there exists a 2digit solution for n if and only if there exists a 3digit solution for n.
Given the nice correspondence between 2 and 3digit solutions described by Sutcliffe and Kaczynski, it is natural to ask if there exists such a correspondence for higher digit solutions. In this paper, we will explore the relationship between 4 and 5digit solutions. Unfortunately, there is not a bijection between these solutions, but there is a nice family of 4 and 5 digit solutions which have a natural onetoone correspondence.
A second extension of Sutcliffe and Kaczynski’s results is to ask, “Is there any value of n for which there is a 5digit solution but no 4digit solution?” We will answer this question in the negative; and, furthermore, we will show that there exist 4 and 5digit solutions for every n >3.
An attempt at generalization
In the case of 3digit solutions, Kaczynski proved that if n + 1 is prime and k(a, b, c)n = (c, b, a)n is a 3digit solution for n, then k(a, c)n = (c, a)n is a 2digit solution. Thus, we consider the following:
Question 1. Let k(a, b, c, d, e)n = (e, d, c, b, a)n be a 5digit solution for n. If n + 1 is prime, then is k(a, b, d, e)_{n} = (e, d, b, a)_{n} a 4digit solution for n?
First, following Kaczynski, let p = n + 1. We have
k(an^{4} + bn^{3} + cn^{2} + dn + e) = en^{4} + dn^{3} + cn^{2} + bn + a. (1)
Reducing this equation modulo p, we obtain
k(a — b + c — d + e) = e — d + c — b + a = a — b + c — d + e mod p.
Thus, (k — 1)(a — b + c — d + e) = 0 mod p, and
p  (k — 1)(a — b+ c — d + e). (2)
Ifp  (k—1), then k—1 > p, which is impossible because k < n. Therefore, p  (a — b + c — d + e). But —2p < —2n < a — b + c — d + e < 3n < 3p, so there are four possibilities:
(i) a  b + c  d + e = p,
(ii) a  b + c  d + e = 0,
(iii) a  b + c  d + e = p, (iv) a  b + c  d + e = 2p.
Write a — b + c — d + e = fp, where f G {—1,0,1,2}. Substituting c = a + b + d  e + fp into equation 1 gives:
k[n^{2} (n^{2} — 1)a + n^{2}(n + 1)b + fpn^{2} + n(n + 1)d — (n^{2} — 1)e]
= n^{2}(n^{2} — 1)e + n^{2}(n + 1)d + fpn^{2} + n(n + 1)b — (n^{2} — 1)a.
After substituting for p, dividing by n + 1, and rearranging, one sees that k [an^{3} + (b — a + f)n^{2} + (d — e)n + e] = en^{3} + (d — e + f)n^{2} + (b — a)n + a. Indeed, this is a 4digit solution for n if f = 0, b — a > 0, and d — e > 0, but not necessarily a 4digit solution of the form conjectured in Question 1.
As in Kaczynski’s proof for 2 and 3digit solutions, it would be ideal if three of the four possible values for f lead to contradictions and the fourth leads to a “nice” pairing of 4 and 5digit solutions. Unlike Kaczynski, we now have the added advantage of exploring these cases with computer programs such as Maple. Experimental evidence suggests that the cases f = —1 and f = 2 are impossible. The cases f = 0 and f = 1 are discussed below.
A counterexample
Unfortunately, Kaczynski’s proof does not completely generalize to higher digit solutions. Most 5digit solutions do, in fact, yield 4digit solutions in the manner described in Question 1, but for sufficiently large n there are examples where (a, b, c, d, e)n is a 5digit solution but (a, b, d, e)n is not a 4digit solution.
A computer search shows that the smallest such counterexamples appear when n = 22:
7(2, 8, 3, 13, 16)22 = (16, 13, 3, 8, 2)22, 3(2, 16, 11, 5, 8)22 = (8, 5, 11, 16, 2)22.
However, there is no integer k for which k(2, 8, 13, 16)22 = (16, 13, 8, 2)22 or k(2, 16, 5, 8)22 = (8, 5, 16, 2)22. Note that 2 + 8 + 13  16 = 3 and 2 + 16 + 5  8 = 11; that is, both of these counterexamples to Question 1 occur when f = 0. The next smallest counterexamples are
3(3, 22, 15, 7, 11)30 = (11, 7, 15, 22, 3)30, 8(2, 13, 8, 16, 9)30 = (9, 16, 8, 13, 2)30,
which occur when f = 0 and n = 30.
A family of 4 and 5digit solutions
Although Kaczynski’s proof does not generalize entirely, there exists a family of 5digit solutions when f = 1 that has a nice structure.
Theorem 1. Fix n >2 and a > 0. Then
k(a, a  1, n  1, n  a  1, n  a)n = (n  a, n  a  1, n  1, a  1, a)n
is a 5digit solution for n if and only if a  (n  a).
Proof. We have
(n  a)n^{4} + (n  a  1)n^{3} + (n  1)n^{2} + (a  1)n + a an^{4} + (a  1)n^{3} + (n  1)n^{2} + (n  a  1)n + (n  a)
(n  a)(n^{4} + n^{3}  n  1) n  a a(n^{4} + n^{3}  n  1) a ^{,}
and the result is clear. □
Notice that
(a+ (a  1)) + ((n  a  1)  (n  a)) +p = 1 + 1 + (n+ 1) = n  1.
That is, this family of solutions occurs when f = 1. Moreover, this family follows the pattern described in Question 1; that is, for each 5digit solution described in Theorem 1, deleting its middle digit gives a 4digit solution.
Theorem 2. If
k(a, a  1,n  1, n  a  1,n  a)n = (n  a,n  a  1,n  1, a  1, a)n
is a 5digit solution for n, then
k(a, a  1, n  a  1,n  a)_{n} = (n  a,n  a  1, a  1, a)_{n}
is a 4digit solution for n.
Proof. By Theorem 1, na G N. Now
(n  a)n^{3} + (n  a  1)n^{2} + (a  1)n + a
an^{3} + (a  1)n^{2} + (n  a  1)n + (n  a)
(n  a)(n^{3} + n^{2}  n  1) n  a
a(n^{3} + n^{2}  n  1) a ^{.}
□
These 4digit solutions were first described by Klosinski and Smolarski [4] in 1969, but their relationship to 5digit solutions was not made explicit before now.
It is also interesting to note that 9801 and 8712, the two integers in Hardy’s discussion of reversals, are included in this family of solutions.
We conclude with the following corollary.
Corollary 1. There is a 4digit solution and a 5digit solution for every n > 3.
Proof. Let a = 1 in the statements of Theorem 1 and Theorem 2 above. □
Some open questions
We have shown that there is no n for which there is a 5digit solution but no 4digit solution. More specifically, we know that there are 4 and 5digit solutions for every n > 3.
Although Kaczynski’s proof does not generalize directly to 4 and 5digit solutions, it does bring to light several questions about the structure of solutions to the digit reversal problem.
First, it would be interesting to completely characterize 4 and 5digit solutions for n. Namely,
1. All known counterexamples to Question 1 occur when f = 0. Are there counterexamples for which f 6= 0? Is there a parameterization for all such counterexamples?
2. Theorems 1 and 2 exhibit a family of 4 and 5digit solutions for f = 1 with a particularly nice structure. To date, no other 4 or 5digit solutions are known for f = 1. Do such solutions exist?
More generally,
3. Solutions to the digit reversal problem have not been explicitly characterized for more than 5 digits. Do there exist analogous results to Theorems 1 and 2 for higher digit solutions?
A Maple package for exploring these questions is available from the author’s web page at http://www.math.rutgers.edu/~lpudwell/maple.html.
Acknowledgment
Thank you to Doron Zeilberger for suggesting this project.
References
Ted's Work as a Michigan PhD Student
1. June 1964  Another Proof of Wedderburn’s Theorem
Original PDF: 1. June 1964  Another Proof of Wedderburn’s Theorem.pdf
Kaczynski, T.J. 1964. Another proof of Wedderburn's theorem. Am. Math. Month. 71:652653.
Another Proof of Wedderburn's Theorem
Author(s): T. J. Kaczynski
Source: The American Mathematical Monthly, Vol. 71, No. 6 (Jun.  Jul., 1964), pp. 652653
Published by: Taylor & Francis, Ltd. on behalf of the Mathematical Association of America
Stable URL: https://www.jstor.org/stable/2312328
The author holds a NSF Coop fellowship. The preparation of this manuscript was partly supported by the NSF (G 24335).
References
1. E. A. Coddington and Norman Levinson, Theory of ordinary differential equations, McGrawHill, New York, 1955.
2. Witold Hurewicz, Lectures on ordinary differential equations, Wiley and MIT Press, New York, 1958.
ANOTHER PROOF OF WEDDERBURN’S THEOREM
T. J. Kaczynski, Evergreen Park, Illinois
In 1905 Wedderburn proved that every finite skew field is commutative. At least seven proofs of this theorem (not counting the present one) are known. See [1], [2], [5] (Part Two, p. 206 and Exercise 4 on p. 219), [6] (two proofs), and [7]. Unlike these proofs, the proof to be given here is grouptheoretic, in the sense that the only nongrouptheoretic concepts employed are of an elementary nature.
Lemma. Let q be a prime. Then the congruence Z^{2}+r^{2}= — 1 (mod q) has a solution t, r with t^O (mod q).
Proof. If —1 is a quadratic residue, take r = 0 and choose t appropriately. Assume — 1 is a nonresidue. Then any nonresidue can be written in the form — s^{2} (mod q) with s^O. If t^{2}+r^{2} is ever a nonresidue for some t, r, set t^{2}+r^{2} s— s^{2}, and we have (/5~^{1})^{2} + (r5”^{1})^{2} = — 1. (Throughout this note, x^{1} denotes that integer for which xx^{1} = l (mod q).) On the other hand, if t^{2}+r^{2} is always a residue, then the sum of any two residues is a residue, so —l=g—1 = 1 + 14 • • • + 1 is a residue, contradicting our assumption.
Proof of the theorem. Let F be our finite skew field, E* its multiplicative group. Let 5 be any Sylow subgroup of F*, of order, say, p^{a}. Choose an element g of order p in the center of 5. If some h^S generates a subgroup of order p different from that generated by g, then g and h generate a commutative field containing more than p roots of the equation x^{p}=l, an impossibility. Thus 5 contains only one subgroup of order p and hence is either a cyclic or a generalized quaternion group ([3] p. 189).
If S is a generalized quaternion group, then 5 contains a quaternion subgroup generated by two elements a and &, both of order 4, where ba — a~^b. Now a^{2} generates a commutative field in which the only roots of the equation x^{2} — 1 or (x+l)(x—1) =0 are ±1, so since (a^{2})^{2} = l, we have
(1) a^{2} =  1.
Hence a^ — a^{2} — —a_{t} so
(2) ba = — ab.
This content downloaded from 82.46.120.253 on Thu, 21 Feb 2019 17:17:06 UTC All use subject to https://about.jstor.org/terms
Similarly,
(3) 5^{2} =  1.
Taking q — characteristic of F (#l = 0), choose t and r as specified in the lemma. Using relations (1), (2), (3), we have
(/ + ra + 5)(r^{2} + 1 + rta + tb) = r(/^{2} + r^{2} + l)a + (/^{2} + r^{2} + 1)5 = 0.
One of the factors on the left must be 0, so for some numbers u, v, w, u 0 (mod g), we have w+^a+^5 = 0, or b= u^wau^w. So b commutes with a, a contradiction. We conclude that 5 is not a generalized quaternion group, so 5 is cyclic.
Thus every Sylow subgroup of F* is cyclic, and F* is solvable ([4], pp. 181— 182). Let Z be the center of F* and assume Z^F*. Then F*/Z is solvable, and its Sylow subgroups are cyclic. Let A/Z (with ZC^4) be a minimal normal subgroup of F*/Z. A/Z is an elementary abelian group of order (F prime), so since the Sylow subgroups of F*/Z are cyclic, A/Z is cyclic. Any group which is cyclic modulo its center is abelian, so A is abelian. Let x be any element of F*, y any element of A. Since A is normal, xyx~^{x}^A, and (l+x)y = z(l+x) for some zf^A. An easy manipulation shows that y — z — zx — xy = (z — xyx~^{l})x.
If y — z = z — xyx^{1} = 0, then y = z = xyx~^{1}, so x and y commute. Otherwise, x= (z — xyx~^{1}')~^{1}(y — z). But A is abelian, and 2, y, xyx^{1}(E^4, so x commutes with y. Thus we have proven that A is contained in the center of F*, a contradiction.
References
1. E. Artin, Uber einen Satz von Herrn J. H. M. Wedderburn, Abh. Math. Sem. Hamburg, 5 (1927) 245.
2. L. E. Dickson, On finite algebras, Gottingen Nachr., 1905, p. 379.
3. M. Hall, The theory of groups, Macmillan, New York, 1961.
4. Miller, Blichfeldt and Dickson, Theory and applications of finite groups, Wiley, New York, 1916.
5. B. L. van der Waerden, Moderne Algebra, Ungar, New York, 1943.
6. J. H. M. Wedderburn, A theorem on finite algebras, Trans. Amer. Math. Soc., 6 (1905) 349.
7. E. Witt, Uber die Kommutativitat endlicher Schiefkorper, Abh. Math. Sem. Hamburg, 8(1931)413.
A NOTE ON PRODUCT SYSTEMS OF SETS OF NATURAL NUMBERS
T. G. McLaughlin, University of California at Los Angeles
In this note, we apply a slight twist to a trick exploited about twelve years ago by J. C. E. Dekker ([2 ]), our purpose being to expose a couple of elementary facts about nonempty, countable "product systems” of infinite sets of natural numbers which are, at the same time, "finite symmetric difference systems.” We proceed in terms of the following definitions.
Definition. By a product system of subsets of N (N the natural numbers), we mean a collection of subsets of N which contains, along with any two of its members, their intersection.
2. 1964  Distributivity and (−1)x = −x (Advanced Problem 5210)
Original PDF: 2. 1964 Distributivity and (−1)x = −x (Advanced Problem 5210).pdf
Kaczynski, T.J. (1964). “Distributivity and (−1)x = −x (Advanced Problem 5210)”. The American Mathematical Monthly. 71(6), pp. 689.
ADVANCED PROBLEMS
All solutions of Advanced Problems should be sent to J. Barlaz, Rutgers  The State University, New Brunswick, N.J. Solutions of Advanced Problems in this issue should be submitted on separate, signed sheets and should be mailed before December 31, 1964 .
5210. Proposed by T. J. Kaczynski, Evergreen Park, Illinois
Let K be an algebraic system with two binary operations (one written additively, the other multiplicatively), satisfying:
1. K is an abelian group under addition,
2. K  {O} is a group under multiplication, and
3. x(y + z) == xy + xz for all x,y,z EK.
Suppose that for some n, 0 = 1 + 1 + ... + 1 ( <em>n</em> times). Prove that, for all <em>x</em> eK, (l)x =
x.
3. 1964  Distributivity and (−1)x = −x (Advanced Problem 5210, with Solution by Bilyeu, R.G.)
Original PDF: 3. Distributivity and (−1)x = −x (Advanced Problem 5210, with Solution by Bilyeu, R.G.).pdf
Kaczynski, T.J. (1965). “Distributivity and (−1)x = −x (Advanced Problem 5210, with Solution by Bilyeu, R.G.)”. The American Mathematical Monthly 72(6), pp. 677–678.
Distributivity and ( 1 )x == x
5210 [1964, 689]. Proposed by T. J. Kaczynski, Evergreen Park, Illinois
Let K be an algebraic system with two binary operations (one written additively, the other multiplicatively), satisfying:
1. K is an abelian group under addition,
2. K  {O} is a group under multiplication, and
3. x(y + z) == xy + xz for all x,y,z EK.
Suppose that for some n, 0 = 1 + 1 + ... + 1 ( <em>n</em> times). Prove that, for all <em>x</em> eK, (l)x =
x.
Solution by R. G. Bilyeu, North Texas State University. The last part of the hypothesis is unnecessary. If z denotes 1, then z + z + z z = <em>z</em> (1 + 1 + <em>z)</em> =
z, so z(?) = 1. Now <em>z(x</em> + <em>zx)</em> =
zx + x = <em>x</em> + <em>zx,</em> so either <em>x</em> + <em>zx</em> =
0 or z = 1. In either case <em>z(?)</em> =
x.
Also solved by Carol Avelsgaard, Richard Bourgin, Robert Bowen, Joel Brawley, Jr., F. P. Callahan, M. M. Chawla (India), R. A. CunninghameGreen (England), M. J. DeLeon, M. Edelstein, N. J. Fine, Harvey Friedman, Anton Glaser, M. G. Greening (Australia), A. G. Heinicke, Sidney Heller, G. A. Heuer, Stephen Hoffman, K. G. Johnson, A. J. Karson, Max Klicker, Kwangil Koh, C. C. Lindner, C. R. MacCluer, H. F. Mattson, C. J. Maxson, R. V. Moddy, Jose Morgado (Brazil), W. L. Owen, Jr., P. R. Parthasarathy (India), Harsh Pittie, Kenneth Rogers, Toru Saito (Japan), Camilio Schmidt, Leonard Shapiro, Frank A. Smith, George Van Zwalenberg, W. C. Waterhouse, Kenneth Yanosko, and the proposer.
4. 1965  Boundary Functions for Functions Defined in a Disk
Original PDF: 4. 1965 Boundary Functions for Functions Defined in a Disk.pdf
Kaczynski, T.J. 1965. Boundary functions for functions defined in a disk. J. Math. and Mech. 14(4):589612.
MR0176080 Kaczynski, T. J. Boundary functions for function defined in a disk. J. Math. Mech. 14 1965 589.612. (Reviewer: C. Tanaka) 30.62
Explanation by John D. Bullough
Let D denote the unit disk z < 1, C its boundary, and let f(z) be any function that is defined in D and takes its values in some metric space S. Then a boundary function for f is a function t on C such that for every x ( C there exists an arc v at x with
lim f(z) = t(x).
z > x
z ( v
The author proves several theorems on boundary functions in the following four cases: (1) f(z) a homeomorphism of D onto D, (2) f(z) a continuous function, (3) f(z) a Baire function and (4) f(z) a measurable function. These theorems include answers to two questions raised by Bagemihl and Piranian.
Theorem 1 states that if f(z) is a homeomorphism of D onto D, then there exists a countable set N such that tC  N is continuous.
In the case of continuous functions, one needs some definitions. Let S and T be metric spaces. f is said to be of Baire class 1(S, T) if and only if (i) domain f = S, (ii) range f ( T and (iii) there exists a sequence {f(n)} of continuous functions, each mapping S into T, such that f(n) > f pointwise on S. g is of honorary Baire class 2(S, T) if and only if (i) domain g = S, (ii) range g ( T and (iii) there exists a function f of Baire class 1(S, T) and a countable set N such that fS  N = gS  N. Using these defnitions, Theorems 2 and 3 read as follows. Theorem 2: Let f be a continuous realvalued function in D and let t be a finitevalued boundary function for f. Then t is of honorary Baire class 2(C, R), where R is the set of real numbers. Theorem 3: Let f be a continuous function mapping D into the Riemann sphere S and let t be a boundary function for f. Then t is of honorary Baire class 2(C, S).
In the cases of Baire functions and measurable functions, for the sake of convenience consider the open upper halfplane D^{0}: I(z) > 0, and its boundary C^{0}: I(z) = 0, instead of D and C, respectively. Theorem 4 states that if f is a realvalued function of Baire class a > 1 in D^{0}, and t is a finitevalued boundary function, then t is of Baire class a + 1. As an immediate consequence of Theorem 4, one has Theorem 5: Let f be a realvalued Borelmeasurable function in D^{0} and let t be a finitevalued boundary function for f; then t is Borelmeasurable.
Next, the author proves that for an arbitrary function t on C^{0}, there exists a function f on D^{0} such that f(z) = 0 almost everywhere and t is a boundary function for f. The paper concludes with some remarks concerning extensions of these theorems into three dimensions.
Article by Ted
Boundary Functions for Functions Defined in a DisB
T. J. KACZYNSKI
Communicated by F. Bagemihl
1. Introduction
Throughout this paper D will denote the open unit disk (in twodimensional Euclidean space) and C will denote its boundary, the unit circle. Bagemihl and Piranian [2] have introduced the following definition.
Definition. If x e C, an arc at x is & simple arc y having one endpoint at x such that y — {x} C D. Let / be any function that is defined in D and takes its values in some metric space S. Then a boundary junction for f is a function <p on C such that for every x e C there exists an arc y at x with
lim f (z) = <p(x).
The purpose of this paper is to prove several theorems concerning boundary functions. These theorems include answers to two questions raised in [2] (see Problem 1 and the conjecture on p. 202).
The set of real numbers will be denoted by R, Wdimensional Euclidean space will be denoted by R^{N}, and the Riemann sphere will be denoted by 2. Points in R^{N} will be written in the form {x_{x} , x_{2 }} • • • , x_{N}) rather than (x_{t} , x_{2} , • • • , x_{N}) (to avoid confusion with open intervals of real numbers in the case N — 2). Whenever we speak of realvalued functions we mean finitevalued functions, and whenever we speak of increasing functions we refer to weakly increasing (nondecreasing) functions. The abbreviations “l.u.b.” and “g.l.b.” stand for “least upper bound” and “greatest lower bound” respectively. Finally, it should be noted that our definition of the Baire classes is slightly unconventional (see p. 6 and p.14) in that we consider Baire class a to include Baire class ft for every ft < a.
2. Boundary functions for homeomorphisms.
Definition. If E C D, let acc (E) denote the set of all points on C which are accessible by arcs in E.
^{1}1 would like to thank Professor G. Piranian for his encouragement.
589
Journal of Mathematics and Mechanics, Vol. 14, No. 4 (1965).
Lemma 1. Let A be an arcwise connected subset of D and let B be a connected subset of D. Suppose that A B —</> . Then acc (A) and B have at most two points in common.
Proof. Assume that p_{r} , p_{2} , p_{3} are three distinct points of acc (A) A B and derive a contradiction. Let 7* be an arc joining pi to a point qi z A, with {Pi} A (i = 1, 2, 3). Let y be an arc in A joining and q_{2} . Putting
, 7_{2} and 7 together, we obtain an arc T joining pj to p_{2}, with r — {p_{r}, p_{2}} C A. We can assume F is a simple arc, for if r is not simple, and p_{2} can be joined by some simple arc F^{z} £ T (see [7]). Let L_{x} , L_{2} be the two open arcs of C determined by the pair of points p_{r} , p_{2} . We may assume, by symmetry, that p_{3} z L_{r} . According to [6] (Theorem 11.8, p. 119), D — T has two components Ui and U_{2} , the boundary of Ui being L_{x} U P and the boundary of U_{2} being z_{2}u r.
Let 7' be an arc in A joining q_{3} to a point q z T A. Putting 73 and 7' together, we obtain an arc 5 joining p_{3} to q. Starting at p_{3} and proceeding along 5, let r be the first point of T that we reach. Let A be the subarc of 5 with endpoints at p_{3} and r. Clearly, A — {p_{3}} GZ A. We can assume (according to [7]) that A is a simple arc.
Since p_{3} z L_{x} , p_{3} is not in U_{2} . Since
A  {p_{3} , r} £ D  T = Ui U_{2} ,
A — {ps , r} must have a point in U_{x} . But A — {p_{3} , r} is connected, so A  {p_{3}, r} C Ui. Hence A is a cross cut of U_{r}. Let M_{r} ,M_{2}]oq the two open subarcs of Li with endpoints p_{t} , p_{3} and p_{2} , p_{3} respectively. Let Pi , r_{2} be the two closed subarcs of T with endpoints p_{r} , r and p_{2} , r respectively. According to [6] (Theorem 11.8, p. 119), U_{r} — A has two components Vi and V_{2}, the boundary of Vi being kJ r_{x} kJ A and the boundary of V_{2} being M_{2} kJ r_{2} kJ A.
Since P U A C 2I, Vi^J V_{2}\J U_{2}. Recall that p_{3} 4 U_{2} . It follows that since p_{3} z B, B has a point in common with Vi^J V_{2} . But B is connected, so B £ Vi yj V_{2}. We see that p_{r} $ V_{2}, and therefore that B Vi 4=</> (because Pi z B). Hence B £ , so p_{2} z W . But, since the boundary of V_{t} is M_{x}
I\ kJ A, p_{2} Vi . This contradiction proves the lemma.
Lemma 2. There exists a countable family 8 of open disks such that every open set U Q R^{2} can be written in the form U — S_{n} , where S_{n}z § and S_{n} £ U.
Proof. Let {p_{n}} be a countable dense subset of R^{2}, and let 8 be the family of all open disks of rational radius having some p_{n} as center. 8 is clearly countable. If U is an open set it is easy to show that for each x z U there exists an S_{x} z 8 with x z S_{x} £ S_{x} £ U. Obviously
u = \J s_{x}.
xt.U
Theorem 1. Let f be a homeomorphism of D onto D, and let <p be a boundary function for f. Then there exists a countable set N such that <p _{C}^ is continuous.
Prooj. Take an arbitrary aS £ S. It is easily shown that DPS and D — S are both connected, so f^{1}(D (P 8) and — 8) are both connected. Given x_{0} e C, let y be any arc at x_{Q} . If
x_{Q} $ acc (f^{1}(D iP aS)),
then we can choose points on y arbitrarily close to x_{Q} which are not in f~^{1} (D (P 8), so
x_{0} £ D  r\D CPS') = r\D aS).
This shows that
(1) C Q acc aS)) kJ r~\D aS).
Let
F = acc (f\D fP aS)) H f~\D aS).
By Lemma 1, F contains at most two points, and from (1) we see that acc (rXD S)) = FU (C  r\b  aS)).
Thus we have shown that for each aS £ S we can write
acc (T\D aS)) = F$ kJ G_{s} ,
where F_{s} is finite and G_{s} is open (relative to C).
For any arc y at a point x on C, the cluster set C(f, t) of f along y is defined by
C(/, t) = {w £ 2?^{2} kJ { oo }  there exists a sequence {z_{n}} QyPD such that z_{n} —> x and f (z_{n}) —> w}.
Let
E = {x £ C  there exist arcs y_{x}, y_{2} at x such that C(/, yj IP C(J, y_{2}) =</> }•
A theorem of Bagemihl [1] states that E is countable. Let
N = E \J F_{s} .
St s
N is countable. Let <p_{Q} denote the restriction of <p to C — N.
If U is any open set, write U = S_{n} , where S_{n} £ S, S_{n} CZ U. Suppose x £ (pQ^{1} (U). Then (x) = <p(x) £ S_{n} for some n, which implies that a; £ acc (/~^{1} (S_{n}C\D)). Thus
¥>0 W) C \J acc (f\S_{n} n Z>))  N. n
On the other hand, suppose x £ acc IP D)) for some n, and x 4 N. Choose an arc y in fXSn (P D) with one endpoint at x. Clearly,
C(f, y) C S_{n} (P D CS_{n}CU.
Since x 4 E,
?.() = p() _{£} C(j, y) c u,
so x t ^’(G). Thus
\J acc (T^{l}(S_{n} r\D))NC ^\U), n
SO
<Po'(U) = U acc (r\S„ f~\ D)) — N = \J (F_{s}, U G_{s}„)  N n n
= VGs. n = (U^_{s}.) n (C  N). n n
Thus, for each open set 17, <Po^{l}(U) is an open set relative to C~N. Therefore
<p_{0} is continuous. Q.E.D.
3. Boundary functions for continuous functions.
Definition. Let 8 and T be metric spaces. We will say the function / is of Baire class 1 (8, T) ij, and only if,
(i) domain j = 8,
(ii) range IQT, and
(iii) there exists a sequence {/„} of continuous functions, each mapping 8 into T, such that f„—>f pointwise on 8.
We will say the function g is of honorary Baire class 2(8, T) if, and only if, (i) domain g = 8, (ii) range g C T, and
(111) there exists a function / of Baire class 1(8, T) and a countable set N such that f Isjv = Isat .
Lemma 3. Let f be a continuous realvalued junction in D and let <pbe a finitevalued boundary junction jor j. Let r and t be real numbers with r < t. Then
(A) there exists a G_{s} set G and a countable set N such that
^([r, 4oo)) o G 2 ^([t, +oo))  AT, and
(B) there exists a G_{s} set H and a countable set M such that
t]) 2 H 2 <^{1}((°°, r])  M.
Prooj. Let
t — r ^{e} ~ 2 ’
C_{n} = Lit^{1}  z = 1  4 , ( n]
A_{n} = \ztR^{2}  1   < > < 4 ,
( n J
E_{n} — {x e C  there exists an arc y at x having one endpoint on C_{n} , with y — {x} Q f^{1}((— ^{00}, r)),
K = {x e C  there exists an arc y at x with y — {$} c f^{1}^ — _{c}, +«>))}.
Observe that
^((co.r)) C \jE_{n} , n«l
and
<\(/  e, +«>)) C K.
For the time being, let n be a fixed integer. If x £ K, we can find an arc y_{9} at x such that
y*  M C A_{n}C\f\\t  _{c}, +«>)).
Since an arc at x is by definition a simple arc, y_{x} — {x} is a connected set. It follows that y_{x} — {rc} must be contained entirely within one component of the open set
A_{n}nn(/e,+)).
We denote this component by U_{x} . U_{x} is a nonempty open connected set._ Let T be the set of all points of K which are twosided limit points of E_{n}.
Assertion. If x, y £ T and x 1
y, then U_{x} C\ U_{y} = <j>.
To prove this assertion we assume that z is a point of U_{x} A U_{v} and we derive a contradiction. Choose points x^{r} and y' in y_{x} — {x} and y_{v} — {?/} respectively. Join x to x^{f} by an appropriate subarc of y_{x} . Join x^{f} to z by an arc in U_{x} . Join z to y* by an arc in U_{v}. Join y' to y by a subarc of y_{v}. Putting these arcs together, we obtain an arc a with endpoints at x and y such that
a  {x, y} C A_{n}C\ f~\(t  e, + co)).
We can assume that a is a simple arc, for if a is not a simple arc we can replace a by a simple arc a' CZ a having endpoints at x and y (see [7]). a is a crosscut of D. Let Li and L_{2} be the two open arcs of C determined by x and y. According to [6] (Theorem 11.8, p. 119), D — a has two components, Pi and V_{2} , whose boundaries are L_{r} kJ a and L_{2} a respectively. From the fact that C_{n} is connected and does not intersect a it follows that C_{n} is contained entirely within one component of D — a. By symmetry, we may assume C_{n} £ V 2 •
Since $ is a twosided limit point of E_{n} , Lr must contain a point of E_{n} , and hence a point of E_{n} . Say w £ A E_{n} . There exists a simple arc ft joining w to some point on C_{n} , with
ft — {w} C ^{00} , r)).
ft — {w} cannot have a point in common with a, because
a ~ {x, y} C f~\(t  _{€}, +^{00})), and
e,+)) =</> .
Thus C_{n} U (0 — {w}) is a connected set not meeting a. C_{n} U (/3 — {w}) meets y_{2},soC_{n}VJ (0 — {w}) C V_{2} . Consequently, w is in the boundary of V_{2} . But this is a contradiction, because w e L_{x} and the boundary of V_{2} is L_{2} U a. This proves the assertion.
From the assertion it follows immediately that T is countable; for any family of disjoint nonempty open sets is countable. We know that the set >S of all points of E_{n} which are not twosided limit points of E_{n} is countable.
K C\ E_{n} = [K C\ S] U [KH (E_{n}  S)] = (KC\ 8) T.
This shows that (for any ri) K C\ E_{n} is countable. So if we let
N = K C\ \JE„ =. Q (K C\ E_{n}), n = l n=l
then N is a countable set. Let
G = C — \_)E_{n} . n = l
G is a G_{s} set. Using the fact that
<*((«>, r)) C \jE_{n} C \jE_{n} , n=l n—1
we find that
00
C  <*(( oo, r)) 2 c  \jE_{n} = G 2 K  N.
n=l
But
=^([r,+))
and
K O ^((/  _{€}, +«>)) o +«>)), so
<X[r, + +oo))  N.
This proves (A). To prove (B), simply replace / and by —f and and apply (A).
Theorem 2. Let f be a continuous realvalued function in D, and let <p be a finitevalued boundary function for f. Then is of honorary Baire class 2(C, R).
Proof. For each pair of rational numbers r and t with r < t, choose G& sets G(r, t), H(r, t) and countable sets N(r, i), M(r, f) such that
+«>)) O G(r, t) O + <»))  N(r, f), and
<*(( ~, t]) 2 H(r, f) 2 «>, r])  M(r, Z).
Let
N = {J [N(r, t) V M(r, /)],
where the union is taken over all pairs of rationals r, t with r < t. N is countable. Let (p_{0} denote the restriction of to C — N, and let G*(r, t) = G(r, t) — N. Since every countable set is an F_{ff} set, G*(r, t) is a G_{8} set. Observe that
(2) ^([r, +)) = ^([r, +«.)) N 2 G*(r, f)
2 +))  N = ^([Z, +))•
If t is a fixed rational number, let {r_{n}} be a strictly increasing sequence of rational numbers converging to t. Then, by (2),
C\ <PoX[r. , +»)) 2 O G*(r. , f) 2 <p~_{0}\[t, + «>)) = ^\[r_{a} ,+«)),
n=l n=l n=l
SO
^’(R,+°°)) = C\G*(r_{n},f). n = l
This proves that for every rational t, ^([J, + ^{00})) is a G_{8} set.
If u is any real number, choose a strictly increasing sequence {£_{n}J of rational numbers converging to u. Then
+^{00})) = fWoWn , +^{00})), n = l
SO <p_{Q}\[u, + oo)) is a G_{8} set. By a similar argument, we find that ^o^{1}((— ^{00}, u]) is a Ga set for every real u. So
>, + )) = (C  N) n (C  , u]y)
is the intersection of an F_{ff} set with C — N. By a theorem stated on p. 309 of Hausdorff’s paper [5], <p_{0} can be extended to a realvalued function on C such that for every real u, +^{00})) is a G_{8} set and +<»)) is an F_{a} set. By Theorem IX of the same paper, is of Baire class 1(C, 7?). Since <p(x) = pxtx) except for xz N, <p is of honorary Baire class 2(C, 7?). Q.E.D.
Corollary. Let f be a continuous junction mapping D into R^{N}, and suppose <p : C —» R^{N} is a boundary junction jor j. Then <p is of honorary Baire class 2(C, R^{N}).
Proof. We simply write our functions in terms of their components, say
f — (fl ! f'2 J * ’ • , and (p — (epi , <p2 i • • • , <Pn) '
Obviously <Pi is a boundary function for /,• , and so is of honorary Baire class 2(C, jR). We choose a function of Baire class 1(C, R) which agrees with <pi except on a countable set M i . Setting
0 = <0i , 02 , • • • , 0AT>,
it is clear that g is of Baire class 1(C, R^{N}), and that g agrees with <p except on the countable set VJii Mt • Hence <p is of honorary Baire class 2(C, R^{N}).
Q.E.D.
Lemma 4. Let g be a continuous junction mapping C into R^{3}. Let q be a point of R^{3} and let e be a positive real number. Then there exists a continuous function g* : C —> 7?^{3} such that q does not lie in the range of g*, and for all x v C,
\g(x)  q\ e => g(x) = g*(x).
Proof. Let
S = {yvR^{3}  \y  g < e}.
If 0(C) CZ S, let g* : C —> R^{3} be any continuous function whose range does not include q. Otherwise, 0~^{1}(/S) is a proper open subset of C and hence can be written in the form
g~XS) = UA, k
where
Ik = {e’‘  at < t < b_{k}], and
k 1
I =>I_{k} Ii =</> .
Since 0~^{1}({0}) is a closed (and therefore compact) subset of 0~^{1}(>S), 0~^{1}({^}) is covered by a finite number of I_{fc}’s. Say
0’^{1}({0}) ... \JI_{n}.
The endpoints e^{iak} and e^{t6A} of I_{k} are not in 0~^{1}({<?}), so we can construct, for each k, a continuous function g_{k} : I_{k}+ R^{3} such that
Sk(e^{iai}) = g(e^{iai}), = g^*),
and q is not in the range of g_{k} . Define
0*(x) = 0(a:), if o;_{£}C~ (AUZ_{2}U ... U/_{n}),
0*(x) = g_{k}(x), if x e I_{k} , k = 1, • • • , n.
It is easy to show that g* has the desired properties.
Theorem 3. Let f be a continuous function mapping D into the Riemann sphere 2, and let <p be a boundary function for f. Then <p is of honorary Baire class 2(C, 2).
Proof, Since 2 is a subset of J?^{3}, the corollary to Theorem 2 shows that <p is of honorary Baire class 2((7, 2?^{3}). Let g be a function of Baire class 1(C, 2?^{3}) which differs from <p only on a countable set N, Then g(C) — 2 is countable, so there exists a point q inside of 2 (that is, in the bounded open domain determined by 2) which is not in the range of g. Let {</„} be a sequence of continuous functions converging to g. By Lemma 4 we can find (for each ri) a continuous function g*_{n} : C —» R^{3} such that q does not lie in the range of g* , and for all x £ C,
 g g_{n}(x) = g*(x). a
It is easy to show that g*_{n} —» g.
We define a function P as follows. If a e R^{3} — {q}, let I be the unique ray with endpoint at q that passes through a, and let P(a) be the intersection point of I with 2. Obviously, P is a continuous mapping of R^{3} — {<?} onto 2, and P fixes every point of 2. Therefore
F(0(z)) = if x^N,
P(g*_{n}(x)) is a continuous function from C into 2, and
F0(x)) as n oo.
This shows that <p is of honorary Baire class 2(C, 2). Q.E.D.
4. Boundary functions for Baire functions.
In this section we concern ourselves only with realvalued functions. We shall prove that a boundary function for a function of Baire class a 1 is of Baire class a + 1. It is convenient to prove this theorem for functions that are defined in the (open) upper halfplane and have boundary functions defined on the rraxis rather than for functions defined in D, Once the theorem is proved in this form it is a routine computational matter to show that it also holds for functions defined in D, The reader should be familiar with the results of Hausdorff [5] before reading this section. Unfortunately, we must begin with some tedious preliminaries.
Let
We will regard C° as being identical with R.
Suppose 8 is a metric space. Let g^ be the class of all open sets of $ and let be the class of all closed sets of 8.
A function / : >8 —> R is of Baire class 0 if and only if it is continuous. For any ordinal number a > 0, f is of Baire class a if and only if / is the pointwise limit of a sequence of functions each of Baire class less than a.
Let denote the class of all sets M C S such that
M = r\(r, +«>)),
for some real rand some function / of Baire class a on >S. Let 912 denote the class of all sets N C & such that
N = r\[r, +)),
for some real r and some function / of Baire class a on 8. It is easily shown that 9TCj = and 9l£ = .
Let
9 — 9c«> = g«,
gj = ,
9E^{a} = ,
91* = 9lco = 912 , If 0 is any class of sets, let 0_{a} denote the class of all countable unions of members of 0, and let 0_{8} denote the class of all countable intersections of members of 0. Each of the following facts is either explicitly stated in [5], or can be easily deduced from statements found in [5], or is obtained by a routine transfinite induction argument.
I. 9TC2 = (\J 9lX , 9i? = (U 9TtX •
X<a X<a
II. Let A be any subset of the metric space 5. If / is a function of Baire class a on S, then / 1^ is a function of Baire class a on A.
III. Let f be a function of Baire class a whose domain contains {(x, b)  x e 2?}. Then j((x, b)) is a function (of x) of Baire class a.
IV. If A C 8, then
9R? = {M r\ A  M e Oil?}, 9Z“ = {N C\ A  Ve9l?}.
V. If / is of Baire class a on 8, then for each real r,
and
VI. If a 2, then (g_{s})_{s} U C gn“ C\ .
VII. E £ °fC^{a}_{s}o 8  E e .
VIII. ^{c}JTCs and are closed under finite unions and intersections. is closed under countable unions and 91^ is closed under countable intersections.
IX. Let f be a realvalued function on S. Suppose that for every real r
and
r>, +«))e^ .
Then / is of Baire class a.
Definition. If A and B are two sets, we will call A and B equivalent, and write A ~ B, if and only if A — B and B — A are both countable. It is easily verified that ~ is an equivalence relation.
Lemma 5. If A ~ E, then 8 — A ~ 8 — E for any set 8. If A_{n} ~ E_{n} (for all n in some countable set V), then
)jA_{n}~\jE_{n} and ~ •
ntN ntN ntN ntN
The proof of this lemma is routine.
Definition. An interval of real numbers will be called nondegenerate if it contains more than one point.
Lemma 6. Any union of nondegenerate intervals is equivalent to an open set.
Proof. Let 4 be a family of nondegenerate intervals and let H = For any x and y let
ZO, y) = [x, y}, if x g y,
and let
i(%>y) = [y, ^1,
Define a relation (R on H by
x(Sly <=>I(x, y) CZ H,
(x, yzH).
It is easy to show that (R is an equivalence relation on H. In view of the fact that a set A of real numbers is an interval if and only if
x, y e A =>I(x, y) C A,
it is obvious that each equivalence class is an interval. For each x £ H, there exists an I £ $ with x £ I. Every member of I is equivalent to x. Thus each equivalence class contains more than one point, and hence is a nondegenerate interval. Let {J_{a}}be the family of equivalence classes. Any disjoint family of nondegenerate intervals is countable, so there are only countably many J_{a}’s. Let E be the set of all endpoints of the various J_{a}’s. Then E is countable and
H = \J J_{a}~ IJ J_{a}  E = \Jj*, a a a
where <7* is the interior of J_{a} . This proves the lemma.
Lemma 7. Let h be an increasing realvalued function on a nonempty set E CZ R, Suppose that \x — h(x)\ 1 for every x s E. Then h can be extended
to an increasing realvalued function h_{x} on R.
Proof. Let e ~ g.l.b. E (e may be — oo). For each x_{Q} e (e, + °°) set
hi(x<f) = l.u.b. {h(x)  x e (— oo , x_{Q}\ C\ E}.
Since Jx — h(x)\ g 1 for all x e E,
x e (— oo, x_{0}] TA E h(x) g x_{0} + 1, so is finitevalued. If e = — oo we are done. If e > — oo_{}} then x e E implies
h(x) e — 1, so A is bounded below. For x_{Q} s (— °°, e] set
Ai(^o) = g.l.b. {A(x)  xtE],
It is easily verified that h_{r} has the desired properties.
Lemma 8. Let f be a realvalued function of Baire class a on R. Let h be an increasing realvalued function on R. Set g(x) = Then there exists
a countable set N such that g tv is of Baire class a.
Proof. It is well known that an increasing function has at most countably many discontinuities. Let M be the set of discontinuity points of h. If f is of Baire class 0, then g is continuous at all points of I? — M, so g is of Baire class 0. This proves the lemma for the case a = 0.
We now proceed by transfinite induction. Suppose the lemma holds for every ordinal X < a. If f is of Baire class a we may choose a sequence of functions {f_{n}} converging to f, where f_{n} is of Baire class X_{n} < a. If we set g_{n}(x) — f_{n}(h(%y) it is clear that g_{}!}(x) —» = {/W By the induction hypothesis we may
choose (for each n) a countable set N_{n} such that g_{n} It?^ is of Baire class X_{n} . Let N = Nn . Then g_{n} _{R}__{N} is of Baire class X_{n} , and since g_{n} t?_at —» g Uat, g  at is of Baire class a. This proves the lemma.
Theorem 4. Let f be a realvalued function of Baire class a 1 on D°_{}} and let <p be a finitevalued boundary function for f. Then <p is of Baire class a + 1.
Proof. Let r and t be two real numbers with r < t. r and t will remain fixed throughout the first part of the proof. Set
Q = +<»)),
E = P\J Q,
t — r e =  •
Observe that P C\ Q = <t>. For each xtE, choose an arc y_{x} at x such that lim f(z) = <p(x), y_{x} C {z  \z — x g 1}, z—>x
and
(a) /(?*) £ (— ^{00} , r + e), if x s P
(b) f(_{7}») £ (J  _{€}, +«>), if xtQ.
(This is accomplished by cutting the arc off sufficiently close to a;.) We remark that if x e P and y £ Q, then y_{x} C\ y_{y} =</> .
We will say that y_{x} meets y_{y} in A^{Q}_{n} provided that y_{x} and y_{y} have subarcs y' and y' respectively such that x £ y' GZ A°, y £ y' C A°, and y' P y_{y} 4
</> . Let
L_{o} = {x £ P  (Vn)(3?/ 4= x)(y_{x} meets y_{y} in A°_{n})},
L_{r} = {x £ Q I (Vn)(3?/ 4= x)(y_{x} meets y_{y} in A®)},
M_{o} = £ P  (ln)(y_{x} meets no y_{v} (y 4= x) in A®)},
{x £ Q  (?ri)(y_{x} meets no y_{y} (y 4= x) in A®)},
L = Lo V Lt ,
M = M_{0}\J M_{r}.
Observe that L_{o} , L_{r} , M_{o} , Mt are pairwise disjoint, and that P = L_{o} M_{o} and Q = L_{x} U M_{r} .
For each x £ M, let n_{x} be an integer such that y_{x} meets no y_{y} (with y 4= x) in A®* . Notice that n n_{x} implies y_{x} meets no y_{y} in A® . Let
K_{n} = {x £ E  y_{x} meets C®, and if x £ M, n_{x} n}.
Clearly E = VJni K_{n} . Moreover, K_{n} C K _{n+1} for each n.
Take any fixed integer n. For each x £ L_{o} we can find a y 4= x such that y_{x} meets y_{v} in A® . Let I^{n}_{x} be the nondegenerate closed interval between x and y. We shall show that I^{n}_{x} C L_{o} U (C®  K_{n}\ If t £ I^{n}_{x} , either t £ C^{Q}  K_{n} or t £ K_{n} . Suppose t £ K_{n} . Then y_{t} meets C®, and (if t £ M) n_{t} n. It is clear from Figure 1 that y_{t} must meet either y_{x} or y_{y} in A® . (This can be rigorized by means of Theorem 11.8 on p. 119 in [6].)
Consequently, t % M. Now x iL_{0} Q P, so since y_{x} intersects y_{y} , y $ Q. So y e E — Q = P. Similarly, since y_{t} meets y_{x} or y_{y}, t e E — Q = P. Thus t e P — M = Lq . We have shown that t e I^{n}_{x} implies that t e C° — K_{n} or t e L_{q} , so I^{n}_{x} £ Lq V (C° — K_{n}). It follows that (for each ri)
L_{0}Q(\Jr^r\E q[l_{0}\j (c°  xj] n e. XzLq
Let W„ = kJxet, I^{n}_{x} . By Lemma 6, W_{n} is equivalent to an open set. l_{o}q(c\w}c\e 'n==l /
C M [L_{o} kJ (C°  K_{n})]} H E = \l_{0} v f} (.0°  K„)} n E
= {L_{o} n E] H (C°  K„) n e\ = L_{o} V = L_{a} .
Therefore L_{Q} = W_{n}) A E. Since each W_{n} is equivalent to an open set there exists a G_{Q} e g_{8} such that
L_{o} — Go n E.
Similar reasoning shows there exists a G_{x} e Q_{§} such that
E.
Next we study the properties of M_{Q} , It is convenient to define a function 7T : R^{2} —» R by ir({x, y)) = If x e M C\ K_{n} , then, starting at x and proceeding along y_{x} , let (?_{n}(x) be the first point of C^{Q}_{n} reached. Set h^(x) = ir(a_{n}(x)) (for x e M C\ Kn).
is an increasing function on M C\ K_{n} ; for if Xi, x_{2} £ M K_{n} and x_{x} < x_{2}, then, since cannot meet y_{Xi} in A^{Q}_{n}, it is evident (see Figure 2) that tt^Xxj)) < ^(<r_{w}(£_{2})). (The argument can be rigorized by means of Theorem 11.8 on p. 119 in [6].) Since
y_{x} £ {z  z  1}, x  h°_{n}(x)\ g 1.
So by Lemma 7 can be extended to an increasing function h_{n} on C°.
Let
Gn(x) = •
For x e M C\ K_{n} ,
gn(x) =
If x £ M, then for all sufficiently large n, x £ M C\ K_{n} , so lim g_{n}(x) = lim f(<r„(a;)) = <p(x). n—*a> n—♦<»
Thus g_{n} k —» cp k . By III, /(($, 1/n)) is a function (of x) of Baire class a, so by Lemma 8 we can choose, for each n, a countable set N_{n} such that g_{n}  _{C}*N_{n} is of Baire class a. Let N = VJZi N_{n} . Then g_{n} k# is of Baire class a. But g_{n} Iatjv —> <p kw , so <p k^ is of Baire class a + 1.
Now
p (M  N) = & kk^{1}(( r]) = Try (m n), where T s 9l^{a+1} (by IV and V). Clearly P C\ M ~ T C\ M.
We have
L = P_{o} U L, ~ (G_{o} n F) VJ (G_{t} n E) = (Go U GJ H E,
so L ~ G C\ E where G e g_{8} . Also
M_{0}=PKM~TC}M = T C\(E  L)
~tc\[e  (g n P)] = [t n (c°  g)] n e.
Since G e g_{5} , G°  G e , so by VI and VIII, T C\ (C°  G) e 9T^{+1}. Thus
M_{o} — T_{o} n E,
where T_{o} e 9l“^{+1}. Now we can examine the properties of P.
P = (G_{q}KE)U (T_{o} n E) = (Go U To) n E,
so, again by VI and VIII,
p ~ 7\ n p,
where T_{x} e 9l^{a+1}. Since a countable set is in and the complement of a countable set is in g_{8} , it is easy to show (using VI and VIII) that
P = T_{2} H P,
where T_{2} e 9l^{a+1}. Since P C\ Q = <f>_{}}
P £ T_{2} C C°  Q.
Remembering the definitions of P and Q_{}} and observing the fact that C° — ^^{1}(k +^{00})) = ^^{1}((” ^{00}, 0)> ^{we can} summarize the results of the first part of the proof as follows.
For each pair r, t of real numbers with r < t, there exists a set T(r, t) e 3l“^{+1} such that
^((~ r]) C T(r, t) C a,, t».
Given any real r, let {Z_{n}} be a strictly decreasing sequence of real numbers converging to r. Then
<’((—^{00}, »•]) = °°, /„)). n = l
So
<*(( »,r]) £ H T(r, /„) C f\ <^{1}(( «>, i_{n})) = _{v}~\( «, r]), n“l n—1
and hence
^((__{ro},_{r]}) = C\T{r,Q. n~l
By VIII,
<^{1}((~°o,r]) e9T^{+1}.
Since / is an arbitrary function of Baire class a in Z)° and <p is an arbitrary boundary function for /, we can replace /, $>, r by — /, — cp, ~r to find that
^([r, +«>)) e9l^{a+1}.
Also,
^((r, +00)) = C°  _{£}snr^{+1}.
By IX, <p is of Baire class a + 1. Q.E.D.
5. Boundary functions for measurable functions.
Theorem 5. Let f be a realvalued Borelmeasurable function in D^{Q} and let <p be a finitevalued boundary function for f. Then <p is Borelmeasurable.
Since every Borelmeasurable function is of some Baire class a, this theorem is an immediate consequence of Theorem 4. We now show that a boundary function for a Lebesguemeasurable function need not be Lebesguemeasurable.
Let u denote Lebesgue measure on R and let /z^{2} denote Lebesgue measure on R^{2}. Let denote exterior Lebesgue measure on R] that is,
Me(S) = g.l.b. {/z(G) I G is open and E C G},
for any set E C R.
Lemma 9. Let h be an increasing realvalued function on a set E C R. Then there exists an open interval I 2 E such that h can be extended to an increasing realvalued function on I.
Proof, If E is unbounded below, set a — — co. If E is bounded below, set a = g.l.b. E, if (g.l.b. E) 4 E,
a = (g.l.b. E)  1, if (g.l.b. E) £ E.
If E is unbounded above, set b ~ + ^{00}. If E is bounded above, set
b = l.u.b. E, if (l.u.b. E) 4 E,
b = (l.u.b. E) + 1, if (l.u.b. E) e E.
Let I = (a, b). Clearly E c: Z. Let e = g.l.b. E (e may be — <»). For x_{0} £ (e, b) set
ftxtf) ~ l.u.b. {h(x)  x c (a, $_{0}] Pi E}.
If e = a we are done. If e >a then e £ E. For x_{Q} £ (a, e] set f(x_{0}) = A(e). It is easily verified that f is finitevalued and increasing, and is an extension of h.
Lemma 10. Let E C R be a set of measure 0 and let h be an increasing function on E. Suppose h(E) has measure 0. Then {x + h(x)  x £ E} has measure 0.
Proof. Extend h to an increasing function g on an open interval I = (a, b) 2 E. Set g(a) = — 00 and g(b) = + 00. Take any e > 0. Choose an open set G such that I and utG) < e/2. Choose an open set H Z) h(E) with /z(ZZ) < e/2.
Say
G = U I_{n} , and H = \J J_{m} , mN mtM
where {I_{n}  n £ N} and {J_{m}  ms M} are countable families of disjoint open intervals. Let I_{n} = (a_{n} , b_{w}), and observe that a_{n} , b_{n} £ [a, b]. Set
s = (J {g(a„), g(b_{n})}  { 00 , + co }.
mN
Notice that S is countable. Set
K_{n} = (g(a„), g(&n)).
One can easily verify that k 1
n implies K_{k} P K_{n} = 0.
If A and B are two subsets of 2?, let
A + B = {a + b \ as, A,b sB}.
It is easy to show that for any two intervals J and <7', g_{e}(<7 +</') g + ^(J'). Let W = {x + h(x)  x £ E].
Assertion.
w c (E + S) KJ \J \J [(Z_{n} n gW) 4 (J_{n} r\ K.)]. mN imM
To prove this, let w be an arbitrary point of W. Write w = x + h(x)_{f} where x sE. For some n, x £ I_{n} . Since g is increasing,
h(x) = g(x) £ [#«), g(b_{n})].
If h(x) equals g(a_{n}) or #(6_{n}), then h(x) e S, so w = x + h(x) e E + 5. On the other hand, suppose h(x) 4= #(a_{n}), <?(&„). Then h(x) £ K_{n} . Also, g(x) = h(x) £ J_{m} for some m. Thus h(x) £ J_{m}C\ K_{n} and x £ I_{n} C\ g~\J_{m})_{}} so that
w = x + h(x) £ (I_{n} C\ g~\J_{m})) + (J_{m} C\ K_{n}).
This proves the Assertion.
Since g is increasing, g^J^ is an interval, so both I_{n} C\ g~\J_{m}) and J_{m} C\ K_{n} are intervals. Also note that m 4= I implies g~\J„^ g~\Ji) —</> . By the Assertion,
M.W g + s) + S E n g~\j_{m}y) + (J_{w} n #_{n})] mN mtM
g n_{e}(E + s) + E E [m(/_{b} n n o
mN mzM
= m/U (s + t?)) + E [ E n g~\j_{m}y) + £ »(j_{m} c\ K_{n})] szS mN mtM mtM
< E <em>+ E)</em> + E W.) + E mGA. n 2Q]
utS mN mtM
= 0 + ju(G) + E E c\ K_{n}) nzN mzM
= m(g) + E E m(A. n K_{n})
mtM nzN
m(G) + E mW = m( <?)+ 1EH) < e. mt M
Since e is arbitrary, jn_{e}(W) = 0.
Lemma 11. Let L — {(x, a)  x £ E} and M = {{x_{y} b)  x £ R} be two horizontal lines in R^{2}. Let E be a set of (linear) measure 0 in L and let F be a set of (linear) measure 0 in M. Let £>be a set of closed line segments such that
(a) , $2 £ <£, S2 ■—Si F\ S2 ^{=}
(b) $£ <£=>one endpoint of s lies in E and the other endpoint lies in F.
Let S = \J_{kt&} s. Then u^{2}(S) = 0.
Proof. Assume without loss of generality that b >a. For any (x, y) £ R^{2} let ^({x, y)) = x. For any y £ R let l_{y} = {{x, y)  x £ R}. Let
E_{o} = {z £ E  z is the endpoint of some s £ <£},
and observe that E_{o} has linear measure 0. For any set A C R^{2} we of course set tt(A) == {x £ R  (xj y) £ A for some y £ R}.
We define a function h on ir(E_{0}) as follows. If $ £ ^(Eq), then {x, a} zE_{0} , so we can choose a (unique) segment s £ <£ with one endpoint at (x, a). If the other endpoint of s is p, we set h(x) = 7r(p). Clearly h maps ir(E_{0}) into tt(F).
Since the segments in £ cannot intersect each other, h must be an increasing function.
Take any y_{Q} with b >y_{Q} > a. Let c = b — y_{0}, d = y_{0} — a. A simple computation shows that if q s l_{Vo} C\ 8, then
tt(q) =
ex + dh(x) c + d
for some x s tt(£?_{0}) So
7r(l_{yo} A 8) C
ex + dh{x) . e + d
X £ 7r(Fo)
Now (d/e)h(x) is an increasing function mapping tt(E_{q}) into (d/c)7r(F), so by Lemma 10
x + h(x)
X £ Tt(Eq)
has measure 0. Hence
X £ Tt(Eq)
ex + dh(x) . e + d
X £ Tt(E_{0})
has measure 0, so A SY) = 0. But A 8)) = 0 also when y_{Q} 4 (a, b), so A 8)) = 0 for every y. If we knew that 8 were measurable, the lemma would follow immediately from the Fubini theorems. But since we have, as yet, no guaranty of the measurability of 8, a more complicated argument is necessary. At several stages in the argument the reader will find it useful to draw diagrams to help him visualize the situation.
For any y_{Y} , y_{2} s R, let
U(,yi > 2/2) = {{x, y) \ x, y t R, y_{t} < y < y_{2}}.
A set of the form U(yi , y_{2}) will be referred to as a horizontal open strip.
For each positive integer n, let <£(n) denote the set of all segments s e £ such that s has a point in common with {{Xj b)  x e (—n, ri)}. Let
S(n) = [ U _{S}]H + i bi).
Since l_{a} and l_{b} have (plane) measure 0, and since
sc I.U4U Q S(n), n = l
it is sufficient to show that each S(ri) has measure 0.
Let n be a fixed positive integer. Set a* = a + 1/n and b* = b — 1/n. Take any e > 0. Choose €_{0} so that 2e_{0} + e* < e/(b — a). Let y_{0} be any member of [a*, b*]. For the time being, y_{Q} will be held fixed.
For each s e <£, let p_{8} be the endpoint of 8 on l_{b} , let q_{t} be the intersection point of s with l_{V9} , and let r_{a} be the endpoint of 8 on l_{a} .
Choose an open set G £ R such that tt(1_{Vq} A S(n)) C G and g((?) < e_{0} • Say G = VJ, Ij, where Z, = (a_{f} , b^ and the Z, ’s are pairwise disjoint. We may assume that each Z,« contains a point of tt(1_{V9} S(n)). For each let
Ci = g.l.b. {7r(p_{a})  s e £(n), 7r(g_{a}) e ZJ,
dj = l.u.b. {^(p,)  s e £(n), 7r(g«) e Z,}, c< = g.l.b. {?r(r_{a})  8 e <£(n), ?r(^) e Z,}, d'i = l.u.b. {?r(r_{a})  8 s £(n), ir(q_{8}) e ZJ.
Note that c, g d, and c< g d< . Since the segments in £ cannot intersect each other, it is easily seen that the intervals (c, , d,) are all pairwise disjoint. It is also clear (from the definition of £(n)) that each (c,, d,) is a subset of (~ n, ri). Hence, if we set a, = d,« c_{7} , we have 22, a, g 2n.
For each j, let s(j) be the line segment joining the two points (c< , a), , b), and let t(j) be the line segment joining the two points (d< , a), (d, , b). Let A_{f} be the closed subset of U(a, b) which is enclosed by the two line segments 80), ^0). Let Hj denote the intersection of A, with the horizontal open strip
V — LI max
€o I • J 7 I ^{€}0 I I ^{o}’^^p^{min}V’^ ^{+} 2^/r
Note that H_{f} is measurable. Setting ZZ = H_{f}, it is clear from the definition of the A/s that
S(ri) C\V QH.
Take any y s R. We wish to show that
_{M}0(ZZ A l_{y})) <
__ e__ b — a
We can, of course, assume that
l J ^{€}o I • J 7 . €o I 
y e I max \a, y_{a} — , mm S b, y_{0} + ^(r
An elementary computation, using the geometrical properties of Hj, shows that
n i,)) g (i + 4^
\ o y_{Q} / o y_{Q}
Therefore
n z„)) g E n w
 G+^{n} ^)^{e}°+^{2n2}
” 2c_{0} + €_{0} < t , b — a
so v(ir(H l_{v}Y) < e/(b — a) for every y.
We have shown that for each y_{Q} e [a* &*] there exists a horizontal open strip
V(y_{Q}) containing l_{Vo} , and there exists a measurable set H(y_{Q}) C y(?/_{0}), such that
S(n) H V(y_{0}) £ H(y_{0})
and (for every y) ir(H(y_{Q}) l_{v}) is measurable and
y«H(y_{0}) Cy Z_{y})) <
The various open strips VG/o) (i/_{0} « [a*, 6*]) clearly cover the compact set {(0, y} I y ^{e} f^{a}> &]} • Choose a finite subcovering V(y_{2}\ • • • , V(y_{m}). Set
tm— 1 /
m \
U V(y,)]
J“»+l /
n I7(a*, 6*).
H(y_{m}) U \J IW) 
X
Obviously K is measurable, and for each y, ir(K A Z„) is measurable and C\ l_{v})) < e/(b  a). Moreover, S(n) C K. We have
\K) = £ ^(K C\ 1,)) dy g £ dy = (6  a*) < e.
Since e is arbitrary, this shows that
g.l.b. {/(X)  K measurable, S(n) C K} = 0.
Therefore S(n) has measure 0.
Lemma 12, For every e > 0 there exists a strictly increasing function h on R such that h(R) has measure 0, and for every x, I# — A(x) e
Proof. For each (not necessarily positive) integer n, let I_{n} = [ne, (n + l)e]. Then I_{n} = R. There exists a strictly increasing function f : [0, 1] [0, 1]
such that m(/([0, 1])) = 0. For example, such a function may be defined as follows. Any number in [0, 1) may be written in the form
.a_{1}a_{2}a_{3} • • • a_{n} • • • , (binary decimal),
where the decimal does not end in an infinite unbroken string of l’s. Set
f(.a!a_{2}a_{3} • • • a_{n} • • •) = bib_{2}b_{3} • • • b_{n} • • • , (ternary decimal), where bi = 0 if = 0 and 6, = 2 if a, = 1. Set f(l) = 1. f maps [0, 1] into the Cantor set, so m(/([0, 1])) = 0. It is easily shown that / is strictly increasing.
For each n, it is easy to obtain from / a function f_{n} :I_{n}> I_{n} such that j_{n} is strictly increasing and /z(fn(In)) = 0. Set
h(x) = f_{n}(x) for x e (ne, (n + l)e].
There is no difficulty in proving that h has the desired properties.
Theorem 6. Let be an arbitrary junction on C^{Q} = {{x, 0)  x e R}. Then there exists a junction j on D° — {{x, y)  y > 0} such that j(z) = 0 almost everywhere and <p is a boundary junction jor j.
Proof. For each positive integer n let h_{n} be a strictly increasing function on R such that y(h_{n}(R)) = 0, and for every x, ]x — h_{n}(x)j ~ 1/n. Let and E_{n} has linear measure 0. For each n, x let s_{n}(x) be the line segment joining (h_{n}(x), 1/n) and (A_{n+1}(a;), l/(n + 1)). Since
E_{n} is a subset of
h_{n}(x) > h_{n}(x') x > x^{f} => h_{n+1}(x) > h_{n+1}(x'\
we find that x 4= x^{f} implies s_{H}(x) C\ s_{n}(x^{f}) = 0. Since each s_{n}(x) has one endpoint in E_{n} and the other in E_{n+1} , Lemma 11 shows that for each n
= 0.
x&R
Hence
M^{2}(0 Us_{n}(a:)) = 0. 'n=^{s}l xtR •
Set
j(z) = £>(($, 0)), if z e s_{n}(x) for some n,
f(z) = 0, if z is not in any s_{n}(x).
j(z) = 0 almost everywhere. Let
y(x) = {{x, 0>} U 0s_{n}(x). 74 = 1
Since the endpoints of s_{n}($) are at (A_{n}(z), 1/n) and (h_{n+1}(x)j l/(n + 1)), and since (h_{n}(x), 1/n) —> (x, 0) as n —> oo, it is clear that y(x) is an arc at (x_{}} 0). Obviously
lim j(z) = <p({x_{t} 0)).
This proves the theorem.
Corollary. There exists a measurable function in D° having a nonmeasurable boundary function.
6. Concluding remarks. Our theorem on boundary functions for continuous functions could have been proved by a small modification of the argument in Section 4, but the proof in Section 3 is shorter and neater.
The reader may wonder whether Theorem 4 holds true for functions taking values on the Riemann sphere as well as for realvalued functions. The theorem does, in fact, remain true in the spherevalued case. If we regard the Riemann sphere 2 as a subset of R^{3} and apply Theorem 4 to each component of f and <p, we find that <p is of Baire class a + 1 with R^{3} as the universal range space. It is then easy to show by means of Satz 2 in Banach’s paper [3] that <p is of Baire class a + 1 with 2 regarded as the universal range space. A similar procedure shows that Theorem 5 also remains true for functions taking values on the Riemann sphere.
The results of Sections 2, 3 and 4 cannot be extended to three dimensions—at least not in the most obvious way. We can show this as follows. Let K be an open cube in R^{3} and let F be one face of K. If f is defined in A, then we say <p (defined on F) is a boundary function for f provided that for each x e F there exists an arc y with one endpoint at x such that y — {x} C K and
lim f(v) = (p(x). v—>® vzy
Lemma 13. Suppose that every point of F is an ambiguous point of the function f : K —» R^{3}. Then f has a nonmeasurable boundary function.
Proof. Let E be a nonmeasurable subset of F. Since each point of F is an ambiguous point we can choose, for each x e F, two distinct points ^(x), <p_{2}(x) e R^{3} such that there exist arcs y_{{} at x with
lim f(p) = Pitx), (i = 1,2). v—*x vsy i
Let
<p(x) = <pi(o:), if x vE,
cp(x) ~ if x e F — E.
Then
<p(x) — ^(x) =0, if xtE,
<p(x) — Vi(x) 1
0, if x s F — E.
Therefore (<p — ^i)^{1}({0}) = E, so — <p_{x} is not a measurable function. Hence either <p or <p_{x} is a nonmeasurable function. Since <p and <p_{Y} are both boundary functions for /, the lemma is proved.
P. T. Church [4] has constructed an example of a homeomorphism / from K onto K such that every point of F is an ambiguous point for f. By Lemma 13, f has a nonmeasurable boundary function <p. Theorem 1 is therefore false in three dimensions. Write f and <p in terms of their components; say / = (f i, f_{2}, fa) and <p = , <?2, ^3). Since <p is nonmeasurable, one of its components, say <?, ,
is nonmeasurable. But is a boundary function for the continuous realvalued function f,« , so Theorem 2 and Theorem 4 must be false in three dimensions.
References
The University of Michigan
5. 1966  On a Boundary Property of Continuous Functions
Original PDF: 5. 1966. On a Boundary Property of Continuous Functions.pdf
Kaczynski, T.J. 1966. On a boundary property of continuous functions. Michigan Math. J. 13:313320.
MR0210900 Kaczynski, T. J. On a boundary property of continuous functions. Michigan Math. J. 13 1966 313.320. (Reviewer: D. C. Rung) 30.62
Explanation by John D. Bullough
The author generalizes the result of McMillan (1966) to the effect that the set of curvilinear convergence of a continuous function f from D into Z is of type F(sd). The generalization considers f as a continuous function from D into a compact metric space E. Topologizing the set of closed sets C(E) of E with the Hausdorff metric and letting E be any closed set in C(E), it is shown that the set of all x ( C such that there is a boundary path v at x with the cluster set of f along v contained in some set of E is of type F(sd). Taking E to be the set of all singletons {y}, y ( E (which is closed in C(Z)) McMillan's theorem is obtained.
Various other corollaries are given by selecting appropriate closed sets E ( C(E).
Article by Ted
On a Boundary Property of Continuous Functions
T. J. Kaczynski
Let D be the open unit disk in the plane, and let C be its boundary, the unit circle. If x is a point of C, then an arc at x is a simple arc y with one endpoint at x such that y  {x} c D. If f is a function defined in D and taking values in a metric space K, then the set of curvilinear convergence of f is
{x e C there exists an arc y at x and there exists a point p e K such that lim f(z) = p} .
Z —> X zey
J. E. McMillan proved that if f is a continuous function mapping D into the Riemann sphere, then the set of curvilinear convergence of f is of type F_{a}$ [2, Theorem 5]. In this paper we shall provide a simpler proof of this theorem than McMillan’s, and we shall give a generalization and point out some of its corollaries.
Notation. If S is a subset of a topological space, S denotes the closure and S* denotes the interior of S. Of course, when we speak of the interior of a subset of the unit circle, we mean the interior relative to the circle, not relative to the whole plane. Let K be a metric space with metric p. If x_{0} e K and r > 0, then
S(r, x_{0}) = {x e K p(x, x_{0}) < r} .
An arc of C will be called nondegenerate if and only if it contains more than one point.
LEMMA 1. Let bea family of nondegenerate closed arcs of C. Then Uie^ I " Uje/Z I* countable.
Proof. Since U_{Ie}^r I* is open, we can write I* = U_{n} J_{n}, where {J_{n}} is a countable family of disjoint open arcs of C. If
X_{0} € U I  U I* , Ie# Ie#
then for some I_{o} e #, x_{0} is an endpoint of I_{o}. For some n, Iq c J_{n}, so that x_{0} e • But x_{0} / J_{n}, so that x_{0} is an endpoint of J_{n}. Thus Uj I  Uj I* is contained in the set of all endpoints of the various J_{n}; this proves the lemma. ■
In what follows we shall repeatedly use Theorem 11.8 on page 119 in [3] without making explicit reference to it. By a crosscut we shall always mean a crosscut of D. Suppose y is a crosscut that does not pass through_the point 0. If V is the component of D  y that does not contain 0, let L(y) = V Cl C. Then L(y) is a nondegenerate closed arc of C.
Received February 8, 1966.
I
Suppose n is a domain contained in D  { 0}. Let r denote the family of all crosscuts y with y n D c n. Let
1(0) = U L(y), I_{0}(n) = U L(y)* .
yE r yE r
Let ace (n) denote the set of all points on C that are accessible by arcs in n.
The following lemma is weaker than it could be, but there is no point in proving more than we need.
LEMMA 2. The set ace (n)  I0(n) is countable.
Proof. By Lemma 1, I(n)  I0(n) is countable; therefore it will suffice to show that ace (n)  I(n) is countable. If ace (n) has fewer than two points, we are done. Suppose, on the other hand, that ace (n) has two or more points. If a E ace (n), then there exists a' E ace (0) with a' =f. a. Let y, y' be arcs at a, a' , respectively, with
ynncn , y'n n cn.
Let p be the endpoint of y that lies in n, p^ the endpoint of y^{1} that lies in n. Let y" c n be an arc joining p to p' . The union of y, y', and y^{11} is an arc o joining a to a'. By [4], there exists a simple arc 0^{1} co that joins a to a^. Clearly, 0^{1} is a crosscut with 0^{1} n D c n and a, a^ e L(0 '). Thus a e I(n), and so ace (n) c 1(0). •
LEMMA 3. Suppose 01 and Oz are domains contained in D  {O}. If
(1) IqCOj) A acc(O_{1}) and Iq(Q_{2}) A ace (Q_{2})
are not disjoint, then n1 and O_{z} are not disjoint.
Proof. We assume n1 and Oz are disjoint, and we derive a contradiction. Let a be a point in both of the two sets (1). Let Yi be a crosscut with Yi n D cni such that a E L(yi)* (i = 1, 2). Let Ui and Vi be the components of D  Yi, and (to be specific), let Ui be the component containing 0. Note that y_{1}n D and y_{2}n D are disjoint.
Suppose yi nD cVz and yz A D c Vj. Then, since yi n D c Ui, Ui has a point in common with Vz. But O E u_{1} nUz, so that U _{1} has a point in common with Uz also. Since U_{i} is connected, this implies that U_{i} has a common point with Yz n D, which contradicts the assumption that Yz nD cVp Therefore yi nD ^ V2 or Yz n D ^ Vi • We conclude that either y 1 nD c U_{2} or y_{2} nD c U1 . By symmetry, we may assume that Yz n D cU_{i}.
It is possible to choose a point b E L(yi) * that is accessible by an arc in Oz, because a is in the closure of ace (Oz). Let y be a simple arc joining b to a point of y_{4} n D, such that y  { b} c Oz. Then y  { b} and y_{i} are disjoint. Also, y — i_b} contains a point of Ui (namely, the point where y meets y2 n D); therefore y  {b} c Ui . Hence b e Ui. Since b e L{y 1 )*, this is a contradiction. •
THEOREM 1 (J. E. McMillan). Let K be a complete separable metric space, and let f be a continuous function mapping D into K. Let
X = {x E CI there exists an arc y at x for which lim f(z) exists} . z^x
zEy
Then X is of type F_{a} $ .
Proof Let {p_{k}}_{k=1} be a countable dense subset of K. Let {Q(n, be
a counting of all sets of the form
where 0 is a rational number. Let {U(n, m, k, £)}^_{=1} be a counting (with repetitions allowed) of the components of
(We consider 0 to be a component of 0.) Let
A(n, m, k, £) = acc[U(n, m, k, £)].
Set
co co co co
_{Y}= n u u u I_{0}(U(n, m, k, £)) Cl A(n, m, k, £).
n=l m=l k=l f = 1
Since I_{0}(U(n, m, k, £)) is open, it is of type F_{a} . It follows that Y is of type F_{a<}5 .
I claim that Y c X. Take any y e Y. For each n, choose m[n], k[n], f [n] with
(2) y € I_{0}(U(n, m[n], k[n], f[n])) A A(n, m[n], k[n], £[n]) (n = 1, 2, 3, •••).
For convenience, set U_{n} = U(n, m[n], k[n], f[n]). By (2) and Lemma 3, U_{n} and U_{n+}i have some point z_{n} in common. For each n, we can choose an arc y_{n} c U_{n+}i with one endpoint at z_{n} and the other at z_{n+1} . Then y_{n} c Q(n + 1, m[n + 1]). Also,
and therefore
r r
P<Pk [n] ’ Pk [n+r] ^{}} P^{(}Pk[n+il]’ P_{k}[n_{+}i]^{)} < < ^2’
Thus {p_{k}[_{n}]} is a Cauchy sequence and must converge to some point p g K. Because
r_{n}^{C U}n+1 ^{C f}'^{1}(^{S}(^iT>Pk[n+l])) ^{and} Pk[n]^P>
lim f(z) = p. It is possible that y is not a simple arc, but by [4] we can replace y z> y zG y
by a simple arc y’ c y. Thus y e X, and we have shown that Y c X.
Suppose x g X. Let y_{0} be an arc at x such that f approaches a limit p ^{1} along
y_{0}. Take any n. Choose k with p’ g
p_{k} j . Choose m so that x is in the
interior of Q(n, m) A C. Then y_{0} has a subarc yjj, with one endpoint at x, such that
t'q  {x} c Q(n, m) n flfs/X p_{k}) j .
Hence, for some f, x g acc[U(n, m, k, £)] = A(n, m, k, f). This shows that
00 00 00 00
_{X}c n u u u A(n, m, k, £).
n=l m=l k=l f = l
By Lemma 2, the set
A(n, m, k, J?)  I_{0}(U(n, m, k, $.)) = A(n, m, k, £)  [l_{0}(U(n, m, k, £)) Cl A(n, m, k, £)] is countable. It follows by a routine argument that
A U A(n, m, k, jO  A U [l_{0}(U(n, m, k, £)) A A(n, m, k, I) ] n m,k,£ n m,k,£
is countable. Because
A U [l_{Q}(U(n, m, k, £)) A A(n, m, k, £)] = Y c X C A U^{1} A(n, m, k, £), n m,k,j£ n m,k,f
the set X  Y is countable, and therefore X is of type F_{a} 5 . ■
Before stating our generalization of the foregoing theorem, we must say a few words about spaces of closed sets. If K is a bounded metric space with metric p, let ^(K) denote the set of all nonempty closed subsets of K. Hausdorff [1, page 146] defined a metric p on MK) by setting
p(A, B) = max{ sup dist(a, B), sup dist(b, A)} ,
a GA b€B
where dist(x, E) denotes inf p(x, e). If K is compact, then $?(K) is a compact
e € E
metric space with p as metric [1, page 150].
If f maps D into K and if y is an arc at a point x e C, we let C(f, y) denote the cluster set of f along y; that is, we write
C(f, y) = {p 6 K there exists a sequence {z_{n}} c y AD such that z_{n} » x and f (z_{n}) —> p} .
THEOREM 2. Let K be a compact metric space, and let 8 be a closed subset of ^(K). Let f: D —>K be a continuous function. Then
{x e c there exists an arc y at x and there exists
E e 8 such that C(f, y) c E}
is a set of type F_{a} $ .
Proof. If s > 0 and E e ^{r}^(K), let
Z/^{?}(8, E) = {a e k there exists be E with p(a, b) < 8} .
Note that <^(8, E) is open and that
F e ^(K), p(E, F) < 8 =>Fc <^(8, E).
Let {P(k)}k_{=1} be a countable dense subset of 8 (such a subset exists, because every compact metric space is separable). Let
X = {x e c there exist an arc y at x and an E e 8
such that C(f, y) c E} .
r t ^{00}
Let {Q(n, m)}_{m=}j be defined as in the proof of the preceding theorem. Let {U(n, m, k, 4)}£_{=}i be a counting (with repetitions allowed) of the components of
^{p}<^{k}>)) ^{n} Q(^{n}> ^{m}>
Let A(n, m, k, £) = acc[U(n, m, k, £)], and set
CO CO CO OO
_{Y}= n u u u I_{0}(U(n, m, k, £)) n A(n, m, k, f).
n = l m=l k = l £ = 1
Since I_{0}(U(n, m, k, £)) is open, it is of type F_{Q} . It follows that Y is of type F^ .
I claim that Y c X. Take any y e Y. For each n, choose m[n], k[n], f[n] so that
(3) y e I_{0}(U(n, m[n], k[n], £ [n])) n A(n, m[n], k[n], £[n]).
Set U_{n} = U(n, m[n], k[n], £ [n]). Since 8 is compact, there exist a P € 8 and some strictly ascending sequence {n;}°°_{=i} of natural numbers such that
J J
P(k[_{nj}]) _{7} P. ^{J} J
By (3) and Lemma 3, U_{n>} and U_{n}. have some point Zj in common. For each j,’ J i
choose an arc yj c with one endpoint at Zj and the other at Zj_{+1}. Then
7j c Q(nj_{+1}, m[nj_{+1}]). Also,
y e A(n_{j+1}, m[n_{j+1}], k[n_{j+1}], £[n_{j+1}]) c U c Q(n_{j+1}, m[n_{j+1}]), J
and therefore each point of y. has distance less than — from y. Now
J Hj + l
2tt + 1  °°
one endpoint at y.
I claim that C(f, y) c P. Take any p e C(f, y). There exists a sequence {w_{s}}^_{=1} in y  {y} such that w_{s} y and f(w_{s}) p. Let e be an arbitrary positive number. Choose j_{0} so that p(P(k[nj]), P) < e/3 for all j > j_{0} . Choose so that j > jj implies l/nj_{+i} < e/3. We can choose an s such that w_{s} e y^ for some i > j_{0} , jj and such that
(4) p(f(w ), p) <L
o
Then
f(w_{s}) e f(y.) c f(U_{n}._{H}) c P(k[n_{i+1}])) ,
and therefore we can choose a point q e P(k[n_{i+1}]) with
(5)
P(f(w ), q) < — » n. ।
E
3*
Moreover, because p(P(k[n_{i+1}]), P) < e/3, there exists some q’ € P with
(6)
p(q, q') <
Together, (4), (5), and (6) show that p(p, q’) < e. Since P is closed and e is arbitrary, this proves that p € P. Hence C(f, y) c P e 8. By [4], we can if necessary replace y by a simple arc y^{1} c y; it follows that y 6 X. Thus Y c X.
Now suppose x e X. Choose an arc y_{0} at x such that C(f, y_{0}) c P_{o} for some P_{o} e 8. Take any n. Choose k with p(P_{0}, P(k)) < 1/n. Then
^{p}o^{c}^(^ w) >
hence C(f, y_{0}) ^{c} P(k) j .
Choose m so that x is in the interior of Q(n, m) A C.
If for each natural number t there exists a point z’_{t} e y_{0} A S x j AD with ^{p(k)})) ’^{then}
f(z') e K S (i P(k)\,
and since K  SP ( P(k) j is compact, there exist some a e K  SP P(k) j and a subsequence {f(z’ )}.°\ such that f(z’ ) => a. But then a € C(f, y_{0}), contrary to ^{t}i । \ ^{r}i
the relation C(f, y_{0}) ^{c} , P(k) ) . We conclude that there exists a natural number t for which
y_{0} n s(,x) o C P(k)) ) •
It follows that y_{0} has a subarc y^ with one endpoint at x such that
y'_{Q} {x} c P(k)) ) nQ(n, m).
Hence there exists an £ such that
x e acc [U(n, m, k, £)] = A(n, m, k, £).
This shows that co oo oo co xcn u u uA(n, m, k, £). n=l m = l k=l £ = 1
By Lemma 2, the set
A(n, m, k, £)  I_{0}(U(n, m, k, £)) = A(n, m, k, J?)  [l_{0}(U(n, m, k, £)) A A(n, m, k, {)] is countable. It follows easily that
A U A(n, m, k, £)  A U [l_{0}(U(n, m, k, £)) A A(n, m, k, £)] n m,k,£ n m,k,£
is countable. Since
A U [I_{0}(U(n, m, k, £)) A A(n, m, k, 4)] = Y c X c A U A(n, m, k, £), n m,k,£ n m,k,£
X  Y must be countable. Thus X is the union of an F_{a}$ set and a countable set, , and hence it is of type F_{a} $ . ■
In each of the following four corollaries, let f denote a continuous function mapping D into the Riemann sphere.
COROLLARY 1 (J. E. McMillan). Let E be a closed subset of the Riemann sphere. Then the set
{x e C  there exist an arc y at x and a point p e E such that lim f(z) = p} z —>x z € y
is of type F_{a 6} .
COROLLARY 2. Suppose d > 0. Then the set
{x e C there exists an arc y at x such that [diameter C(f, y)] < d}
is of type F_{Q 6} .
COROLLARY 3. Let E be a closed subset of the Riemann sphere. Then the set
{x e C  there exists an arc y at x with C(f, y) c e}
is of type F_{a 6} .
COROLLARY 4. The set
{x e C there exists an arc y at x such that C(f, y) is an arc of a great circle }
is of type F_{Q 6} .
We can obtain all these corollaries by taking 8 to be a suitable family of closed sets and applying Theorem 2. To prove Corollary 4, we need the fact that C(f, y) is always connected. One could go on listing such corollaries ad infinitum, but we refrain.
It is interesting to note that in Corollary 1 it is not necessary to assume that E is closed. By combining Corollary 1 with Theorem 6 of [2], one can prove that the conclusion of Corollary 1 holds even if E is merely assumed to be of type .
REFERENCES
1. F. Hausdorff, Mengenlehre, Zweite Auflage, Walter de Gruyter & Co., Berlin und Leipzig, 1927.
2. J. E. McMillan, Boundary properties of functions continuous in a disc, Michigan Math. J. 13 (1966), 299312.
3. M. H. A. Newman, Elements of the topology of plane sets of points, Cambridge University Press, 1961.
4. H. Tietze, Uber stetige Kurven, Jordansche Kurvenbogen undgeschlossene Jordansche Kurven, Math. Z. 5 (1919), 284291.
The University of Michigan
6. 1967  PhD thesis at University of Michigan  Boundary Functions
Original PDF: 6. 1967  PhD thesis at University of Michigan  Boundary Functions.pdf
BOUNDARY FUNCTIONS
KACZYNSKI, THEODORE JOHN
ProQuest Dissertations and Theses; 1967; ProQuest
This dissertation has been ■ — —
microfilmed exactly as received 6717,790
KACZYNSKI, Theodore John, 1942 BOUNDARY FUNCTIONS. 
The University of Michigan, Ph.D„ 1967 Mathematics
University Microfilms, Inc., Ann Arbor, Michigan
BOUNDARYFUNCTIONS
by
Theodore John Kaczynski
A dissertation submitted in partial fulfillment of the requirements for the degree of Doctor of Philosophy in the University of Michigan
1967
Doctoral Committee:
Professor Allen L. Shields
Assistant Professor Peter L. Duren
Associate Professor Donald J. Livingstone Professor Maxwell O. Reade
Professor ChiaShwi Yiih
BOUNDARY FUNCTIONS
By Theodore John Kacijnski
Abstract
Let H denote the set of all points in the Euclidean plane having positive ycoordinate, and let X denote the xaxis. If p is a point of X, then by an arc at p we mean a simple arc v, having one endpoint at p, such that v  {p} ( H. Let f be a function mapping H into the Riemann sphere. By a boundary function for f we mean a function t defined on a set E ( X such that for each p ( E there exists an arc v at p for which
lim f(z) = t(p).
z > p
z ( v
The set of curvilinear convergence of f is the largest set on which a boundary function for f can be defined; in other words, it is the set of all points p ( X such that there exists an arc at p along which f approaches a limit. A theorem of J.E. McMillan states that if f is a continuous function mapping H into the Riemann sphere, then the set of curvilinear convergence of F is of type F(sd). In the first of two chapters of this dissertation we give a more direct proof of this result than McMillan's, and we prove, conversely, that if A is a set of type F(sd) in X, then there exists a bounded continuous complexvalued function in H having A as its set of curvilinear convergence. Next, we prove that a boundary function for a continuous function can always be made into a function of Baire class 1 by changing its values on a countable set of points. Conversely, we show that if t is a function mapping a set E ( X into the Riemann sphere, and if t can be made into a function of Baire class 1 by changing its values on a countable set, then there exists a continuous function in H having t as a boundary function. (This is a slight generalization of a theorem of Bagemihl and Piranian.) In the second chapter we prove that a boundary function for a function of Baire class e > 1 in H is of Baire class at most e + 1. It follows from this that a boundary function for a Borelmeasurable function is always Borelmeasurable, but we show that a boundary function for a Lebesguemeasurable function need not be Lebesguemeasurable. The dissertation concludes with a list of problems remaining to be solved.
TABLE OF CONTENTS
LIST OF ILLUSTRATIONS
INTRODUCTION
Chapter
I. BOUNDARY FUNCTIONS FOR CONTINUOUS FUNCTIONS
II. BOUNDARY FUNCTIONS FOR DISCONTINUOUS FUNCTIONS
SOME UNSOLVED PROBLEMS
REFERENCES
LIST.OF ILLUSTRATIONS
Figure Page
1. 27
2. Trap (J, _{e}, 0)
3. Tri (J, 0)
4. S(x_{0}, e, 0)
5. 62
INTRODUCTION
1. Preliminary Remarks
Let H denote the upper halfplane, and let X denote its frontier, the xaxis. If x£X, then by an arc at x we mean a simple / 
arc y with one endpoint at x such that y  Suppose that f
is a function mapping H into some metric space Y. If E is any subset
I of X, we will say that a function <p:E >Y is a boundary function for f if, and only if, for each x€E there exists an arc Y at x such that
lim f(z) = ip (x) . z x z € y
The study of boundary functions in this degree of generality was initiated by Bagemihl and Piranran [2]. A function defined in H may have more than one boundary function defined on a given set E S X, but it follows from a famous theorem of Bagemihl [1] that any two such boundary functions differ on at most a countable set of points.
If f is defined in H, then the set of curvilinear convergence of f is the set of all points x£ X such that there exists some arc alj x along which f approaches a limit.. Evidently, this is the largest set on which a boundary function for f can be defined.
J. E. McMillan [10] discovered that the set of curvilinear convergence of a continuous function is always of type F and in this paper we show that every set of type in X is the set of curvilinear convergence of some continuous function. Next, we show that if is a function defined on a subset E of X, then <p is a boundary function for some continuous function if and only if if can be made into a function of the first Baire class by changing its values on at most a countable set of points. (This solves a problem of Bagemihl and Piranian [2, Problem ,1].) We then consider functions that are not assumed to be continuous, and we prove that a boundary function for a function of Baire class 5 1 is of Baire class at most 5+1 (thus proving another conjecture of Bagemihl and Piranian [2]). It follows from this that a boundary function for a Borelmeasurable function is always Borelmeasurable, and in the last section we show that a boundary function for a Lebesguemeasurable function need not be Lebesguemeasurable.
Most of the results appearing here have already been published ([8] and [9]). At the time I published these papers I did not expect to have to make use of this material for a dissertation.
2. Notation
R will denote the field of real numbers, n
R will denote ndimensional Euclidean space.
Points in R^{n} will be written in the form {x.,..., x \ rather than (Xp..,, x_{n}) (to avoid confusion with open intervals of real numbers in the case n = 2).
If v€R^{n}, then v denotes the length of the vector v. 2*3'2
S denotes {vgR : v = 1}. S^{z} will be referred to as the Riemann sphere.
. Let
H = ’ { <x,y>6 R^{}[2] : y > 0} ? 1
H_{n} = { <x,y>£R^{Z} : £ > y > 0} X = { <x,0>: x€R} < ^{X}n= ^{{}<^{x}4> ^{: x€R}} •
We consider X as being identical with R. Thus, for example, <x,0>< <x’ ,0^ means x £ x', and for p, q € X, the notations [p,q], [p,q), etc. refer to the obvious intervals on X.
If E is a subset of a topological space, then E" denotes the i i
it closure of E, E denotes the interior of E, and E’ denotes the
* complement of E. Of course, if E is a subset of X, then E means the interior of E relative to X, not relative to the whole plane. In Section 7, we often denote two line segments by s and s'. Since the prime notation is never used for complementation in that section, there is no danger of confusing s' with the complement of s.
2
If f is a function defined in a subset of R , then f(x,y) means f(<x,y^). Thus we write f(z) for z£R and f(x,y) for x,y£R interchangeably. ^{1 1}
3. Baire Functions
In this section we review the main facts concerning Borel sets and Baire functions, and we prove some results that will be needed later.
If C is any family of sets, let C_{g} be the family of all sets that can be written as a countable intersection of members of C, and let be the family of all sets that can be written as a countable union of members of C.
Suppose M is a metrizable topological space. Let p\m) be the family of all open subsets of M and let q\m) be the family of all closed subsets of M. If ? is an ordinal number greater than 1, let
P^{5}(M) = (^Jq^(M)) n<C
Q^{C}(M) = (UP^{n}OT)_{s} n<€ ^{9}
For any C, E € Q^{?}(M) <=>E’ 6 P^{?}(M) .
For any subset L of M. E€ P^(L) (respectively Q^(L)) if and ^{1} F F
only if there exists a set D€ P (M) (respectively Q (M)) such that e = dal.
P^(M) and Q^(M) are closed under finite unions and finite intersections. P^(M) is closed under countable unions and Q^(M) is closed under countable intersections.
If n<5, then P^{n}(M) U Q^{n}(M) P^{?}(M)A Q^{?}(M).
Let F$(M) be the class of all F^ sets of M, and let Gg(M) be the class of all G$ sets of M.
2 2
P (M) = F (M) and Q (M) ' = G$(M).
Let Y be a metric space. For any family C of subsets of M we will say that a function f : M > Y is of class (C) if and only if f"\u)€C for every open set U — Y.
The following definition of the Baire classes is somewhat different from the classical definition, but it seems more.convenient for our purposes. A function f : M > Y is said to be of Baire class 0(M, Y) if and only if it is continuous. If 5 is an ordinal number greater than or equal to 1, then f is said to be of Baire class • 00
£J(M, Y) if and only if there exists a sequence of functions mapping M into Y, f being of Baire class n_{n}(M,Y) for some n_{n} < £, such that f ■* f pointwise.
If f : M + Y is of Baire class £(M, Y) and if L is a subset of M, then f_{T} is o'f Baire class 5(L, Y).
If K is a metric space, if g : K •* M is continuous, and if ■ f : M > Y is of Baire class ?(M, Y), then the composite function f og is of Baire class €(K, Y).
If Y is separable and if f : M > Y is of Baire class £(M, Y), then f is of class (P^^{+}\m)) [4, page 294].
If Y is separable and arcwise connected, if g 1, and if f : M> Y is of class (P^^{+}\m)), then f is of Baire class ?(M, Y) [4].
For any 5, if f : M > R is of class (P^^{+}\m)), then f is of Baire class £(M, R) [6].
If l€Q^^{+}\m) and f:L ■* R is of Baire class 5(L, R), then f can be extended to a function ? : M + R of Baire class 5 CM, R) [6].
We say that a function f : M •* R is Borel measurable if, and only if, for every open set u£r, f’^CU) is a member of the oring of subsets of M generated by the open sets.
If f : M ■> R is of some Baire class ?C^{M}, R), then f is Borel measurable, and, conversely, if f : M > R is Borelmeasurable, then f is of Baire class £i(M, R) for some countable ordinal number 5 [7, page 294].
The proofs of Lemmas 1 through 6 are based on standard techniques in the study of Baire functions.
Lemma 1. Let M be a metric space, and let E and F be two F^ sets in M. Then there exist two disjoint F$ sets A and B £ M such that
EFGA and F  E Q B.
00 oo
Proof. Let E = E and F = I J F , where E and F are closed, n n n n
n=l n=l
Then
B_{n}, F_{n} £ f_{o}(m)a g_{6}(M).
It is easy to check that F^fM) AG^fM) is an algebra (i.e., is closed under complementation, finite unions, and finite intersections). We inductively define a sequence of pairs of sets (A_{n}, B_{n}) as follows. Let
A_{x} = E_{x} , B_{x} = F_{x} A A’ .
For n > 1, let
n1 n
A = E n C\ B! , B = F nfl A! . n n^{n} ’ n n'' j
By induction, A_{r}, B_{n} € F$(M) A G$(M). Let , CO 00
A =U A_{n} , B =U B .
n=l n=l
Then A and B are F sets. Notice that
(j ^{B}  £ ^{F} O A. g e ,
j=l ^{3} j=l ^{3}
from which it follows that nl ,
A = E a ( U B.) O E A F’ , n n ^{z v} — n *
1=1 ^{J}
and n ,
^{B}n ■ y 2 ^{P}n^{AB}' •
3=1.
Therefore
00
A 5 LJ CE_{n}A F’) = E  F n=l
and
CO
^{B} “ U CF_{n}AE’) = F  E. n=l
It only remains to show that A A B = <j>. Suppose x 6 A A B. Choose
£, m with x £ A and x € B . If £ > m, then £ > 1, so that
J6 in
Hence x € B’ —a contradiction. On the other hand, if St, < m, then m —
m
f.nA A! a a; , 1=1 ^{J}
so that x€£A} —another contradiction. We conclude that AAB =
If E is a subset of a space M, we let x_{E} denote the characteristic function of E.
Lemma 2. Let L be a. subset of a metric space M, and suppose that
co
E G F^CL) /A GgfL). Then there exists a sequence {f_{n}}_{n}_^°f continuous realvalued functions on M such that f^ > x_{E} pointwise on L.
Proof. Both E and L  E are in F^CL), so there exist sets E^, F^ £ F^fM) such that
E = E^a L and . L  E = F^n L.
By Lemma 1, there exist A, B € F$(M) such that A AB = <>and
E^  F^ G a, F^  E^ G B. We have
E = AAL and . L  E = BAL.
00,00..
Write A =LR. bU. ®n* where A^, are closed and A_{n} ^_{n}+i’ n=l n=l
B^ B for each n. By Urysohn's Lemma there exists a continuous function f_{n} : M »■ [0,1] such that
f fx) = 1 when x £ A
n^{1, J} • n
f f x) = 0 when x £ B .
n n
CO
{f_{n}}_{n=1} is the desired sequence.®
Lemma 3. Let L be a subset of a metric space M, f : L> R a function
of class (F_{a}(L)) that takes only finitely many different values.
Then there exists a sequence {f } .of continuous realvalued ^{n} n n=l
functions on M such that f_{n} > f pointwise on L.
Proof. From Banach’s Hilfssatz 3 [4], we see that there exist real
numbers a_{n},..., a and sets
1 _{;}q .
Ep..., E_{q}€F_{o}(L)r»G_{6}(L)
such that
. q
f = E a. Xn •
. . i E.
J=1 ^{J} J
■ 1 00
If we choose for each j a sequence {f_{n}^{J}J_{n}_^ of continuous realvalued.
functions on M such that f > x_{c} pointwise on L, and if we set n _ _{n} c j
■ 00
then is the desired sequence.®
Lemma 4. Let L be a metric space, f a bounded realvalued function
• . co
on L of Baire class 1(L, R). Then there exists a sequence tf_{n}^_{n}_j of realvalued functions on L converging uniformly to f, such that each f is of class (F^fL)) and takes only finitely many different values. '
Proof, f is of class (F^CL)) and the range of f is totally bounded, so an obvious modification of the proof of Banach’s Hilfssatz 4 [4] gives the desired result.®
Lemma 5. Let M be a metric space, L a subset of M, f : L + Ra /
’ ■ co
function of Baire class 1(L, R). Then there exists a sequence
of continuous realvalued functions on M such that f + f pointwise on L.
Proof. We first* prove the lemma under the assumption that f is bounded. For any bounded realvalued function h, let •
Uh II = sup { h(x)  : x£ domain of h} .
By Lemma 4 we can choose, for each n, a function g_{ri} : L + Rof class (F^CL)) such that g_{n} takes only finitely many different values and
^{Let}
^{h}l «1 ■ ^{h}n ^{=}  «nl ^{for} ” ^{>K}
Then, for ri > 1,
UM <sup>=</sup>  <sup>f + f</sup>  «nlll
i it it *
Each h n
values,
is of class (F_{o}(L)) and takes only finitely many different
* *1 00
so by Lemma 3 we can choose (for eachn) a sequence {h_{n}^{J}}^_^
of continuous functions on M such that h^ h_{n} pointwise on L.
Set
k^fx) =   h_{n} if h_{n}^{J}'(x) <_   hjl
^{k}n^{j}« = H^{h}nH ^{i£ h}_{n}^{j}W> H^{h}nll
^{k}n^{j W = h}n^{j Cx) i£} " II ^{h}nll ^{< h}n^{j (x) <} H ^{h}nH *
Then is continuous, k^ j h_{n} pointwise on L, and  k^ ' £ II ^_{n}ll < —. Therefore, if we set 2^{n}“^{2} CO
f. = Z k j ,
J i n
^{J} n=l
then the series converges uniformly and fj is continuous on M. We claim that ff pointwise on L. Take any x £ L and any e > 0. Choose m large enough so that —< 4 ^{e}* For each ^{n}> choose j (n) so 2^{m} that
Let i = max {j (1) ,..., j (m) }. Then j >_ i_{Q} implies that
f (x)  f(x)  < f.(x)  E k ^{j}(x) +  E ^{3 3} n=l ^{n} n=l
m
_{+} I z n=l
“ ii ^{m}
< Z II k_{n}^{3} II + E k_{n}^{3}(x) h_{n}(x)l +  g_{m} n=m+l n=l
n m , _{n}
< —iy + ( E —i=)+ —= e.
< _{o}m2 ^{k} . „n+l ' 3 _m 3
z n=l 2 2
Thus f. fx) ■* f(x) for each x € L, and the lemma
m k \x)  E h (x) n=l
h_{n}«  £(X)
 f II
is proved for bounded ^{J} J f. __ _
If f is not bounded, let
g(x) = arctan f(x)
Then  ^{<} gW < y for every x € L, and g is of Baire class 1(L, R),
00
so there exists a sequence {g_{n}^_{n}_j °f continuous functions on M converging to g pointwise on L. Set
h_{n}M = 14 if . g_{n}(xj <  41
h_{n}(x) =41 if g_{n}(x)> 11
h_{n}(x) = . ^(x) if . ’ . 1< ^(x) < ’ . 1 .
71 7T
Then h_{n} is continuous on M,  < h_{n}(x) < and Ir ■* g pointwise on
L. Let f (x) = tan h (x). Then f is continuous on M and f > f n n n n
pointwise on L.l '
Lemma 6. If L is a subset of a metric space M and f : L ■> R^{m} is a function, then the following are equivalent.
(i) f is of Baire class 1(L, R^{m}) .
(ii) f is of class (F^fL)). 00
(iii) There exists a sequence <^_{n}^_{n}—j of continuous functions mapping M into R^{m} such that f •* f pointwise on L.
This lemma is an easy consequence of Lemma 5.
3
Definition. Let q be any point of R lying inside the bounded open 2 3 2
domain determined by S . By the qprojection of R  {q} onto S we
mean the function P^ defined as follows. If a is any point of 3
R  {q}, let I be the unique ray, having its endpoint at q, that
passes through a, and let P^(a) be the intersection point of I with 2 3 2
S . Pq is a continuous mapping of R {q} onto S that fixes every 2
point of S .
Theorem 1. Let L be an arbitrary subset of R . Then a function 2 2
f : L S is of Baire class 1(L, S ) if and only if it is of class (F_{a}(L)).
Proof. Assume that f : L > S is of class (F_{o}(L)). S $s R , so by
Lemma 6 there exists a sequence ^{of} continuous functions mapping
2 3
R into R such that f > f pointwise on L. Let
^{A}n ^{</sup> V<sup>1 C {v € r3 :</sup> l<sup>V</sup>> <sup>} 7^{}}
11 11 “
^{B}n ^{=} V^{1 C {v€ r3 :} M ' S 7^{}} )
11 11
' ^{C}n ^{=} V^{1 ( {vg :} H
11 11 L*
Let f ° = f_{n}^ • According to [5, Lemma 2.9, page 299], f^^{0} can be n 2 3 1
extended to a continuous function g_{n} : R > {v €. R : v = y}. 2 3
Define h^ : R > R  {0} by setting
h_{n}Cx) = g_{n}(x) if x 6 B_{n}
h_{n}(x) = f_{n}(x) if x € C_{n}.
Since B_{n}, are closed, h_{n} is continuous, and it is easy to verify that h_{n}(x] ■* f(x) for each xG L. Let k_{n} : R^ •* be the composite
function P o h . Then k is continuous, and for each x£L, o n n
2
k_{n}(x) > P_{o}(f(x)) = f(x). Thus f is of Baire class 1(L, S
Definition. Let M and Y be metric spaces. Then a function f : M Y
is said to be of honorary Baire class 2(M, Y) if and only if there exists a countable set N^m and a function g : M ■> Y of Baire class
1(M, Y) such that f(x) = g(x) for every x €. M  N.
2
Theorem 2. Let L be an arbitrary subset of R and let Y be either
2
the real line, a finitedimensional Euclidean space, or S . Then a function f : L •* Y is of honorary Baire class 2(L, Y) if and only if there exists a countable set N C L such that fis of class (F_{q}(LN)).
Proof. Suppose that f : L •* Y is of honorary Baire class 2(L, Y). Then there exists g : L ■> Y of Baire class 1(L, Y) and a countable set N£L such that f_{L N} = gl_{L}__{N}* But gI£__{N} is of class (F_{o}(L  N)). Conversely, suppose that f^ is of class (F_{o}(L  N)), where N is countable. We must show that f is of honorary Baire class 2(L, Y). First consider the case where Y = R^{m}. Write
f(x) = Xf^x), f_{2}(x), ...,
Then f^^ is of class (F^CL  N)) (i=l,..., m), and it follows that fflb jq is of Baire class 1(L  N,*R). Since L  N €G_{g}(L), we can extend f. _{T} to a function g. : L > R of Baire class 1(L, R) . If we set g(x) = <^g^(x),..., g_{m}(x)} , then g is of Baire class 1(L, R^{m}) and g(x) = fCx) for x € L  N, so we have the desired result.
2 2 3
Now consider the case where Y = S . Since SCR, there 3 exists, as we have just shown, a function g : L ■* R of Baire class 3 2
1(L, R ) such that g(x) .= f(x) for all x € L  N. Then g(L)  S is countable, so there exists some point q in the bounded open domain
determined by_S such that q(£ g(L) . Let h be the
2
P o g. Then h maps L into S , and for each x £ L 4
h(x)
P.(gM) = P (f(x)) 4 4
composite function “ N, f(x).
2
If (JC S is open, then
h^{1}(U)
^{1}(P "^U)) € F CL) ,
so h is of class (F (L)) . By Theorem 1, h is of Baire class
2 M
1(L, S ), so we have the desired result.®
CHAPTER I. BOUNDARY FUNCTIONS FOR CONTINUOUS FUNCTIONS
If r is a positive number and.if y_{Q} is a point of a metric space Y having metric p, then
S(r, Y_{o}) denotes ' {y £ Y : p(y, yj < ri.
We will repeatedly make use of Theorem 11.8 on page 119 in [11] without making explicit reference to it. This theorem states that if D is a Jordan domain in R or in R U {“}, if y is the frontier of D, and if a is a crosscut in D whose endpoints divide y into arcs y^ and y£, then Da has two components, and the frontiers of these components are respectively a u y_{1} and a u (The term crosscut is defined on page 118 in [11].)
4. Domain of the Boundary Function
Definition. If f is a function mapping into a metric space Y, then ‘ the set of curvilinear convergence of f is defined to be
{x € X : there exists an arc y at x and there exists y € Y such that lim f(z) = y}. z >x zg y
J. E. McMillan [10] proved that for suitable spaces Y, the set of curvilinear convergence of a continuous function is always of type Ftffi . ^{We} ®i^{ve a more} direct proof of this result than McMillan’s.
(This proof can be modified to give a more general result; see [9].)
. 15
i i t I I 16 ।
* An interval of X will be called riondegenerate if and only if !
it contains more than one point. ।
Suppose y is a crosscut of H. If V is the bounded component
of H  y, let L(y) = V fX. Then L(y) = [c, d], where c and d are the
•  J endpoints of y and c < d. Suppose Q is a domain contained in H. Let r denote the family of all crosscuts y of H for which y H C fl, and let
1(8) = U L(y)*. Y^r
Let acc(fl) denote the set of all points on X that are accessible by arcs in fl.
Lemma 7. Assum^ that acc (fl) is nonempty. Let a be the infimum of acc (fl) and let b be the supremum of acc (fl). Then
I (fl) = (a, b) .
Proof. Suppose x€.I(fl). Let y be a crosscut of H such that 'it *
Y fs H £ fl and x € L(y) • L(y) = [c, d], where c and d are the endpoints of y and c < d. It is evident that c and d are in acc(fl), so a £ c < x < d £ b, and x € (a, b). Conversely, suppose x* (a., b) • Then there exist points c', d’ € acc(fl) with c’ < x’ < d’. Since fl is arcwise connected, it is easy to show that there exists a crosscut y’ of H, with y'^HSa, having c', d' as its endpoints. But then x’ € (c’, d’) = L(y’) , so x' € I(fl).■
Lemma 8. If fl., and fl_ are domains contained in H, and if X M
(1) I (fl) A acc(flj) and I(fl_{2}) H acc(fl_{2})
are.not disjoint, then and Q_{2} are.not disjoint.
Proof. We assume that and fi_{2} ^{are} disjoint and derive a contradic
tion. Let a be a point in both of the two sets (1). Let y^ be a ★ crosscut of H, with A such that a g L(y^) (i = 1, 2). Let
IL and AL be the components of H  y^, where is the bounded component. Observe that y^ A H and y_{2} /'Mi are disjoint.
Suppose yj H H C v_{2} and y_{2} AH G V^. Then, since y^AH $Up has a point in common with V_{2}« But, since U.^ is unbounded, U.^ cannot be contained in V_{2}, so must have a point in common with y_{2} nH. This contradicts the assumption that y_{2} A H £Vp so we conclude that either y^ A H^V_{2} or y_{2} AH^Vj. Hence, either y^ A H U_{2} or y_{2} A H By symmetry, we may assume that
y_{2} HH$U_{r}
fi_{2} does not meet y^, and fi_{2} does meet U^ (because y_{2} A H S Ui A ft?) > so fi_{2} C U^. Since a € acc(fi_{2}), there exists a point b € L(yp such that b € acc(ft_{2}). But then b € fi_{2}£Up and this is * impossible because the frontier of U^ is disjoint from L(y^) . ■ Theorem 3 (J. E. McMillan). Let Y be a complete separable metric space and let f : H > Y be a continuous function. Then the set of curvilinear convergence of f is of type F* .
Proof. Let {p_{v}}_{v}”. be a countable dense subset'of Y. Let {Q(n, m)} " K m=l
be a counting of all sets of the form
' { <x, y>: 0 < y < and r < t < r + i}
where r is a rational number. Let’{U(n, m, k, £)} “ be a counting 1
(with repetitions allowed) of the components of
i •
p_{k}))n Q(n, m) .
(We consider  to be a component of <j>.) Let
A(n, m, k, &) = acc[U(n, m, k, £)] .
Set
CO CO CO • CO
»• n u u u I(U(n, m, k, $,)) n A(n, m, k, £).
n=l m=l k=l £=1
Since I(U(n, m, k, &)) is open in X it is of type F . It follows that
O’
B is of type F . Let C denote the set of curvilinear convergence o 6
of f. I claim that B $C. Take any b <B. For each n, choose m[n],
k[n], £[n] with I
(2) b € I(U(n, m[n], k[n], a[n])) A A(n, m[n], k[n], £[n])
(n = 1, 2, 3,...).
For convenience, set U_{n} = U(n, m[n], k[n], £[n]). By (2) and Lemma 8,
U and U . have some point z in common. For each n, we can choose n n+1 n
an arc y_{T} i  ^{U}n+1 with one endpoint at z_{n} and the other at ^{z}_{n+}^« Then y_{n}CQ(n+l, m[n+l]). Also,
b € A(n+1, m[n+l], k[n+l], £[n+l]) S=U_{n+1} Q(n+1, m[n+l]),
2 and therefore each point of y_{n} has distance less than —from b.
2 l"l
—>0 as n >«; hence, if we set y =' {b} U y_{n>} then y'is an arc n=l
with one endpoint at b.
Since U and U , have a point in common, n n+1 ^{r} ’
11 11
^{£ (St}J? Pk[n]» ^{and £} (^{S}<JST Pk[n_{+}1]»
have a common point, and hence ,
^{SC}3P ^{p}k[np ^{and SC}3irT’ ^{p}k[n+i:P z z
have a common point. Therefore, if p is the metric on Y, then
_{t} ^{p}C^{p}k[n]’ ^{P}k[n+lp  _{9}n ^{+} _{9}n+r ^{<} _{9}nl *
and therefore
f _ y 1 1
^{pCp}k[n]’ ^{P}k[n+rp  ^{p(p}k[n+il]’ ^{P}k[n+ip ^{<} _{2}n+i2 ^{x} ’
Thus {p_{k}[_{n}p i^{s a} Cauchy sequence and must' converge to some point p € Y. Since
v_{n}SVlS^{f}'^{1}(^{s}(^r> Pk{ntl]” ”^{d}
^{p}k[n] ;^{p}
lim f(z) = p. It is possible that y is not a simple arc, but Z*b
z€Y according to [12] we can replace y by a simple arc
y’ C y. Thus b € C, and.we have shown that B^C.
Suppose c € C. Let y_{Q} be an arc at c such that f approaches a limit p’ along y_{Q}. Take any n. Choosek with p’ € P_{k}) • Choose m so that c is in the interior of Q[h, m) A X. Then y_{Q} has a subarc y_{Q}’, with one endpoint at c, such that
Y_{o}'  {c} S Q(_{n}, m) n f"^{1}CS(~, p_{k}B .
Hence, for some £, c £ acc[U(n, m, k, £)] = A(n, m, k, s.) . This
shows that
CO CO 00 co
csQUUUA(n, m, k, £) n=l* m£l k=l «=1
It is easy to deduce from Lemma 7 that the set
A(n, m, k, £)  I(U(n, m, k, £)) = . . . . . . . . . . .
A(n, m, k, £)  [I(U(n, m, k, £))nA(n_{5} m, k, £)]
contains at most two points. It follows by a routine argument that
A(n, m, k, £)  kJ [I(U(n, m, k, £)) n A(n, m, k, £)] n m,k,£ n m_{}}k,£
is countable. Since
[I(U(n, m, k, A)) A A(n,' in," k, £')] = B^c n m,k,£
n m,k,£
C  B is countable, and therefore C is of type F _{g}. B
Next we will show that the foregoing theorem is as strong as possible, in this sense: if A is any set of type F $ contained in X, then there exists a bounded continuous complexvalued function f defined in H such that A is the set of curvilinear convergence of f. The proof is unfortunately quite long. .
Definition. Let E^ and E_{2} be two sets on the real line. A point p on the real line will be called a splitting point for E^ and E? if either
Xj £ p for all x_{x} £ E^ and p £ x? for all x_{2} G E_{2} or x_{2} £ p for all x_{2} € E_{2} p £ x^ f°^{r} a*^{1 x}i € ^{E}^.
We will say that two sets E^ and E_{2} split, or that E^ splits with E_{2}, if and only if there exists a splitting point for E^ and E_{2}.
• co
Lemma 9. Let E be an set in R. Then there is a sequence {E_{n}}_{n}_^
of sets such that
(i) E is bounded and closed n
(ii) if n m, then either E_{n} and E^ are disjoint or E_{n} and
E_{m} split
(ifi)
Proof. We can
n, and .
CO write E = I J A where n=l
A is closed, A CA. for all n n — n+1
Observe that if I is any open interval, then there exists a
CO
countable family {J } .of bounded closed intervals such that ^{n} n=l 00
n m J and J split, and I = J^. Since any open set of
n=l
real numbers is a countable disjoint union of open intervals, it
00
follows that for any open U there exists a countable family H J
of bounded closed intervals
00
such that n i m I and ^{T} n
I split, and m ^{r} ’
n=l
' r Ih ^{00}
For each n, let {ipj_^ vals such that j k ZZ^ and
be a family of bounded 00
I? split, and A = IJ j=l
closed inter
im. Let J
IF = {Ap U {l“ n A_{n+1} : n = 1, 2,...; j = 1, 2,...}.
Then^^{7} is a countable family of bounded closed sets, and CO 00
^{E} ■  ^{A}1^{W}U (A_{n+1}AA_{A})
n=l n=l
= i”) = A_{1U}C{
n=l ]=1 ^{J J J}
■IK
/ J*
Let and F_{2} be any two distinct members of . If either F^ or F_{2} is A^ = $, then F^ and F_{2} are automatically disjoint. If neither F^ nor F_{2} is A^, then we can write
. „ . _{T}n(l) . , „ _{T}n(2). .
^{F}1 " ^{X}j(l) ^{n A}n(.l)+l.^{and F}2 ^{= I}j(2) ^{n A}n(2)+1‘
If n(l) < n(2), then n(l) + 1 ^n[2),_{(}so
p_{2} = l"^2) ^{n A}n(2)+1 — ^{A}n(2) — ^{A}n(l)+1’ ^{and} therefore Fj and F_{2} are disjoint. A similar argument shows that if n(2) < n(l), then F^ and F_ are disjoint. Thus, if F_{q} arid F_{9} are not disjoint, then n(l) = n(2) and we have
^{F}1 ^{= I}j(l)^{AA}n+l ^{311(1 F}2 ^{=} ^(2)^^+!’
where n = n(l) = n(2). But then j (1)  j (2), so I^{1}?,,. and I^{3},,,.. / ^{J T J} ’ j (1) ] (2)
split, and therefore F^ and F2 split. So we have shown that any two distinct members of either split or are disjoint.
If p has infinitely many distinct members, let E_{q}, E„, E„... be a counting of p . If has only finitely many distinct members, let E_{n},..., E be the members of 'p and let E, = d> for k > m. In either case, IE J is the desired sequence.®
If F is a closed subset of the real line, then by a complementary interval of F we mean a component of F'. (If F = R, then $ is considered to be a complementary interval of F.)
Definition. By a special family we mean a family of subsets of R such that
(3) is nonempty
(4) each member of ^p is bounded and closed
(5) there exists a sequence’ {F_{n}J_{n}”j of members of p such that every member of p is equal to some F , and the following condition is satisfied.
(5a) If m > n, then either F_{m} is contained in one of the complementary
intervals of F , or else F splits with F . n’ m ^{r} n
Lemma 10. If E is an F^ set in R, then there exists a special family jP such that E = LJ .
• 00
Proof. By Lemma 9 we can choose a sequence tE_{n}}_{n=}i of bounded closed sets such that if n I m then E and E either split or are disjoint, ^{1} n m ^{47}
00
and E = U E n=l
Let n. = 1 and let F. = E_{n}. Now suppose that n_{q}, n_{9},..., n  t 11 1 s
are chosen and F., F„,..., F are chosen so that 12 n s
(i) 1 = n_{T} < n_{2} < ... < n_{s}
(ii) F. is closed and bounded (i = I,..., n ) 1 o
(iii) if n_{s} >_ r > t >_ 1, then either F^ is contained in one of the complementary intervals of F , or else F^ splits with F^
(iv) if 1 £i £^{n}_{s}, then there exists j s} such
We construct F F as follows. Let be the family of
n_{s}+l ^{n}s+l
complementary intervals of the bounded closed set
We assert that E j meets at most finitely many members of xd^. If this • co
assertion is false, then there exists an infinite sequence {I_{n})_{n}_^ of members of such that n m implies I_{n}<^ I = <>, and there exists ' • '■ co
(for each m) a point x € I HE {x} . is a bounded sequence, ^{47} m m s+1 m m=i
and n i m implies that x„ x . From this it follows that {x_{m}} ^{1 r} n ^{1} m m m=i
has either a strictly increasing or a strictly decreasing convergent
CO subsequence. We will assume that is a strictly increasing
convergent subsequence; the reasoning is similar
in the case of a
strictly decreasing convergent subsequence. Say 1*^ ® b^) .
^{X}m(k)^ we let
< X H X. m(k)
^{I}mCk+l)
< b, , so since x _{v} < x , k’ mfkj m(k+l)
^{b}k+l
we must have x ,, < < a. . < x . m(k) — k+1 m(k+l)
and
Therefore, if
lim
k*» ^{X}m(k)
then x
lim kx»
Moreover, for k >_ 2
finite real
number
so
that a. € \ J E•. Therefore there exists u €{1,..., s}
^{K} £T1 ^{1}
such that a, € E for infinitely many values of k. Consequently KU.
x € E . But since x ,, _{A} £ E u m (.kJ
so that E and E
u
E and E .. u s+1 that are less
_{n} must split s+1 ^{r}
Since infinitely
than x;
and E
, x € E . also. But then x €. E nE ,, S + 1 U S + 1
and x must be a splitting point for
many
also
a^ lie in E^, E_{u} contains points contains points less than x;
therefore E and E .
cannot split
and we have a contradiction. This
u s+1
proves the assertion.
3 = {(J}u{InE . : 16^ and I n E . ± <>}. S+1 S+1
Let n_{s+}i equal n_{g} plus the number of members of ,8. Let F_{n +}^,..., F be all the members Of We must show that conditions fi) ^{n}s+l
through (v) are still satisfied when s is replaced by s+1. Conditions (i)> (ii)_{f} and (iv) are obvious. The verification of (iii) is divided into three parts. Suppose n_{g+}^ >. r > t >_ 1.
Case I. Assume that n$ > r > t > 1. In this case we already know that either F_{r} is contained in one of the complementary intervals of F_{t} or else F_{r} splits with F_{t}.
Case II.
Assume that n
. > r > n s+1 —
There exists v & {1
,s}
such.that F_{t} E_{v}» Either E_{v} and E_{g+}^ are disjoint or they split.
Case Ila. Assume E^. and E_{s+}j are disjoint. Either F*. = <j>(in which case F ' is certainly contained in a complementary interval of F ) or else F 4 I and F = InE for some I€«j?. Let J be the smallest r ^{1} r s+1
— * i^{S} । ’
closed interval containing F . Then J £ I and J £ I C (E.) , so
T • i 1
* ^{1=}1
that J does not meet E . The endpoints of J lie in F <= E ., so v ^{r} r s+1’
neither endpoint of J lies in E_{v>} So J does not meet E_{v} and therefore
J does not meet F ; from which it follows that F^ is contained in a complementary interval of F^.
Case lib. Assume that E and E . split. Since F. E and F £ E . v s+1 ^{r} t ~ v r s+1
it follows that F. and F split, t r
Case III. Assume that n , > r > t > n . If either F or F* is d>, s+1 — s r t ^{T}’
it is clear that F_{r} is contained in a complementary interval of F^.
Otherwise, there exist I,, I^ such that Ij A I2 = <l>and
^{F}r ^{= Z}1 ^{A E}s+1 ^{F}t ^{=} *2^{nE}s+r
Since 1^ and evidently split, F^ and F^ must split.
Thus condition (iii) is verified. /
As for (v), it is clear that , ^{s} V^{S+}1
E U E.C2 F.CE _{q}, s+1 1 i=n+l i s+1’
1=1 ^{J} s '
so that
Hence
Thus we
• ' co
have shown that we can construct sequences {n_.},._p
co k=l
in such
a way that conditions (i) through (v) are satisfied
for every value of s. If we set^= {F. : k = 1, 2, ...}, it is easy
to verify that is a special
family and that E = ■
Definition.
IfJ^ and^{are two} families.of sets, let
^L^{a}^2 ^{= {F}i^{n}
^{F}2 ^{: F}l^ ^1 ^{F}2 ^2^'
Lemma 11. Ifandare two special families, thenJA^is a special family.
Proof. Conditions (3) and (4) in the definition of a special family are clearly satisfied, so we just have to verify (5).
Arrange all pairs of positive integers in a sequence according to the scheme shown in Figure 1. Let (a(k), b(k)) be the kth term of the sequence'’’ (k = 1, 2, ...). Observe that k < I if and
^The reader may find it amusing'to derive the following formulas for (a(k), b(k)). For real t, let t denote the largest integer that is strictly less than t. Then
a(k) = ^{+} _{+} ^{X}]])  k + 1
= (/8k71 +   (1)^{C}(/8k+l +   (1) t['^{/}8kU] ]) __{k}
+1
Hctt/SkTTl] + 3) (MST + 1)  k + 1 if /81E+I is odd
I (’<sup>/</sup>8ic+T + 2)’<sup>/</sup>8lc+T  k + 1 if M+T. is'even, and
(1,2)
(1,3)
a $/■ ■
(2,1)
(2,2)
(2,31
(2,4) ■ • •
Q,l)
(3,3)
(3,4) • • •
L4.1)
(+.2)
(4,3)
(4,4) • • •
Figure 1.
only if either a(k) + b(k) < a(£) + b(£) or else a(k) + b(k) = a(£) + b(£) and b(k) < b(£). Thus k < £ implies that either a(k) < a(£) or b(k) < b(£).
Let b^{e a} sequence of elements of f such that every
member of^is equal to some F_{n} and such that condition (5a) in the
• "*■ 00
definition of a special family is satisfied. Let {F_{n})_{n}_^ be a similar sequence for Set
^{F}k ^{= F}a(k) ^{n F}b(k) '
Then fF.K, is a sequence in such that every member of J? K K= JL
is equal to some F, . We must show that condition (5a) is satisfied.
Suppose that £ > k. Two cases occur.
Case I. a(k) < a(£).
Note that F^^ ^{F}_{a}(k) ^{and F}£ — ^{F}a(£) ’ ^{Either F}_{a}(£) ^{is contained in} one of the complementary intervals of F . (in which case F. is contained in a complementary interval of F^), or else F^^ and ^{F}_{a}(fc) split (in which case F. and F. split).
Case II. b(k) < b(£).
In this case a similar argument shows that either F_{o} is contained in
b(k) = (2!®pJi _ /SF* *. ^{2}) _{+} k
= 4(»<tF] + I  i(l) <sup>[[,/</sup>^<sup>1]</sup>) ([[y^T    (1)U’^{/§E}*^{T}^) +k
 2) + k
if /8k+l is odd
if [[Vsk+l)] is even.
I
a complementary interval of F^ or F$ and F^ split. Thus condition (5a) is satisfied, andJ^A^is a special family. ■
Lemma 12. Let E,, E_{9} be two F , sets in R such that E. C E_{9}, and suppose that and are special families such that E^ = ^{aIi}d ^2 = Then E^ = •
The proof is obvious.
Next we introduce some notation.
Let J be a nonempty interval on X with endpoints a, b (a £ b).
IT
By Trap (J, e, 0)(where 0 € (0, y) and e > 0) we mean the interior z
of the trapezoid shown in Figure 2. That is,
Trap(J, e, 0) = { (x, y> :0<y<e, a + y ctn 0 < x < b  y ctn 0}.
7T
For 0 g (0, ^), let Tri (J, 0) be the closed triangular area shown in Figure 3. That is,
Tri (J, 0) = { <x, y>: y >_ 0 and a + y ctn 0 £ x £ b  y ctn 0}.
7T
If x € X, e > 0, and 0 € (0, y), let S(x , e, 0) denote the open 0 Z O
Stolz angle shown in Figure 4. That is,
S(x_{o}, e, 0) = { <x, y>: 0 < y < e, X_{q} + y ctn (it  0) < x < x_{q} + y ctn 0).
If K is a closed set on a real line, let J(K) be the smallest closed interval containing K. If K is bounded, closed, and nonempty, e > 0, and 0 < B < a < p then we define
B(K, e, a, B) = Trap (J(K), e, a)  U Tri (I, B),
where denotes the set of complementary intervals of K.
Figure 2. — Trap(J,E ,0)
Figure 4.—S(x_{0}>£ ,9)
I > We state without proof the following readily verifiable facts.
(6) B(K, e, a, 3) is an open subset of H.
(7) S(e, e, 0) is an open subset of H.
(8) If Kjl and K_{2} split, then for any e_{2}, a, 3, B(Kp Ej, a, 3) and B(K_{2}, e_{2}, a, 3)
are disjoint. /
(9) Suppose that £ K, e > e^ > 0, and 0 < 3 < 3^ < < a < y.
Then
B(Kp E_{p} (Xp 3p n H C b(K, e, a, 3) .
(10) Suppose Kj is contained in one of the complementary intervals of K, and suppose e, a, 3 are given. Then there exists 6 > 0 such that for every n £ 5,
B(K, e, a, 3) and B(Kp n, a, 8) are disjoint.
IT L *
(11) Suppose that a < 0 < and x_{q} J(K) . Then, for any e, Ep
B(K, e, a, 3) and S(x_{q}, E_{p} 0) are disjoint. ★ 7f
(12, Suppose that x € K n J(K) and 3 < a < 0 < Let e be given.
Then there exists 6 > 0 such that for every h £ 6,
S(x_{Q}, n/ 6) n H S B(K, e, a, 3) .
(13) Suppose that e < e’ and 0’ < 0. Then
stx“TT^TAH £S(x_{q}, s’, 0’)
(14) Suppose x_{q}$K and e, a, 3, 9 are given.. Then there exists 6 > 0 such that for. every h £ 6,
S(x_{q}, n, 0) and B(K, e, a, B) are disjoint.
(15) If x_{q}  x^ and e, 0 are given, then there exists 6 > 0 such that for every n £ 6,
S(x_{q}, e, 6) and S(Xp n, 0) are disjoint.
(16) B(K, e, a, B) n X C K.
(17) S(x_{o}, e, 6) nx = {x_{o}}.
2
Definition. If^ris a special family, let T be the set of all members of 7 that have two or more points.
Definition. LetJ^be a special family, let E be the set of all endpoints of intervals J(F) where F € F , and suppose that O<B<a<0<H. By a pair of special a, 3, 0 functions for we mean a pair (e, 6), where e and 6 are positive.realvalued functions, 2 the domain of e is E, the domain of 6 is , and
(18) for each n > 0, there exist at most finitely many F € such that 6(F) >_ p;
(19) for each n > 0, there exist at most finitely many e e E such that e(e) £ n;
(20) if e, e* € E and e e’, then
S(e, c(e), 0) and S(e', E(e’), 0)
are disjoint;
(21) if F, K ef^{2} and F 4 K, then
B(F, 6(F), a, B) and B(K, 6(K), a, B) are disjoint;
(22) if e € E and F e^^{2}, then
S(e, e(e), 0) and B(F, 6(F), a, B) are disjoint.
Lemma 13. Letbe a special family and suppose that Oc'B<a<0<y. Then there exists a pair of special a, B, 6 functions for
Proof. Let {F_{n}}_{T}7j be a sequence of members of 5^{s}”of the type referred to in condition (5) in the definition of a special family. Let (J^(n) = {F € J®^{2} : F = F^ for some k <_ n}
E = set of all endpoints of intervals J(F) for FGf, F 4 I
E(n) = {e € E : e is an endpoint of J(F^) for some k £n for which F^ 4
If J(Fj) has one endpoint e, set e(e) = 1. If J(F^) has two endpoints e^, 62, then by (15) we can choose e(e^) £ 1 andc(e_{2}) £ 1 so that S(e_{1}} e^), 0) and S(e_{2}, ^{e}C^{e}p> ^{9}) are disjoint. If F_{1}6'f^{;2}, set 6(F_{1}) = 1. In this case, J(F_{n}) hats two endpoints e. and e_{9} and (by (11)) B(F_{1}, 6(F_{1}), a, B), S(e_{p} e^), 0). and S(e_{2}, cfep, 0) are 1 all disjoint.
Now suppose that e(e) and 6(F) have been defined for all
2
e € E(n ) and all F Gy (h) in . such a way that
(i) if e, e' € E(n) and e  e^{1}, then S(e, e(e), 6) and /
S(e’,e(e*), 6) are disjoint;
(ii) if F, K€dP^{2}(n) and F  K, then B(F, 6(F), a, B) and B(K, 6(K), a, B) are disjoint;
2
(iii) if e €E(n) and F€(p (^{n}) > then S(e, e(e), 0) and
B(F, 6(F), a, B) are disjoint;
(iv) if e € E(n) and k <nis the least integer for which
e G E(k), then e(e) £
2
(v) if F £ jF (n) and k < n is the least integer for which F€f*(k), then 6(F) <p
We must extend the definitions of e and 6 to E(n+1) and
2
tF (n+l) in such a way that conditions (i) through (v) are still satisfied when n is replaced by n+l.
2
Case I. If F . = * or if F . = F. for some k < n, then J^{1} (n+l) = n+l n+l k — ^{v}
^(n) and E(n+1) = E(n), so that nothing is required to be done.
Case II. If F , consists of a single point e and if e 6 F, for some n+l k
k < n, then (since F . and F. must split in this case) e is an n+i k
endpoint of J(F.), so that again 7^{s2} (n+l) = ^p^{2}(n) and E(n+1) = E(n), K
and nothing is required to be done.
Case III. Suppose thatF j consists of a single point e_{Q} and that
for each k < n, e F. . By (14), (15), and the fact that E(n) and
(n) are finite, we can choose e(e_{o}) €£ (0, so that S(e_{Q}, e(e_{Q}), 0) is disjoint from S(e, e(e), e) and from B(F, 6(F), a, B) for each e €E(n) and each F^^pfn). The construction is then finished for
E(n+1) and F(n+1).
I
Case IV. Suppose that F_{n+}j contains at least two points and that, for each k < n. F, i F ,. For each k < n, either F . splits with F. , or else F_{n+1} is contained in a complementary interval of F^. Since ^(n) is finite, (8) and (10) show that we can choose ^{5}(F_{n+1}) £(0, ^y) so that B(^{F}_{n+}i, ^{5}(^{F}_{n+}^), a, 6) is disjoint from B(F, 6(F), a, B) for each F€j*(n).
Say e € E(n). Then e is an endpoint of J(F_{k}) for some k <_ n, so (since F_{n+}^ either splits with F^ or is contained in a complementary interval of F_{R}) e ^J(F_{n+1})*. By (11), B(F_{n+1}, 5(F_{n+1}), a, 8) and S(e, e(e), 0) are disjoint.
' Let e , e ^{1} be the endpoints of J(F _{n}).
o o ^{r} n+1
Case IVa. e_{Q}, e • £E(n). ,
In this case the construction is already finished.
Case IVb. e_{Q}€E(n) and e_{Q}* $E(n).
If e ’ £ F. for some k < n, then F , splits with F. , so that o k — ’ n+1 ^{r} k’
e_{Q}’ must be and endpoint of J(F^) which contradicts the assumption that e ’ 4 E(n). Hence, for each k < n, e ’ £ F. . By (14), (15), and the fact that E(n) and J^(n) are finite, we can choose
1 — 
e(©_{0}’) C (0, ^y) so that S(e_{o}’, e(e_{o}')» 0) is disjoint from
S(e, e(e), 0) and from B(F, 6(F), a, B) for each e6 E(n) and each F€ f^{2}(n). By (11), S(e_{Q}’, e(e_{o}’)» 0) and B(F_{n+1}, 6(F_{ft+1}), a, B) are disjoint. Thus the construction is finished for E(n+1) and(n+1).
/
Case IVc. e_{Q} E(n) and e * € E(n).
This case is essentially the same as Case IVb.
Case IVd. e_{Q} E(n) and e_{Q}'
<E(n).
k < n. then F ysnlits with F, . so
If e 6F, for some o K
e_{Q} is an endpoint of J(F_{k}); a contradiction. Thus e^ F_{k} for k £ n, and similarly e ’ F. for k <_ n. Therefore, by (14) and (15), we can choose e(e ) and e(e ’) E (0, —rr) so that S(e , e(e ), 9) and o o n+1 o o
S(e_{o}’, E(e_{o}’), 6) are disjoint and each of S(e_{Q}, e(e_{Q}), 9) and
S(e_{o}’j ^{e}(e_{o}’)> ^{0}) ^{is} disjoint from every S(e, e(e), 9) (e €E(n)) and ~ from every B(F, 6(F), a, 8) (FG'J^n)). By (11), S(e_{Q}, e(e_{Q}), 9) and
S(e_{o}’, e(e_{o}’)> ^{0}) ^{are} each disjoint from B(F_{n+1}, ^{a}» > ^{so t}^e
construction is finished for E(n+1) and(r (n+1).
I
We have shown that we can inductively define e(e) for every e € E and 6 (F) for every F € f^{2} in such a way that (i) through (v) are satisfied for every value of n. Conditions (20), (21) and (22) in the definition of a pair of a, B, 9 special functions are thus automatically satisfied by (e, 6). We must verify that (18) and (19) are also satisfied.
Suppose (19) is false. Then there exists n > 0 and there
00 exists an infinite sequence distinct members of E such that
E(^{e}k) j£ n for every k. Let m(k) be the least integer for which e^ is an endpoint of J(F ,...). Each J(F ) has at most two endpoints, so, m j in ^{1}
since the e^ are all distinct, there exists (for given m) at most two values of k for which m(k) = m. Therefore there exist infinitely
many distinct integers among m(l), m(2), m(3), .... Consequently
there exists j with ^{J} m(j)
< n. But, by (iv), e(e..) £ ^
contradiction. So (19) must be true. A similar argument shows that
(18) is true. ■
_IT
Lemma 14. Let T be a special family, 0 < g <. a < 0 < y, and. let E be the set of all endpoints of intervals J(F) for F €J*. Suppose (e, 6) is a pair of special a, g, 0 functions for If e^, are two real 2
valued functions having domains E and^f respectively, and.if
0 < E(e) £ e(e) for all e £E, and
0 < £ 6(F) for all F^J^{2},
then (Sp 6^) is a pair of special a, 0, 0 functions for^.
Proof. This follows from the fact that
S(X_{o}, E’, 0) <^S(X_{o}, E", 0) I
and B(K, e', a, g) OB(K, e", a, g)
^{1} whenever e' £ e”. 0
Theorem 4. Let A be any set of type F_{ag} in X. Then there exists a . bounded continuous complexvalued function f defined in H such that A
is the set of curvilinear convergence of f.
co
Proof. We can write A = f  A , where each A is of type F and ' J n’ n o
n=l
A_{p+1}^A_{n} for every n. For each n, let be a special family with CXv ^{ut}
“ ^n^{A}^n+l ^{for n} ’
By Lemmas 11 and 12, together with mathematical induction, K is a special family and ^{= A}_{n} Moreover, every member of^_{+}^ is a subset of some member of^.
00
Let be a strictly ascending sequence in (0, $)
. IT
converging to g.
Let {<x be a strictly descending sequence in ($, ■$) • ^{11} converging. to ©o 7T ' ' 31T
Let be a strictly ascending sequence in Q, •—)
converging to g—.
Let E be the set of all endpoints of intervals J(F) for
Let (e(l,0, 5(1,)) be any pair of special o^, 8^, 6_{1} functions for .
Nov; suppose that for each k<nwe have chosen a pair of
special a, , 3,, 6, functions (e(k,), 6(k,)) for ft. in such a way K K K i*
that
' (i) whenever 1 £ k £ n  1, eg ^{E}k_{+}p F g
e € F n J(F) , then
S(e, e(k+l, e), 6_{k+1}) AHClW 6(k, F), <x_{k}, 3_{k});
(ii) whenever 1 £ k £ n * 1, eg
., and e g E. , then k+1* k’
SCe, e(k+l, e), 6_{k+1}) nHCs(e, e(k, e), 6_{k});
(iii) whenever 1 £ k £n  1, KgJ^_{+1}, Fe^^{2}, ^{K} — ^{F}» then
B(K, 6(k+l, K), a_{k+1}, B_{k+1}) AHCB(F, 6(k, F), a_{R}, 3_{k}).
Then we construct (e(n+l,), 6(n+l,‘)) as follows. Let
(e, 6) be any pair of special a_{n+}p ^_{n}+l’ ^{9}n+l ^{:Funct}l°^{ns} ^°^^n+i’ ^2 *
e g E_{n+}i  E_{n}, then for some unique F€ , e £ F A J(F} , so by (12) we can choose 5(e) >0 such that n £ 5(e) implies
S(e, n, 0_{n+1}) nHGB(F, 6(n, F), a_{n}, &_{n}).
We set e(n+l, e) = min {c(e), 5(e)). On the other hand, if e g E_{n+}^ r\ E^, then we set e(n+l, e) = min {e(e), e(n, e)}.
If then there exists a unique K with F £ K.
Set
6(n+l, F) = min {6(F) , ~ 6(n, K)}.
By Lemma 14, (e(n+l,*)> <5(n+l,’)) is a pair of special a p
B ., 6 , functions for^.,, and by (13) and (9), conditions (i), /
n+1 n+i n+i (ii), and (iii) are still satisfied when n is replaced by n+1. Thus we can inductively construct a pair (e(n,)> 6(n,)) of special a_{n}, $_{n}, 0_{n} functions for in such a way that conditions (i), (ii) and
(111) are satisfied for every n.
Let
V_J ^{S}O> ^{e}(^{n}» ^{e}) > ^{0}J  V e € E„ J
VL ^{B}C^{F}> ^{6}<^{n}> ^{F})’“ € t n n
Then U_{n} is open. For fixed n, all the various sets S(e, e(n, e), 0^) (e € E ) and B(F, 6(n, F), a_{n}, &_{n}) (F € 5^) are open and pairwise disjoint, so that every component of U_{n} is contained in one of the sets ^,2
S(e, e(n, e), 0_{n}) (e € E_{n}) or B(F, 6(n, F), «_{n}, &_{n}) (Ffify. It therefore follows from (16) and (17) that if Q is any component of U , then
(23) « AX A_{n}.
From the fact that (e(n,'), 6(n,*)) is a pair of special a_{n}, B_{n}, ©_{n} functions for 5^ together with conditions (18) and (19), it follows that
^{u}_{n}^{nH} = tU S(e,.e(n, e) , 6 ) HH] u e 6 E n ...........
[ U_{2} B(F, 6(n, F), a_{n}, y n H].
Consequently, conditions Ci)> (ii), (iii), together with the fact that e € E ,  E e 6 F n J(F) for some ^{1}
n+1 n ^{1 J} n’
show that U , H U for every n. n+1 n
By Urysohn’s Lemma, there exists a continuous function
g_{n} : H ■+ [0, 1] such that
g_{n}(z) = 1 for z € H  U_{n} and g (z) = 0 for z €. TT' _ n H.
■ n n+1
co
Let g(z) = S — s (z). Then 0 < g(z) <_ 1, and the series converges n=l 2^{n}
uniformly, so g is continuous on H.
If
z € H  U_{n}, then z £ H  for every m >_ n
so
that
and hence
(24)
CO
. g(z) 1 ^{E} m=n
1 _{=} 1
2^{m} 2^{n1}
(z € H  U_{n}) .
Also
if z € U ,, n+1’
then z € U,, U», ,.., U ,, so that 12 n+1
= g_{n}(z) , and
(25)
. i ^{z} ns m=n+l 2^{,u}
1
2^{n}
^{€ U}n+1
We assert that
3ir
(26) for each x_{q} € A, g(z) + 0 as z + x_{q} with z € S(x_{q}, li g—) •
Take any natural number n. Since x C A , = i> either ^{7} o n+1 ^{v}[†]^{z}*’n+l’ case, set n = e(n+l, x_{q}). In the.second case, (12) shows that we can choose n > 0 so that
S(x_{q}, ni J^{1}) (F, 6(n+l, F), a_{n+1}, .
Suppose <x, S(x_{q}, 1, g) and y < n. Then, in the first case, <X, y>€ S(x_{o}, n, ■§) £S(x_{q}, e(n+l, x_{q}) , 6_{n+]}) ^U_{n+1}, and in the second case,
<x, y>£ S(x_{o}, n, £b(F, 6(n+l, F), B_{nU}) £ U^. So, by (25), ( <x, y>6 S(x_{o}, 1, jl) and y < n) =><x, y>£ U_{n+1}
0 5 g(x, y) 2^{11}
This proves (26). ^{1}
Let x_{q} be a point in X and y any arc at x_{q}. Suppose g(z) ■> 0
as z ■* x^ along y. Then y has a subarc y’ with one endpoint at x_{q}
such that y*  {x_{q}} Cg"‘^{l}(( Therefore, by (23), x_{q} € A_{fl}.
1_ £.
2^{n}’ 2^{n} Since
0). By (24), y’  {x^.
n was arbitrary, x_{q}
^{1} i n n=l
= A.
Thus,
(27) if there exists an arc y at x_{q} such that g(z) •* 0 as z approaches x_{q} along y, then x$ € A.
Now define
f(x, y) = . gCx, y) sin + i g(x, y) ( <x, y>€ H).
3tt
If x_{q} e A, then, by (26), f(z) •* 0 as z > x_{q} with z GS(x_{q}, 1, •$—).
Thus every point of A is in the set of curvilinear convergence of f.
Conversely, suppose x_{q} is any point of the set of curvilinear
convergence of f. Let y be an arc at x_{q} such that f approaches the limit c + di along y. Then g approaches the limit d along y. If d is different from zero, then g(x, y) sin — (the real part of f) cannot approach any limit along y — a contradiction. Therefore g approaches the limit 0 along y, and, by (27), x_{q} € A. Therefore A is the set of curvilinear convergence of f.M
5. Boundary Functions for Continuous Functions
Lemma 15. Let E be a metric space, Y a separable metric space. i
Suppose that : E > Y is a function having the following property. For every open set U $=Y there exists an F set L?E and a countable set NO such that
9^{1}(U) CL C <p^{1}(U) u N. ~"
Then there exists a countable set M ^E such that cpL is of class eM
CF_{a} (E  M)).
Proof. Let B be a countable base for Y. For each B€&, let L(B)
be an F^ set and let N(B) Q E be a countable set such that
<^{1}(B) Q L(B) C <p^{1}(B) u N(B)
Let M = VJ N(B). Then M is countable. Let E = E  M and let B€fi °
= <pg . We show that <p_{o} is of class (F (E_{q})) . o .
Let W be any open subset of Y. If p € W, there exists r > 0 such that S(r, p) Q W. Choose B€ ® so that p €BSs( r, p) . Then JBSS(r, p) Cl w. It follows that
N = U B = U
B€dt(W) B € d(W)
where (J.(W) = £B€® : B" ^W}. Therefore
V_{b}^{1}(W) = E A ‘p'M = E A VJ • <p^{1}(B)
^{0 0 0} B€ffl(W)
 E A U L(B) ^{0} Bcacw)
— E U [<p^{1}CB) u N(BJ]
B£ (B(W)
CEO O [CP^{[‡]}® U M]
B€Q(W)
= e n U <p^{1}CB) ° b e <j(w) i i
= E_{o}n<p^{1}CW) = q ^{1} (W). o
Consequently cp^CW) = E_{q} H _{b} L(B), so ^“^{l}(W) is of class W’^{1}
Theorem 5. Let Y be a separable metric space and let f : H •> Y be a continuous function. Suppose that E C X and that cp : E > Y is a boundary function for f. Then there exists a countable set M Q E such thatcpL .. is of class (F (E  M)).
Proof. Let U be any open subset of Y, and let W = (U)’. Set
E_{n} = {x € X : there exists an arc y at x, having one endpoint on X_{n}, such that y  {x} f’\u)}
K = {x € X : there exists an arc y at x such that
Y  {x} C f"^{1}(w)}. '
Ob serve that
^(U) cQ E n=l ^{n} and <jp^{_1}(W) Q K.
For the time being, let n be a fixed natural number. For each x € K we can choose an arc y_{x} at x such that
Y_{x}  {x} £ H_{n} n f^{1}(W) .
Since an arc.at x is by definition a simple arc, y_{x}  {x} is a connected set and hence must be contained within one nonempty component of H n f"\w). Let U denote this component (for each x € K) . n * x
Let T be the set of all points of K that are twosided limit points of E"_{n}. We claim that if x, y € T, then x  y implies
U_{x} n U = 4>. If U_{x} n Uy. <j>, then (since U_{x} and U^. are two components
of the same set) U and U are equal. Let p be the endpoint of y x y x
lying in U and X
let
q be the endpoint of y lying in U^. = U_{x<}
We can
join p and q by an arc y lying in U . X
Putting y , y and y together, x y
we obtain an arc a with one endpoint at x and the other at y, such that a  {x, y}gU'. According to [12] we can choose a simple arc
a’ £ a having one endpoint at x and the other at y. Of course, a’  {x, y) ^{u}_{x}^ A f^{1}(W). Let I be the open interval in X with endpoints at x and y, and let J = X  I. Let B be the bounded component of H  a’ and let A be the other component. Since X_{n} is unbounded and does not meet a’, X £ A. n
Because x is a twosided limit point of E_{n}, we can choose a point w €. I H E_{n}« Let g be an arc at w, having one endpoint on X_{n}, such that g  {w} C f’^CU). Then g does not meet a’ (because a'  {x, y] ^.f \w) and f \w) f \u) = <f>), and therefore (since g.  iw} contains a point of X C A) g  {w} £. A. It follows that
w € A. This, however, is a contradiction
because the frontier of A
(relative to the finite plane) is a’u J.
We conclude that, for x,y € T, countably many components,
x y implies U*. C\ U = <j>.
An open set in the plane has only
so it follows that T must be countable. Let S be the set of all
points of E_{n} that are.not twosided limit.points of E_{n>}
.We know that
S is.countable, so
K A E_{n} = [K n (E  S) ] U [K n S]
= T u [K aS]
is countable. ^{00} ।" ।
Let N = K n \^J E = (K HE ). Then N is countable, and, . n=l n=l
since <p (W) K, . 
00 00
<p^{1}(U) Q EaJ E C E
n=l n=l
00 CO
= (E A K E ) U ((E  K) n E ) n=l n=l
(E n N) U (E  K) C (E nN) U (E  (p"^{1} CW))
= (E A N) U <p“^{1}(U) .
1. 00
Thus <p"l(U) Q E A \^_J E C(EnN) u ^{<}p"''’(U), and the desired result ^{n=}l
follows from Lemma 15. ■
Corollary. Let Y be either the Riemann sphere, the real line, or a finitedimensional Euclidean space. If f : H + Y is a continuous function, if E Q X, and if q>: E + Yis a boundary function for f, then cp is of honorary Baire class 2 (E, Y).
Next we show that the foregoing corollary is as strong as possible in the sense that if E is any subset of X and (p is a function of honorary Baire class 2 mapping E into a suitable space, then there exists a continuous function in H having f as a boundary function. A proof of this result at least for real or vectorvalued functions was outlined by Bagemihl and Piranian [2, Theorem 8], in the case
where E = X. Although the construction given here is carried out much l
more explicitly than the construction given by Bagemihl and Piranian, my treatment differs from theirs in only two aspects that are of any significance. First of all, the proof of the theorem for arbitrary subsets E of X depends on Lemma 6 of the Introduction. Secondly, Bagemihl and Piranian say in the last line of their proof that there is "no difficulty now in extending f continuously to the whole of D in such a manner that <j>is a boundary function for f." While this appears to be all right for real or vectorvalued functions, it is not clear why the extension should be so easy for functions taking values on the Riemann sphere. Theorem 7 of the present paper shows, however, that the result can be obtained for functions taking values on the sphere once it is known for vectorvalued functions.
The following miniature closed graph theorem will be a convenience.
Lemma 16. Suppose that M is a metric space and that u : M > R is a function having the following properties:
(i) if {p_{n}J is a convergent sequence of points of M, then {u(p_{n})} converges neither to + » nor to 
(ii) if {p } Cm, p € M, and y € R, and if p ► p and n n _{n}
u(p_) ■* then u(p) = y.
^{n} n Then u is continuous.
Proof. Suppose that {p^} is a sequence of points in M converging to a point p € M. Using (i) it is easy to show that {u(p_{n})J is a bounded sequence. Suppose that {u(p_{n})} does not converge to u(p). Then there exists a subsequence {u(pthat converges to a real number y u(p). This, however, contradicts (ii). We conclude that
u(p_) ^{u}Cp) • ■ •
n
Lemma 17. Let h : R + R be a strictly increasing function such that h(R) is neither bounded above nor bounded below. Then there exists a 1c ±
continuous weakly increasing function h ; R ■* R such that h (h(x)) = x for every x € R.
Proof. Let Z = h(R). Observe that h'^{4} : Z ■> R is strictly increasing. For any x € R, the set (^{00}, x] r\ Z is nonempty. Also, h^{}\(°°, x] n Z) is bounded above, because if we choose y G Z with x ^y, then
1 1 h ((“> x] Z) is bounded above by h (y) .
We claim that for every x E R
1 1
(27) sup h ((«>, x] n Z) = sup h ((<», x) n Z).
If x £ Z, the equation is trivial. Suppose x G Z. Then y < h”\x) (h(y) < x and h(y) € Z), .
so that h((«>, h”^(x))) C (oo_{J} x) n Z. Hence
(«, h^{4}(x))Ch^{4}((», x) n Z),
so that sup h^{1}((<», x) n Z) >_h^{1}(x) = sup h^{1}((«>, x] n Z). The opposite inequality is trivial, so (27) is established.
We also claim that
(28) inf h~‘^{l}((x, +°°) rv Z) = sup h^{1}((°°, x] n Z).
1 1
Obviously, inf h ((x, +“)nZ) >_ sup h (( <», x] n Z). Take any y >sup h \(<», x] n Z). If h(y) x, then h(y) £ (~, x] n Z, and
so ygh ((«>, x] n Z) a contradiction. Thus h(y) > x and h(y) € (x, +“) n Z. Therefore y^h"^((x, +~) n Z), and so inf 'h ^{1}((x, +°°) n Z) _< y. In view of the choice of y, this implies
that
inf h^{1}((x, +“) n Z). £ sup h"^{1} ((^{00}, x] n Z), and (28) is established.
Define
* 1 h (x) = sup h ((<», x] n Z).
* [§]
It is clear that h is weakly increasing and that h (h(x)) = x for ' * I every real x.. The continuity of h can easily be deduced from the equations
"k ft
sup h ((», x)) = h (x) inf h*((x, +«)) = h*(x),
which are established as follows:
sup h*((~,.x)) = sup sup h"^{1}(C°°, y] r\ Z)
= sup h~^{1}((«% x) n Z)
= sup h"l((“, x] rx Z)
, * ■
= h (x)
inf h ((x, +«)) = . sup h ((<», y] n Z)
= juf ^f h^{_1}((y, +») rx Z)
= inf h~\(x, +«) rx Z)
= sup h”\(°o, x] n Z) ,
* *
= h (x). ■
Theorem 6. Let E be any subset of X and let <p: E >be any function of honorary Baire class 2(E, R^). Then there exists a ontinuous function f : H >R^ such that cp is a boundary function for f.
Proof. Let ip : E ■> be a function of Baire class 1(E, R^) and N a countable subset of E such that <^(x) = ip(x) for every x € E  N. Let {s } . (with n + m implying s 4 s ) be a countable dense subset of X that includes every integer and every point of N. Let
t =1 if s is an integer
t = — if s is not an integer,
n 2^{n} ' n ^{6}
Define
h(x) = E t if x > 0
0<s_{n}<x ^{n}
i. h(x) = .E t if x £ 0.
x<s <0 — n—
Then h is a strictly increasing function from R into R, and h(R) is
* bounded neither above nor below. Let h be the function described in
Lemma 17.
Suppose that 0 < y <1. Then (for fixed x)
*, x  (ly)u , _ „
u  h (  
is a strictly increasing continuous function of u that approaches +°° as u > +~ and » as u > <». Consequently there exists precisely one number u(x, y) that satisfies the equation
(29) u(x, y)  h*( ^{y)} ) = 0.
I claim that u(x, y) is a continuous function on
= { <x, y>: x, y € R and Q < y < 1}. Suppose’ { <x_{n>} y^} C H_{1}
and y_{n}> * <*, y>€ H_{r} If u(x_{n}, y_{n}) + then  ^{x}n  >'n^{)} _ .
y_{n} n ’
and hence
* x  (1y )u(x , y ) u(x , y )  h*( ^{n n} >
n’ ^{7}n' ^{k} y which.contradicts (29) , Thus u(x_{n?} y ) cannot approach +»., A similar argument shows that u(x_{n}, y ) cannot approach ^{w}.. Now assume that
u(x n
Then, by (29)
lim r r s k* r
nx» t^{u}(^{x}n’ C
x: r. (1y )u(x , y )
* x  (ly)u
 h (  
so u o
= u(x
y) •
By Lemma 16, u is continuous.
From
Lemma 6, there exists a sequence °f continuous
functions mapping X into such that g^(x) •* ip(x) for each x € E. n
For n > 2
define
f_{o}(x, y) =
(yn(n+l)  n)g_{n}(u(x, y)) + ((n+1)  yn(n+]))g_{n+]L}(u(x, y))
■ 1 1 when —= < y < —. n+1 — ^{7} — n
Then f is o
continuous
on
we can assume that
is
^{r}n
o
inf x>s
n
2 " “•
defined
h(x)
By the Tietze extension theorem
and continuous on
all
of H. Let
^{v}n
sup x<s n
h(x)
V(^{s}_{n})  Ks_{n})
if s € N n
^{v}n
if s £ N. n T
If x
and y are real numbers, define
x V y = max{x
y>
set
A_{n}(x, y) =
[(1  ny) V 0] [(1  ——^{1} . X A/
n n
I r + A  2s
^{1} n n n
. .s x
2 2— y
Then
^{A}n
is continuous in H. Observe that A_{n}(x, y) = 0 when y >_
Using this fact, it is:easy to show that, if we set
f = f + E A , ^{0} n=l ^{n}
then f is defined and continuous in H. We now show that (pis a boundary function for f.
Let p be any point of 'E. The line
(30) x = (h(p)  p) y + p
passes through (p, 0), and the part of it that lies in is an arc at p. We will show that f approaches ip(p) along this line. If we substitute (h(p)  p)y + p for x in the expression for A (x, y), we i ■ ^{n}
obtain
(31) A_{n}(x, y) =
[(1  ny) V 0] [(l —ij r_ <• 1 4 2(1  1) (s  p) n n ^{3}
 2h(p)  ) V O] v_{n<}
If p £ s , then h(p) £ &_{n}, and one can verify directly that (31) vanishes. If p > s_{n}, then h(p) £ r , and again one can verify directly that (31) vanishes. Thus A_{n}(x, y) vanishes along that part of the line (30) lying in H.
Solving (30) for h(p), we find that, along the given line, h(p) = ,
and hence p = h (h(p)) = h ( —XlXlE) . Therefore, if 0 < y < 1, p = u(x, y) . So, if ^x, y^ satisfies (30) , n £ 2, and £ y £ ~, then
f_{0}(x, y) = (yn(n+l)  n)g_{n}(p) + ((n+1)  yn(n+l))g_{n+1}(p).
Since the coefficients of g_{n}(p) and g_{n+}j(p) in the above expression add up to 1 and since both coefficients lie in [0, 1], f (x, y) lies on the line segment joining g_{n}Cp) to g_{n+}^Cp)» ^{an}^ it follows that f_{Q}(x, y) approaches ip(p) ^{as} y^ approaches p along the line (30). this line lying in H, f(x, y) show that f approaches (s^) that lies in H. Again, we first consider the value of along the
Since each A_{n} vanishes on the part of also approaches ip(p) along the line.
Let s^ be any point of N. We along the part of the line r. A
(32) x = ( x  sjy +
given line. Substituting the value of x given by (32) into the expres
sion for A , we obtain n’
(33) A_{n}(x, y) =
_{[C}lny) V 0j[(l  1 r_{n}  r_{m + +} 2(1  1) (s_{n} ) V 0]v_{n}.
n n
If s < s , then £ < r < £ < r , and one can verify directly that m n m m r. n n’
(33) vanishes. If s < s , then £ < r < £ < r , and again one can n m n n — mm
verify that (33) vanishes. Thus, for n  m, ^(x, y) = 0 when <x, y^ lies on the line (32) and in H.
If we take n = m in (33), we obtain
A_{m}(x, y) = [(1  my) V 0]v_{m}.
Therefore A_{m}(x, y) approaches v_{m} = 9(^{s}_{m})  along the given line. Take any <x, y>€ satisfying (32), and take any a and b satisfying
(34) a < s < b.
m
■^ < r^ £ h(b), so that
Then.h(a). < ft . <  — m
(h(a)  s )y + s m ' m
x < (h(b)  s )y + s ; from which we deduce that m m
h(a).
x  (ly)s
*
Since h
is weakly increasing, * * x  (lry)s
_{:} h (h(a)) <h (  
m *
—). <h (h(b)) = ,b.
Since a
and b were
taken to
be any two numbers satisfying (34), we
conclude that
whence it follows that u(x, y) =
s . Thus m
^{f}o^{(x}>
y) = (yn(n+l)  lOg^s^) +
when
^{f}o^{lx}
1 1
(x, y> lies on the given line and £ y £ y Consequently y) approaches ip(s_{m}) along the line (32). So f approaches <p(s_{m}) +
<p(s ) ’ ^(^{s}_{m}) ^{=} <p(^{s}_{m}) > and the theorem is proved. ■
2
Theorem 7. Let E be any subset of X and let cp: E > S be any
2
function of honorary Baire class 2(E, S ). Then there exists a
2
continuous function f : H > S such ithat <p is a boundary function for
f.
Proof. The proof of this theorem is very similar to that of Theorem 1.
2 3
Since S CR , there exists, by Theorem 6, a continuous function
3
g : H > R having ipas a boundary function. Let
K = g^{1}({v € R^{3} : v =  J)
L = g^{1}({v € R^{3} : v >  })
F = g^{1}({vGR^{3} : v £y }).
Let g_{0} = gl^. H is homeomorphic to R , so by [5,.Lemma 2.9,p. 299], g can be extended to.a continuous function ^{s}o
g_{1} : H > {v G. R^{3} :  v  = . •
3
Define f^ : H > R  {0} by setting <
fjCz) = g(z) if z € L
f_{x}(z) = . g_{x}(z) if z € F.
Then, since F and L are closed, f^ is continuous on H. It is easy to 3 2
verify that <p is a boundary function for f^. Let P_{q} : R  {0} ■* S
2
be the 0projection onto S (see page 11), and let f be the composite
2
! function P_{Q} « f^. ^{1} Then f maps H continuously into S , and P_{o}« <p =
is a boundary function for f. ■
CHAPTER II. BOUNDARY FUNCTIONS FOR DISCONTINUOUS FUNCTIONS
6. Boundary Functions for Baire Functions
It is not known whether the set of curvilinear convergence of a Borelmeasurable function defined in H is necessarily a Borel set. The answer is not known even for functions of Baire class 1. However, a theorem on boundary functions that is similar to the corresponding result for continuous functions in H can be proved for functions of Baire class 5 in H.
Definition. If A and B are two sets, we will call A and B equivalent and write A  B if and only if A  B, and B  A are both countable. It is easy to check that  is an equivalence relation.
Lemma 18. If A s E, then S  A  S  E for any set S. If A  E ' ' n n
for all n in some countable set N, then
LJ A ^{s} Ij E and p A “ P E . n 6N n £ N n € N n e N
The proof of this lemma is routine.
Definition. An interval of real numbers will be called nondegenerate if it contains more than one point.
Lemma 19. Any union of nondegenerate intervals is equivalent to an open set.
Proof. Let be any family of nondegenerate intervals. .It will
suffice to prove that UiUI is countable. We can is4J ie<5
write
I€&
J n * n
where {J } n
is a
countable family of disjoint open intervals.
If
then x o
so that
is
an endpoint of I
for some I € For some n, I o
n*
* I
is an endpoint of J^.
*—
G J o n
Thus
is
contained in the
set of all endpoints of
the
various J . and the n’
lemma is
proved.
Lemma 20. Let h be a weakly increasing realvalued function on a nonempty set E £r. Suppose that x  h(x) . £ 1 for every x € E. Then h can be extended to a weakly increasing realvalued function h^ on R.
Proof. Let e = inf E (e may be »). For each x € (e, +“), set
h^x) = sup h ((<», x] r\ E).
Since t  h(t)  1 for each t € E,
t G (°°,^{;} x] n E r^>h(t) — x + 1,
so h^ is finitevalued. If e =  <» we are done. If e >then x e E implies h(x) >x  1 >e  1, so h is bounded below. For x C C°°, e] set
h^(x) = inf h(E).
It is easy to verify that h.^ has the required properties. ■
Lemma 21. Let Y be a metric space, f : R •* Y a function of Baire class 5(R, Y), and suppose that h : R > R is weakly increasing. Then there exists a countable set N Cr such that the composite function f o h_{D} .. is of Baire class 5(R  N, Y). KIn
Proof. Let N be the set of discontinuities of h. By a wellknown theorem, N must be countable. But then h_{R} is continuous, so that f ® (h_{R N}) = (f <» h) _{R N} is of Baire class ?(R  N, Y). 8
Lemma 22. Let Y be a separable arcwise connected metric space, E any metric space, and let <p: E >Y be a function having the following property. For every open set U *= Y there exists a set T € P^^{+}\e) such that <p^{i}(U) Ct Then, if £ >_2, <p is of Baire class 5(E, Y) .
Proof. The proof is similar to that of Lemma 15. Let (B be a countable base for Y, and suppose that W is any open subset of Y. Let
d(W) = {US® : U C _{W}}.
The argument in the proof of Lemma 15 shows that w = u = L7 u. U€(J(W) U€O(W)
For each U€(fi, let T(U) € P^^{+}\e) be chosen so that «p'^{1}(U)^T(U)G<p'^{1}(ii). Then
<p^{1}(W) = T(U)
Thus <p \w) = T(U) , and since P^^{+}\e) is closed under countable
1 E+1
unions, <p (W) € P^{5,} (E) . Therefore is of Baire class £(E, Y) . ■
Theorem 8. Let Y be a separable arcwise
f : H + Ya function of Baire class g(H
X, and <p: E >Y a boundary function for
C + 1(E, Y).
Proof.
Observe
that
connected metric space,
Y) where
f. Then
Let U be any open subset of
C = AUB.
For each x
and let V =
g >_ 1, E a subset of
ip is of
Y  U.
= V^{1}(V)
choose an arc y_{v}
Baire class
Set
at x such
lim z>x f(z) ^{z €Y}x
^{Y}x
^{Y}x
z  x £ 1}
 ’ {x} c f^{1}(V)
Notice that if x G A and
We will say that
have subarcs y ' and y ' 'x 'y
if
if
x € B.
y G B, then
y„ meets y.
' X
in
respectively such
n’ ^{311(1 Y}x' ^{n Y}y* *
L = {x€ A : (Vn)(3y)(y € C, a
M = {x € a
= {x'€
Let
and
: (Vn).(3y)(y€ C
: (3n)(v meets no x
: (9 n)(y meets no
L ^ULh
M = M_{a} U .
and
y " *•
provided that y and y x y
that x £ y_{x}’ C H_{n},
meets y in H )} y ' x n'
meets y
in_{Hn})}
Y_{y} (with y  x) in H_{n})}
Y_{y} (with y f x) in HJ
Observe that L_{&}, 1^, M_{&}, are pairwise disjoint, and that and B = L^ u M&.
For
meets no y y
meets no y y
each x6M, let n(x) be a positive integer such that y
(with y
x) in . Then n >_n(x) implies that y_{x}
Let
• meets X_{n}, and, if x € M, n > n(x)}.
Then K K . for each n, and C = ( 7 X . n n+1 ’ n
n=l
We next show that for each positive integer n and each x there exists a nondegenerate closed interval I® such that
x € C L_{a} (X  K^) . By the definition
y € C (y 1
x) such that y meets y_{x} in
interval having its endpoints at
x and
y
of L , there exists a’
Let I^{n} be the closed x
Let t be any point of
We must prove that t £ L_{&} u (X  assume t £ K . Then y. meets X n ^{1}1 n
that y. must meet either y or y ’t 'x ’•
K ) . n'
If t i K .we are done. ~ n’
and hence it is clear from
rigorized by means of
Theorem 11.8
in (This argument can ; on p. 119 in [11].) But,
then (because t € K_{n})
n >_ n(t), so
Therefore
t £ M. Now
Hence y
C  B = A.
' I^{n}. x
So
Figure 5
be
if t € M,
that this situation^is impossible.
x £ L £ A, so, since y intersects y , a • ’ 'x 'y*
Similarly, since y intersects y or y
■x 'y
tec
= A.
Thus t £ A  M = L , and we have shown that a
(X 
K ).
Let W n
x€ L
For each n,
a
L c W n C Q [L u (X  K )] A c, Cl 11 d 11
and therefore
Vac n=l co
■ {p [L u (X  K_{n})]} n c '
n=l co
 I^{L}_{a}^{u (X} C[ ^{K}n^{)] nC}
CO.
= (L a C) u (C  U K ) = L u <j>= L . .
d <t 11 CL d
n=l
co
It follows that L = (fA W_{n}) a C. By Lemina 19, each is equivalent n=l .
to an open set, so there exists a ($. set G_{a} Q X such that
I
L  G a c. a a
• • ■
A similar argument shows that there exists a ($ set G^ ?= X such that
^{L}_{b} “ %^{AC}
Next we study the properties of M . In doing this, it is 81 ‘
2 convenient to define a function tt : R ■* R by setting ir(x, y) = x.
If x € M nK , then, starting at x and proceeding along y , let p (x) n * • • x I*
be the first point of X that is reached. Define h° : M AK •* R ^{r} n n n
by setting h°(x) = ir(p_{n}(x)). If x, x' € M a and x < x’, then, since
y cannot meet y ’ in H , it is evident that p fx) must lie to the 'x x n n
left of p_{n}(x’); that is, ir(p_{n}(x)) < Tr(p_{n}(x’)). Thus h° is a strictly
increasing function on M A K^. Moreover, •
x  h°(x)  £ 1 because y_{x} C {z : z  x <_ 1}.
So, by Lemma 20, h° can be extended to a weakly increasing function
h : X > R. Let n
g_{n}Cx) = ,f(h_{n}(x),l) (x € R).
f(x, is a function (of x) of Baire class £(X, Y), so, by Lemma 21,
there exists a countable set N X such that g _{v} is of Baire class n • n^{1}XN
— n
N_{n>} Then is of Baire class
n=l
€(M  N, Y) .
For x € M A K_{n}, g_{n}(x) = f(h°(x), £) = f(p_{n}(x)). If x € M, then
for all sufficiently large n, x € M AK_{n}, so —
_{n}^> f(P_{n}W) = V>(X).
(pl., hence «pL, _{XT} is of Baire class _{n} t imN’ ^{T} ‘MN
Thus g I.. ■ ^{6}n'M
' ? + 1(M 
A
Obviously
L
^{so} sJmn
N, Y). It follows that there exists D G P^^{+2}(X) such that a(mn) = (vl_{M}__{N})^{1}(u) = DA(MN).
A a M ~ DAM. Now,
= L V L “ fG> A Cl U fG. A Cl = fG v G, 1 A C
a n 'a ' ' d ' 'a d'
so
I
M_{a} = A n M■  P n M = D A (C  L)
= D A [C  ((G_{a}U GJ,) A C)]
= Dn [x  (G_{a} u g^] a c.
G_{a} and G^ are G$ , so X  (G_{a} u G^) is , and hence
^{x}  (G_{a}uG_{b})€ P^{2}(X) Cp£^{+2}_{(X}).
Therefore M_{&} « F A C, where F € P^{?+2}(X). Now, G_{a} € G_{g}(X) = Q^{2}(X), and since E, > 1, Q^{2}(X) £ P^{5+2}(X), so G u FGP^{C+2}(X). But a
A = L u M  (G A C) U (F A C) = (G_ v F) A C, AAA d
so A  S a 0, where S € P^^{+2}(X). Since every countable set is F , it o
is now easy to show that
A = T n C
for some T € P^^{+}^(X). From the definition of C it follows that
T — X  B. Thus we have
= ACTHECEB = E  cp^{1}(V) =
THE G P^^{+}2(E), so Lemma 22 shows that is of Baire class g + 1(E, Y) . ■
Corollary. Let Y be a separable arcwiseconnected metric space, f : H > Y a Borelmeasurable function, E a subset of X, and <p: E >Y a boundary function for f. Then vp is Borelmeasurable.
Proof, f is of some Baire class 5CH, Y), hence <p is of Baire class g + 1(E, Y), hence (p is Borelmeasurable. ■
This corollary raises the question of whether a boundary function for a Lebesguemeasurable function is necessarily Lebesgue measurable, which we answer in the next section.
7. Boundary Functions for LebesgueMeasurable Functions
i
Suppose that a_{Q}, b , a^, b^ are extended real numbers, and that a < b , a; < b.. To make the formalism more convenient we let 0—0* 1—1
(«>) (») = 0 and (+<»)  (+00) = 0. In other respects we adhere to the usual conventions regarding arithmetic operations that involve co or +00. Let
^{T(}^{a}0’ ^{b}o’ ^{a}l» ^{b}P = < : 0 £ y £ 1 and
(^{a}l  ^{a}_{o})y + a_{o} £ x £ (b_{x}  b_{Q})y + b_{Q}}.
A set of this form will be called a closed trapezoid. We also consider <>to be a closed trapezoid. A set S will be called a trapezoid if there exists a closed trapezoid T such that T^{1} Q S ST, where T^{1} denotes the interior of T relative to H^. Every trapezoid is Lebesguemeasurable, though not necessarily Borelmeasurable. *
If s, s’ are disjoint line segments having endpoints ^a_{Q}, 0^ <a_{1}, 1> , and <a_{Q}’, 0> , <a_{1}', 1> respectively, where a^ £a^' (i = 0, 1), then let
T(s, s’) = T(s’, s) = T(a_{o},,a_{o}’; a_{p} a_{J}’).
If s = s’, then we let_T(s, s’) = T(s’, s) = s. In what follows we will use the symbol X_{Q} as an alternative designation for the xaxis X. This will enable us to make statements about X^ (1=0, 1) (where Xj, denotes, as before, { <x, O : x€ R}).
We omit the proofs of the following two routine lemmas.
Lemma 23. Let theline segments s', s_, s~, s. each have one endpoint X Z O
on X_{Q} and the other on X^, and assume that i 4 j implies that either
s. n s. = $ or s. = s.. If T(s,, s_{9}) n T(s,, s.) 4 <l>>then
1 J .1 J AZ □ 4
T(s_{1?} s_{3}) Q T^, s_{2}) u T(s_{3}, s_{4}).
Lemma 24. Letbe any set of line segments, each of which has one endpoint on X_{q} and the other on X^, and no two of which intersect.
Then T(s, s’) is a trapezoid.
s,s’€£ _{2}
Let m denote twodimensional Lebesgue measure in R . If E
2 Z
is a measurable subset of some line in R , let m (E) denote the linear
—— £
Lebesgue measure of E. Let m_{g} and m_{g} denote twodimensional exterior measure and linear exterior measure, respectively; i.e., for any E c_{R}^{2};
m (E) = inf {m(U) : E U and U is open); 6
67
t
and if E is.a subset of a line L, then p’ p _
m*(E) = inf {m (U) : E L and U is open relative to L).
Theorem 9. Let «C be any set of line segments, each of which has one endpoint on X_{q} and the other on Xp and no two of which intersect. Let S = . Then
i m (bj =  (mUS n X ) + mHS n X )) . C 4 U U C X
Proof." We may assume that <£ is nonempty. Let e be any positive
2 number. Choose an open set U 5= R such that S <=. U and
m(U)‘ £ m (S) + e.
Let E^ = S n X^ (i = 0, 1). Choose sets <=. X^ that are open relative to X. such that E. Q. G. and i ii
p p
m (G.) < m\E.) + e (i = 0, 1).
1 “ C X
2
Let V be the union of all lines L Cr such that L meets both G and o
Gp It is easy to show that V is an open set. Furthermore, S £V and V n X. = G. (i = 0, 1). Now let W = U n Y. Then i i *• ’
W is open, S C W S U, and 
E. C W n X. C G. (i = 0, 1).
If s, s'€ <£ , define s = s' if and only if T(s, s') Q W.
It is easy to verify by means of Lemma 23 that = is an equivalence relation. Let T be the set of all equivalence classes. We prove that r is countable.
If s € , we let ^a^(s) , i^> be the endpoint of s on X^
(i = 0, 1). Then ,
2
s = { (x, y^^R : 0 _< y 1 and x = (a^s)  a_{Q}(s))y + a_{Q}(s)}.
Since s is compact and contained in W, there is no difficulty in showing that there exists V > O .such that
£ {x, y> € R : 0 £y < 1 and
(a, (s)  a (s))y + a (s)  & . x < (a_{1}(s)  a (s))y + a (s) + 6 } XU U 3 X v v O
Let <K(s) = (a^(s)  &_{s}, a^(s) + S_{g}) (i = 0, 1). A sketch will rapidly convince the reader that if s, s’ €«C , J_{Q}(s) ^{n J}_{0}(^{s}’) 4 and J_{x}(s) n J^(s') 4 ‘I’, then T(s, s’) Q W, so that s = s’. Thus
(J_{o}(s) x j_{i(s})) A CJ_{0}(s’) x J_{x}(s’)) 4 <> =*>. s’ = s’.
For each C € F, choose s(C) € C and let
Q(C) = J_{0}(s(C)) x J^sfC)).
Then C. 4 C_{9} =^Q(C.) n Q(C_{9}) = $. Since each Q(C) is a nonempty 2
open subset of R , this implies that F is countable.
If C6F, let
T(C) = VJ T(s, s’), s, s’GC
By Lemma 24, T(C) is a trapezoid. Also,
(35) CCt(C)$W.
Suppose that C^, C_{2} £F and 4 ^{C}2' ^e claim that
T(C..) T(C_{9}) = <j>. Assume that T(C.)nT(C_{9}) 4 • Then there exist
S,, s_{n} ‘ € C. and s_{9}, s_{9}' £ C_{9} such that T(s_{n}, s') A T(s_{9}, s_{9}') 4 4» •
By Lemma 22,
T(s_{1}, s_{2}) C T(s_{1}, Sj’) u T(s_{2}, s_{2}') C W,
so that s. = s_{9}; a contradiction. Therefore T(C.) A T(C_{9}) = <>. X I X £»
Let . IL (C) = T(C) a X_{±} (i.= 0, 1). . Then IL (C) is an interval and
(36) E. C K. (C) C.W AX. CG (i = 0, 1).
Furthermore, C. C_{o} implies that K.(C.) A K.(CL) = $. Using the X Z XX X £»
formula for the area of a trapezoid, we find that
' 4 [A U K (O) + im^C U K (C))]
= cgr C€r
= s' 4 (m*(K (O) + m^{£}(K (C))) cer ^{z}
= E m(T(C)) = m(U T(C)). cer cer
Let a = 4 [m\U ^{+ K}iCC))] ^{z} cer cer
= iucI^Jtcc)).
cgr
According to (35), S Q \) T(C) Q W Q U, so that cer
(37) m (S) < a < m(U). < in (S) + e. e — — c
By (36),
(38)  (m*(E_{o}) + ^(Ep) < a <  (m^{£}(G_{Q}) + m^{£}(G_{1}))
<7 CmhE ) + m*(E )) + _{e}.
Since e is arbitrary, inequalities (37) and (38) imply that
m (S) =  (m*(E ) + m*(E )). ■ ^{:}
One wonders to what extent a result resembling the foregoing theorem might be obtainable without the hypothesis that no two of the line segments intersect. The’following example is relevant to this question. Let M_{q} be a residual set of measure zero in/X_{Q} and let be a.residual set of measure zero in Let <x_{q}, y\ be.any point of Hj. We claim that there is . a line segment passing through ^x_{Q}, y^ that has one endpoint in and the other in M^. For 0 € (0, it), let
Fife) = <(1  y_{0}) ctn 6 + x_{o}, 1> and
f_{0}(q) = <^{x}_{0}  y_{0} ^{ctn e}> ^{0} >•
Then F^ is a homeomorphism of (0, it) onto X^, so F_{q}^{1}(M_{o}) and F^^(M^) are both, residual sets in (0, it). Choose a £ ^{F}o^M_{o}) HF^^(M^). Let L be the line whose equation is
x = x_{q} + (y  y_{o}) ctn a.
Then L passes through the points ^x_{q}, yj) , F_{Q}(a) and F_{1}(a), so that LaH^ is the desired line segment. Let be the set of all line segments having one endpoint in M_{q} and the other in M^. Let S = Then S r\ X_{q} and S f\ X^ both have measure zero, but, as we have just
1  '
shown, S, so that S has infinite measure. See Problem 5 at the end of this paper.
Lemma 25.
For every e > 0 there exists a strictly increasing real
valued function h on R such that h(R) has
measure zero, and, for every
real x, x  h(x) £ e.
I ^{+}” I
Proof. For each integer n, let I = [ne, (n + 1)e]. Then I_{n} = R« n^ioo
There exists a strictly increasing function f : [0, 1] •* [0, 1] such that m\f([0,1]))= 0. For example, such a function may be defined as follows. Any number in [0, 1) may be written in ti& form
•$,a_{9}a,.. .a. ... (binary decimal),
A bi O IX
where the decimal does not end in an infinite unbroken string of 1’s.
Set.
f (a..a„a,.. .a ...) = •b.bb .. .b ... (ternary decimal), ■ x z n ’ x z o n
where b. = 0 if a. =0 and b. = 2 if a. =1. i i i i e
Set f(l) = 1. Then f maps [0, .1] into the Cantor ternary set, so
£
m (f([0, 1])) = 0. It is easily shown that f is strictly increasing.
For each n, it is easy to obtain from f a function f : I ■* I ^{9 J} n n n
£
such that f is strictly increasing and m (f_{n}(I_{n})) = 0. Set
h(x) = f^fx) for x C(^{ne}> C^{n +} 1)^{£}]
There is no difficulty in proving that h has the required properties.®
Theorem 10. There exists an indexed family £y_{v}}_{v} _{Y} simple arcs X X C A
such that
(i) for each x g X, y is an arc at x X
(ii) x 4 y =^y_{v} ny_{v}M
(iii) I J y is a set of measure zero, x € X ^{X}
Proof. For each natural mumber n, let h_{n} : R •* R be a strictly increasing function such that h_{n}(R) has measure zero and, for every x, x  h_{n}W  For every x € R, let s_{n}(x) be the line segment joining the point ^h_{n}(x), to the point ^h_{n+}^(x), Since
< ^{h}n^{tx}2^{]} ><sup>x</sup>l < <sup>X</sup>2
^^{h}n_{+}l^{(X}l^{) < h}_{n+}l^{tx}2^{)}’
we see that x^ implies s_{n}(x^) n Sn^) = <>. Let S_{n} = VJts_{n}(x) : x£ R}. Then
S n X C{(x, : x€ h (R)}
n n — ' ’ n ' n^{k} '
and S n X . C. {/x, A*5 : x g h _{n} (R)}, n n+1  x » n+1 / n+l^{k J} *
g £
so m'(S n X ) = m (S n X .) = 0. It is easy to deduce from n n n n+1
Theorem 9 that _{;}
1'1 If 2
V^{S}n> • < K  nJT ’ T CV^{S}n " V * V^{S}n ^{n X}n_{+}1» ' ’•
For x € X, let y = {x} U I ) s (x). Since \h (x), — / •* x, y is x n n n / _{n} x
an arc at x.
00
®_{e}(U Y_{x}) < m_{e}W + m_{e}(U ^{S}J
^{e} x£X ^{x e e} n=l ^{n} . co
£m (X) + E m (S ) = 0, ^{e} n=l ^{e n}
so y is a set of measure zero. ■ x6X ^{x}
Corollary. Let <p be an arbitrary function mapping X into any topologi
cal space Y having an element called 0. Then there exists a function
f : H > Y such that f(z) = 0 almost everywhere and ipis a boundary
function for f.
Proof. If {y } _{Y}th® family of arcs described in Theorem 10, let X X £ a f(z) =0 if z is in no y
X
f(z) = <f>(x) if z €Y . A
Then f is the desired function.®
Corollary. There exists a realvalued Lebesguemeasurable function f defined in H having a nonmeasurable boundary function defined on X.
SOME UNSOLVED PROBLEMS
1. If A is an arbitrary set of type F_{a5} in X, does there necessarily exist a realvalued continuous function f defined in H having A as its set of curvilinear convergence? If <p is an arbitrary realvalued function of honorary Baire class 2 on A does there exist a continuous realvalued function f defined in H having A as its set of curvilinear convergence and (p as a boundary function?
2. (First proposed by J. E. McMillan [10]). If A is any set of type 2
F_{a}g in X and if (pis any function of honorary Baire class 2(A, S ), 2
does there necessarily exist a continuous function f : H > S having A as its set of curvilinear convergence and <p as a boundary function?
3. If f is a realvalued Borelmeasurable function defined in H, is the set of curvilinear convergence of f necessarily a Borel set? What if f is assumed to be of Baire class 1?
• 3 
4. Let S = { <x, y, z) g R : Z >0}. If f is a function defined in S, we define the set of curvilinear convergence of f in the obvious way. If f is continuous, is its set of curvilinear convergence necessarily a Borel set? Is it necessarily of type F ?
5. Let <£ be a set of line segments each having one endpoint on X_{q} and the other on X^, and let SUz . Assume that S is a Borel set.
£ £
If m (S n X_{Q}) and m (S n X^) are known, what lower bound can be given for m(S)? . A solution to this problem might be helpful in attacking a problem of Bagemihl, Piranian, and Young [3, Problem!].
REFERENCES
1. F. Bagemihl, Curvilinear cluster .sets of arbitrary functions, Proc. Nat. Acad. Sci. U. S. A., ! (1955) 379382.
2. F. Bagemihl and G. Piranian, Boundary functions for functions defined in a disk, Michigan Math. J.; 8 (1961) 201207.
3. F. Bagemihl, G. Piranian, and G. S. Young, Intersections of cluster sets, Bul. Inst. Politehn. Iasi, 5 (9) \1959) 2934.
<sup></sup>.. ,.
4. s. Banach; Uber analytisch darstellbare Operationen in. abstrakten .Raumen, Fund. Math,, .!I_ (1931) 283295.
5. S. Eilenberg and N. Steenrod; Foundations of.'algebraic topology, Princeton University Press, Princeton, NewJersey, 1952.
6. F. Hausdorff, Uber halbstetige Funktionen und deren Verallgemeinerung, Math. Z., 5(1919) 292309/ ^{. 7}
7. F. Hausdorff; Set Theory, Second edition, Chelsea Publishing Company, New York, N. Y., 1962.
8. T. J. Kaczynski, Boundary functions fo:r functions defined in a disk, J. Math. Mech., 14 (1965) 589612.
9. T. J. Kaczynski, On. a boundary property of continuous functions, ' Michigan Math. J., 13 (1966) 313320. ^{.}
10. J.E. McMillan, Boundary properties of functions continuous in a disc, Michigan Math. J.,'3 (1966) 299312. ^{.}
11. M. H. A. Newman; Elements of the topology of plane sets'of points, Cambridge University Press, 1964.
12. H. Tietze, Uber stetige Kurven, Jordansche Kurvenbogen'und geschlossene Jordansche Kurven, Math. Z.; 5 (1919)' '284291,
Ted's Work as an Assistant Professor of Mathematics at the Uni. of California
7. 1968  Note on a Problem of Alan Sutcliffe
Original PDF: 7. 1968  Note on a Problem of Alan Sutcliffe.pdf
MR0228409 Kaczynski, T. J. Note on a problem of Alan Sutcliffe. Math. Mag. 41 1968 84.86. (Reviewer: B. M. Stewart) 10.05
Reprinted from the Mathematics Magazine
Vol. 41, No. 2, March, 1968
NOTE ON A PROBLEM OF ALAN SUTCLIFFE
T. J. KACZYNSKI, The University of Michigan
If n is an integer greater than 1 and ah, • • , a_{b} ao are nonnegative integers, let
(ah, • • • , ai, ao)_{n} denote ahn^{h} + • • • + ain + ao.
Thus if O;;;ai;;;n1 (i = O, • • •, h), then a_{h},^^, a_{1}, ao are the digits of the number (ah, • • •, a_{b} ao)n relative to the radix n. Alan Sutcliffe studied the prob ,
lem of finding numbers that are multiplied by an integer when their digits are reversed (Integers that are multiplied when their digits are reversed, this Magazine, ^{?}
1964] MATHEMATICAL NOTES 653
Similarly,
(3) 6* =  1.
Taking q = characteristic of F (gl=O), choose t and r as specified in the lemma. Using relations (1), (2), (3), we have
(Z + ra + i)(r» + 1 + ria + lb) = r(Z^{2} + r* + l)a + (Z^{2} + r^{2} + 1)6 = 0.
One of the factors on the left must be 0, so for some numbers u, v, w, u 0 (mod q), we have w+va\ub = Q, or b = — u~^{l}va — u~^{l}w. So b commutes with a, a contradiction. We conclude that S is not a generalized quaternion group, so 5 is cyclic.
Thus every Sylow subgroup of F* is cyclic, and F* is solvable ([4], pp. 181— 182). Let Z be the center of F* and accnme ^{7}F*. Then F*/Z is solvable, and its Sylow subgroups are cyclic. Let A/Z ("fit.. ZC^) be a minimal normal subgroup of F*/Z. A/Z is an elementary abelian group of order p^{k} (p prime), so since the Sylow subgroups of F*/Z are cyclic, A/Z is cyclic. Any group which is cyclic modulo its center is abelian, so A is abelian. Let x be qny element of F*, y any element of A. Since A is normal, xyx~'(E.A, and (1 +x)y = z(l +x) for some z£Z. An easy manipulation shows that y — z = zx — xy = (z — xyx~')x.
If y — z = z — xyx~' 0, then <em>y = z = xyx~<sup>l</sup>,</em> so <em>x</em> and <em>y</em> commute. Otherwise, <em>x
(z — xyx~^{l})~'(y</em> — z). But A is abelian, and z, y, xyxr'^A, so x commutes with y. Thus we have proven that A is contained in the center of F*. a contradiction.
References
1. E. Artin, Uber einen Satz von Herrn J. H. M. Wedderburn, Abh. Math. Sem. Hamburg, 5 (1927) 245.
2. L. E. Dickson, On finite algebras, Gottingen Nachr., 1905, p. 379.
3. M. Hall, The theory of groups, Macmillan, New York, 1961.
4. Miller, Blichfeldt and Dickson, Theory and applications of finite groups, Wiley, New York, 1916.
5. B. L. van der Waerden, Moderne Algebra, Ungar, New York, 1943.
6. J. H. M. Wedderburn, A theorem on finite algebras, Trans. Amer. Math. Soc., 6 (1905) 349.
7. E. Witt, Uber die Kommutativitat endlicher Schiefkorper, Abh. Math. Sem. Hamburg, 8(1931)413.
Reprinted from the American Mathematical Monthly
Vol. 71, No. 6, JuneJuly, 1964
y e A(n + 1, m[n+ 1], k[n + 1], f [n 4 1]) C U_{n+1} c Q(n 41, m[n+ 1]),
2tt i 1
and therefore each point of y_{n} has distance less than from y. Now
^+~F 0 ^{a}s n * °°; hence, if we set y = {y} U U_{n=}i y_{n}, then y is an arc with
one endpoint at y.
Since U_{n} and U_{n+1} have a point in common,
^{f1}(^{* S}(^> ^{p}k[n])) ^{and rl}(^{S}(^Tl’ Pk[n+1]))
have a common point, and hence
^{S}(^>Pk[n]') ^{and S}(^+l’ Pk[n+1])
have a common point. Therefore, if p is the metric on K, then
< \ ^{1} . ^{1} / 1 , ^Pk[np Pk[n+1] — 2^{n} 2^{n}”^
8. March 1969  Boundary Functions for Bounded Harmonic Functions
Original PDF: 8. March 1969  Boundary Functions for Bounded Harmonic Functions.pdf
Kaczynski, T.J. 1969. Boundary functions for bounded harmonic functions. Trans. Am. Math. Soc. 137:203209.
MR0236393 Kaczynski, T. J. Boundary functions for bounded harmonic functions. Trans. Amer. Math. Soc. 137 1969 203.209. (Reviewer: J. E. McMillan) 30.62 (31.00)
Explanation by John D. Bullough
A function p(e) defined on the unit circle is a boundary function for a function f(z) defined in the unit disk provided for each e, f(z) has the limit p(e) at e along some curve lying in the unit disk and having one endpoint at e. Any two boundary functions for the same function f differ at only countably many points by the ambiguouspoint theorem of Bagemihl; and a boundary function for a continuous function differs from some function in the first Baire class at only countably many points. In answer to a question of Bagemihl and Piranian, the author constructs a bounded harmonic function having a boundary function that is not in the first Baire class. He shows that nevertheless the set of points of discontinuity of such a boundary function is a set of the first Baire category.
Article by Ted
BOUNDARY FUNCTIONS AND SETS OF CURVILINEAR CONVERGENCE FOR CONTINUOUS FUNCTIONS
BY
T. J. KACZYNSKI
Let D be the open unit disk in the complex plane, and let C be its boundary, the unit circle. If x e C, then by an arc at x we mean a simple arc y with one end point at x such that y{x}^D. If /is a function mapping D into some metric space M, then the set of curvilinear convergence of f is defined to be
{x e C: there exists an arc y at x and there exists a point pEM such that /(z) > p as z x along y}.
If is a function whose domain is a subset E of the set of curvilinear convergence of /, then</> is called a boundary function for / if, and only if, for each x e E there exists an arc y at x such that /(z) ></> (x) as z > x along y. Let S be another metric space. We shall say that a function</> is of Baire class 1(S, M) if
(i) domain ^ = S,
(ii)range and
(iii) there exists a sequence of continuous functions, each mapping S' into M, such that <f>_{n} > <£ pointwise on S.
We shall say that is of honorary Baire class g 2(S, M) if
(i) domain <£ = S,
(ii)range</> ^Af, and
(iii) there exists a countable set N^S and there exists a function 0 of Baire class 1(5, Af) such that <£(x) = ^(x) for every xeSN.
It is known that if/is a continuous function mapping D into the Riemann sphere, then the set of curvilinear convergence of / is of type F_{a69} and any boundary function for/is of honorary Baire class ^2(C, Riemann sphere). (See [3], [4], [5], [6], [9].) J. E. McMillan [6] posed the following problem. If A is a given set in C of type F_{ad}, and if is a function of honorary Baire class ^2(A, Riemann sphere), does there always exist a continuous function / mapping D into the Riemann sphere such that A is the set of curvilinear convergence of / and is a boundary function for /? The purpose of this paper is to give an affirmative answer to McMillan’s question. However, the corresponding question for realvalued functions remains open. (See Problems 1 and 2 at the end of this paper.) In proving our result, we first give a proof under the assumption that</> is a bounded complexvalued function, and we then use a certain device to transfer the theorem to the Riemann sphere. As we shall indicate in an appendix, the same device can be
Received by the editors September 20, 1968.
107
used to transfer certain results concerning realvalued functions of the first Baire class to the case of functions taking values on the Riemann sphere.
Our proof is divided into several major steps, which are labeled (A), (B), (C), etc. The proofs of some of the major steps are divided into smaller steps, which are labeled (I), (II), (III), etc. The results (A) and (B) are taken from the author’s doctoral dissertation [5].
Throughout this paper we shall use the following notation. R denotes the set of real numbers, S^{2} denotes the Riemann sphere, and R^{n} denotes wdimensional Euclidean space. Points in R^{n} will be written in the form <x_{b} x_{2},..x_{n}> (rather than (x_{15} x_{2},..., x_{n})) in order to avoid confusion with open intervals of real numbers in the case n = 2. The empty set will be denoted by 0. When we speak of a complexvalued function, we mean a function taking only finite complex values. The closure of a set E will be denoted either by E or by Cl E. If I is an interval of real numbers, then Z* denotes the interior of I. If p is a point of some metric space and r e (0, 4oo), then S(r, p) denotes the set of all points of the space having distance (strictly) less than r from p.
We define
Q = {(x,y)eR^{2} : 1 x 1,0 < y 1}, X = { <x,0>: 1 < x < 1}, H = { <x, y>e R^{2} : y > 0}.
It will be convenient to identify <x, 0>with the real number x, and X with (—1, 1). If f is a complexvalued function defined in Q, then we shall understand the set of curvilinear convergence of f to mean the set of all x e X for which there exists an arc y at x (contained in the interior of Q except for its end point at x) such that f approaches a finite limit along y. If a e X, e > 0, and 0 < 0 < then we let
s(a, e, 0) = {<x, y) e R^{2} : 0 < y < e, a—y ctn 0 < x < a+y ctn 0}.
Thus s(a, e, &) is the interior of an isosceles triangle in H with apex at a.
(A) If A^X is a set of type F_{a6i} then there exists a bounded continuous realvalued function g defined in Q such that
(i) for each xe A, g(z) > 0 as z approaches x through s(x_{9} 1, tt), and
(ii) if xe X, and if there exists an arc y at x such that g(z) > 0 as z approaches x along y, then xe A.
(I) Let E_{±} and E_{2} be two sets on the real line. A point pE R will be called a splitting point for E_{x} and E_{2} if either
(i) Xi ^p for all x_{±} e E_{y} and p x_{2} for all x_{2} e £_{2}, or
(ii) x_{2}^p for all x_{2} e E_{2} and p^x_{1} for all x_{r} e E_{±}.
We will say that E_{±} and E_{2} split if and only if there exists a splitting point for E_{t} and E_{2}.
(II) By a special family we mean a family of subsets of X such that
(i) & is nonempty,
(ii) & is countable,
(iii)each member of & is compact,
(iv)if E, Ftthen either E=F, Ec\ F=0, or E and F split.
(III) If £■£ JTis an F_{a} set, then there exists a special family & such that E= (J
Proof. We can write E= (Jn= i where Ai = 0, A_{n} is closed, and ^4_{n}s^_{n+1} for all n. Observe that if I is any open interval contained in X, then there exists a countable family {J_{n}}"= i of compact intervals contained in X such that 1= IJ "= i A, and n/m implies that J_{n} and J_{m} split. Since XA_{n} is a countable disjoint union of open intervals, it follows that we can choose (for each ri) a family
i
of compact intervals such that X— A_{n}— U"i and <em>j
£k</em> implies that I_{nJ} and Z_{n},_{fc} split. Let
= {Ai} u {I_{nJ} n ^_{tt+1} : n = 1, 2,...; j = 1, 2,...}.
Then SF is a countable family of compact sets, and
E = u A = Ai V U A_{+}1 n (X—A_{n}) n=1 n=1
= Ai u (J (J ^4_{n+1} ri I_{n}j n=l 1=1
=
Let Fi and F_{2} be any two distinct members of If either F_{x} or F_{2} is A_{x} = 0, then Fi and F_{2} are automatically disjoint. If neither F_{r} nor F_{2} is A, then we can write
Ei = 4i(i),«i> 24_{n}(i)_{+}i,
E_{2} — fn<2)J(2) A_{nm + 1}.
If n(l)<n(2), then n(l)+1 ^n(2), so
Ez </em> fn<2),/(2) ^n(2) + l £
X— ^4_{n}(2) — X— ^n(l) + l — X— Fi,
and therefore F_{±} and F_{2} are disjoint. If n(2)<«(l), a similar argument shows that Fi and F_{2} are disjoint. Now suppose n(l)=n(2). Then, since Fi/F_{2}, we have j(l)/j(2). So /_{n}(i).«2)=A(2),/(2) and Lxd.xi) split, and consequently F_{r} and F_{2} split. We have shown that any two distinct members of & either split or are disjoint, so
is a special family.
(IV) Let A £ X be a set of type F_{a!>}. Then there exists a sequence of special families {^}”=i such that
(0 ^An^JU^n),
(ii) if 1 and Ee^_{+1}, then there exists Fewith E^F.
Proof. There exist sets Ai 2^22^32 • • • such that A </em> A"=i <em>A<sub>n</sub>.</em> By (III), we can choose (for each n) a special family <em>A<sub>n</sub></em> such that <em>A<sub>n</sub>
(J <^_{n} Let For 1, let
•^+1 = {fn E : and £e«f_{n+1}}.
By induction on n, one can show that each is a special family and that A_{n} = (J It is clear that the other conditions are satisfied.
(V) Suppose that J is a nonempty interval with X, and let a, b (a b) be the end points of J. By Trap (J, e_{9} 0) (where 0 e (0, ^r) and e > 0) we mean the trapezoidshaped open set defined by
Trap (J, e, 0) = {<x, : 0 < y < e, a+y ctn 0 < x < by ctn 0}.
For 0 e (0, i^) let Tri (J, 0) be the closed triangular area defined by
Tri (J, 0) = {<x, : y 0, a+y ctn 0 x b—y ctn 0}.
If K is a nonempty compact subset of X, let J(K) be the smallest closed interval containing K. If e > 0 and 0</?<a<77, then we define
B(K_{9} e_{9} a, 0) = Trap (/(£), e, a)  U Tri (Z, 0)_{9}
leS
where</ denotes the (possibly empty) set of disjoint nonempty open intervals whose union is J(K)—K.
We state without proof the following readily verifiable facts ((VI) through (XVIII)).
(VI) s(x_{9} e_{9} 0) is an open subset of H.
(VII) Cl [s(x, s, 0)] n y={x}.
(VIII) If e<e and 0'<0, then Cl [s(x, e, 0)] n H^s(x_{9} e_{9} 0').
(IX) If x/j and c, 0 are given, then there exists 8>0 such that, for every 77 8, j(x, e, 0) and s(y_{9} t?, 0) are disjoint.
(X) B(K_{9} e_{9} a, 0) is an open subset of H.
(XI) If K_{±} and K_{2} split, then, for any e_{2}, a, and 0_{9} B(K_{19} s_{19} a, £) and Z?(/C_{2}, ^{e}2, 0) are disjoint.
(XII) If K+ and K_{2} are disjoint compact subsets of X_{9} and if e_{9} a, 0 are given, then there exists 8 > 0 such that for every 77 8, B(K_{19} e_{9} a, 0) and B(K_{29} 77, a, 0) are disjoint.
(XIII) Cl [B(K_{9} e, a, 0)] n X^K.
(XIV) Suppose that K^K, c>£_{1}>0, and 0<j3<^_{1}<a_{1}<a<7r/2. Then Cl [B(^, 8_{19} a_{19} 0,)] n H^B(K_{9} 8, a, 0).
(XV) Suppose that a<0<fyr and x^J(X)*. Then, for any e_{9} e_{19} and 0_{9} B(K, e, a, 0) and s(x_{9} e_{19} 0) are disjoint.
(XVI) Suppose that x$K and that e, a, 0_{9} 0 are given. Then there exists 8>0 such that for every 8, s(x_{9} 77, 0) and B(K_{9} e_{9} a, 0) are disjoint.
(XVII) Suppose that x$ K and that 8, a, 0_{9} 0 are given. Then there exists s > 0 such that for every s(x_{9} 8, 0) and B(K, ^_{9} a, 0) are disjoint.
(XVIII) Suppose that xe Kc\J(K)* and O<0<a<0<7r. Let e be given. Then there exists 8>0 such that for every 77 8, Cl [s(x, rj_{9} 0)] n B(K_{9} e_{9} a, 0).
(XIX) If & is a special family, let «^^{2} be the set of all members of & that have two or more points, and let E(^) be the set of all end points of intervals J(F)_{9} where F e & and f / 0.
Suppose that O</3<a<0<w, and that SF is a special family. By a pair of special a, fl, 0 functions for I mean a pair («, 8), where e and 8 are positive realvalued functions, the domain of e is E(&), the domain of 8 is &^{2}, and
(i) for each > 0, there exist at most finitely many F e &^{2} such that 8(F) q;
(ii) for each q > 0, there exist at most finitely many x e E(&) such that e(x) t];
(iii) if x, x' e E(^) and x#x', then s(x, «(x), 0) and s(x', «(x'), 0) are disjoint;
(iv) if F, Ke&^{2} and F*K, then B(F, 8(F), a, fl) and B(K, 8(K), a, fl) are disjoint;
(v) if x e E(&) and FeF^{2}, then s(x, e(x), 0) and B(F, 8(F), a, fl) are disjoint.
(XX) Let & be a special family and suppose that O</3<a<0<?r. Then there exists a pair of special a, fl, 0 functions for
Though a formal proof of this statement is lengthy, it requires no originality, so we omit the details. The idea is to arrange the members of & in a finite or infinite sequence F_{x}, F_{2}, F_{s},..., and then define e and 8 inductively. One makes use of statements (IX), (XI), (XII), (XV), (XVI), (XVII).
(XXI) Let & be a special family, and suppose O</?<a<0<£n. Let («, 8) be a pair of special a, fl, 0 functions for If e_{1(} 8j are two realvalued functions having domains E(&) and &^{2} repectively, and if
0 < ej(x) g «(x) for all x e E(&’),
0 < 8_{X}(F) 8(F) for all Fe&^{2},
then («!, 8J is a pair of special a, fl, 0 functions for 3F
The proof of this statement is trivial.
(XXII) We now proceed to the proof of statement (A) itself. Let A be our given F_{06} set. By (IV), we can choose a sequence of special families such that A = (1“_{=}i (U ^n), and for each Ke _{+ 1} there exists Fe with K^F.
Let {j8_{n}}“=i be a strictly increasing sequence in (0, jw) coverging to ?r.
Let {a_{n})„=i be a strictly decreasing sequence in (in, Jw) converging to in.
Let {^_{n}}n=i be a strictly increasing sequence in (in, jw) converging to
Let E_{n}=E(^_{n}).
Let (e(l, •), 8(1, •)) be any pair of special aj, fl_{lf} 0_{Y} functions for
Now suppose that for each k^n we have chosen a pair of special «k, Pk, Qk functions (e(k, •), 8(k, •)) for in such a way that
(i) whenever 1 ^k^n — 1, x e E_{k+1}, FeF_{k}, and x e Fc\ J(F)*, then
Cl [s(x, e(k+1, x), 0_{k+}i)] n H £ B(F, 8(k, F), a_{k}, fl_{k});
(ii) whenever l^k^n1, xe E_{k+1}, and x e E_{k}, then
Cl [s(x, e(k+1, x), fl_{k+1})] H £ s(x, e(k, x), 0_{k});
(iii) whenever KKn1, Xe («^+i)^{2}, Fe (^)^{2}, and K^F, then
CI [B(K, 8(k+l, K), a_{k + 1}, fl_{k + 1})] B(F, 8(k, F), a_{k}, fl_{k}).
Then we construct (s(n+l, •), 8(«+l, •)) as follows. Let (e, 8) be any pair of special a_{n+1}, £_{n+1}, 0_{n+1} functions for «^_{+1}. If x e E_{n+1} — E_{n9} then for some unique Fe(^,)^{}[6], xe F C\J(F)*. By (XVIII), we can choose f(x)>0 so that ??^f(x) implies
Cl [s(x, rj_{9} e_{n+1})] B(F, 8(n, F), a_{n9}
We set e(n+1, x) = min {c(x), f(x)}. On the other hand, if x e E_{n+1} n E_{n9} then we set e(n+1, x) = min {e(x), ^e(n_{9} x)}.
If Ke («^_{+}i)^{2}, then there exists a unique Fe(^_{n})^{2} with K<^F. Set
8(«+1, K) = min {8(F), 8(n, F)}.
By (XXI), (e(n+1, •), 8(n+1, •)) is a pair of special a_{n + 1}, £_{n + 1}, 0_{n + 1} functions for &n _{+ 1}, and, by (VIII) and (XIV), conditions (i), (ii), (iii) are still satisfied when n is replaced by « + l. Thus we can inductively construct a pair (e(n_{9} •), 8(n, •)) of special a_{n}, ft_{n9} 0_{n} functions for in such a way that (i), (ii), and (iii) are satisfied for every n.
Let
U_{n} = r u s{x_{9} 8{n_{9} x), 0_{n})i U r U b(f_{9} 8(_{w}, f), a_{n}, p_{n})]. L^{xe£}n J LFe^n)^{2} J
Then U_{n} is open. For fixed n_{9} all the various sets s(x_{9} e(n_{9} x), 0_{n}) (x e F_{n}) and B(F, 8(n, F), a_{n}, 0_{n}) (Fe(^)^{2}) are open and pairwise disjoint, so that every component of U_{n} is contained in one of the sets s(x_{9} e(n_{9} x), 0_{n}) (x e F_{n}) or B(F, 8(n_{9}F)_{9}a_{n9}p_{n}) (Fe(^_{n})^{2}). It therefore follows from (VII) and (XIII) that if W is any component of U_{n9} then
(1)
From conditions (i) and (ii) in the definition of a pair of special a, £, 0 functions, it follows that
Un n H = r U Cl [s(x, e(n_{9} x), 0_{n})] n 771 u [ J Cl [B(F, 8(w, F), a_{n}, &)] n 771.
l*G^n J Lre(^_{n})2 J
Consequently, conditions (i), (ii), (iii) in our inductive construction of (e(n_{9} •), 8(«, •)) (together with the fact that x e F_{n+1} — E_{n} implies x eF J(F)* for some Fe («^)^{2}) imply that U_{n+1} n U_{n} for every n.
By Urysohn’s Lemma there exists a continuous function g_{n}: H > [0, 1] such that g_{n}(z) = 1 for z e H— U_{n} and g_{n}(z)=0 for z e l/_{n+1} n 77. Let
g& = 2 n=l
Then 0^g(z)^ 1, and the series converges uniformly, so g is continuous in H.
If z e 77 U_{n9} then z e 77 U_{m} for every m n_{9} so that 1 =g_{n}(z)=g_{n +} i(z)=g_{n+2}(z) = • • •, and hence
Also, if z e t/_{B+1}, then z e U_{u} U_{2}, ■ ■U_{n+1}, so that 0=gi(z)=g_{2}(z) = • • • =g_{n}(z), and
(3) g(z) 2 ®^{m} = ®^{n} (zel/_{n+x}).
m = n + l
Let x_{0} e A be given. We must show that g(z) > 0 as z approaches x_{0} through s(x_{0}, L i^{77}") Take any natural number n. Since x_{0} g (J^_{+1}, it follows that either x_{0}GF_{n+1} or else x_{0} g FnJ(F)* for some Fe(J_{n+1})^{2}. In the first case, set rj = e(n +1, x_{0}). In the second case, (XVIII) shows that we can choose rj>0 small enough so that
s(x_{0}, 7], >77) c B(F, 8(h+ 1, F), a_{n +} i, _{+}
Suppose <x, y>e s(x_{0}, 1, far) and y<rj. Then, in the first case,
<X, y) 6 s(x_{0}, 7), >77) Q s(x_{0}, e(n+ 1, X_{o}), 0_{n +} l) ^_{n + 1},
and, in the second case,
<x, y) g s(x_{0}, 7J, <77) C B(F, 8(n +1, F), a_{n+1}, p_{n+1}) <= U_{n+1}.
Thus, referring to (3), we see that g(x, y) ()^{n} whenever <x, y) e 5(x_{0}, 1, $77) and y <t]. Therefore g(z) > 0 as z approaches x_{0} through s(x_{0}, 1, $77).
Let Xi be a point of X, and assume there exists an arc y at x_{Y} such that g(z) > 0 as z approaches x_{±} along y. Then y has a subarc y with one end point at Xi such that By (2), /{x^^^. Therefore, by (1), _{X1} e
U «^. Since n is arbitrary,
*1 6 A (U ^n) = A. n = 1
Thus, by restricting g to Q we obtain the desired function.
(B) Let A be a subset of X of type F_{ad}, and let be a bounded complexvalued function of honorary Baire class ^2(A, F^{2}). Then there exists a bounded continuous complexvalued function h defined in Q such that, for each x e A, there exists an arc y at x with y—{x}^s(x, 1, $77) and
lim h(z) = <£(x).
(I) Let Z be a bounded open interval in R, and let/: 7> 7? be a bounded, strictly increasing function. Then there exists a continuous, weakly increasing function /: R^ R such that f(f(x))=x for every x e I. (This result is probably not new, but I do not know of a reference for it, so I am obliged to prove it here.)
Proof. Let Z=f(I), let c=inf Z, and let <7=supZ. Observe that Zs(c, d), and that /^{1}: Z> I is strictly increasing. I assert that for each x e (c, d)
(4) sup/^{} \(c, x] n Z) = sup/^{} \(c, x) n Z).
If x £ Z, the equation is trivial. Suppose x e Z. Then
c < y < /^{1}(x) => (f(y) < x and f(y) eZ),
so that (c,/^{1}(x))£/^{1}((c, x) Ci Z). Hence
sup/^{1}((c, x)r\Z) f~\x) = sup/^{1}((c, x] n Z).
The opposite inequality is trivial, so (4) is established.
I also assert that for each x e (c, d)
(5) inf/^{X}((x, d) n Z) = sup/ ‘((c, x] n Z).
Obviously,
inf/'XCx, d) n Z) sup/^{x}((c, x] n Z).
Take any y>sup/^{1}((c, x] nZ). If /(y)^x, then f(y) e (c, x] n Z, and so y^{e}f~K{c, x] H Z), a contradiction. Thus f(y)>x and f(y) e (x, d) n Z. Therefore y 6/^{1}((x, d) r> Z), so that inf/^{1}((x, d) c\Z)^y. In view of the choice of y, this implies that
inf/’^x, d) O Z) sup/^{x}((c, x] n Z),
and (5) is established.
Define/* on (c, d) by
/♦(x) = sup/^{1}((c, x] n Z) (x e (c, d)).
It is clear that /* is weakly increasing and that f*(f(x))=x for every xeZ. The continuity of/♦ can easily be deduced from the equations
sup/((c, x)) = /(x), inf/*((x, d)) = /♦(x),
which are established as follows:
sup/*((c, x)) = sup sup/^{1}((c, y] n Z) c<y<x
= supx) Cl Z)
= sup/^{1}((c, x] n Z)
= /*(x),
inf/*((x, d)) = inf sup/^{1}((c, y] n Z) x<y<d
= inf inf/^{x}((y, J) nZ) x<y<d
= sup/^{1}((c, x] n Z) = /*(x).
We now extend/* to all of R by setting
/*(x) = infd)) if x c,
/(x) = sup/((c,</)) if x >d, and we are finished.
(II)Suppose that M is a metric space and that u: M >R is a function having the following property. For every sequence {p_{n}} of points of Af, every pe M, and every y e R u {oo, +oo}, if p_{n}>P and u(p_{n}) > y as n > oo, then y g R and u(p)=y. Under this hypothesis, u is continuous.
Proof. Let {p_{n}} be any sequence of points in M converging to a point pE M. We have only to show that u(p^ > u(p). But suppose u(p_{n}) t/(p). Then there exists a subsequence {u(/?_{n(fc)})} and there exists y e R\J {—as, +oo} such that y± ^{u}(p) and w(/*n(k)) y as k > oo. Since p_{n(fc)} >p as k > oo, this contradicts our hypothesis.
(III)Let A^(1,1) be of type F_{ad9} and let be a complexvalued function of Baire class 1(?1, R^{2}). Then there exists a sequence {g_{n}} of continuous functions, each mapping R into R^{2}_{9} such that for each x e A, g_{n}(x) > i/j(x) as n > oo.
Proof. This can be proved in a more general context, as shown in [5]. For a quick proof of the special case stated above, we can refer to a theorem of Bagemihl and McMillan .[1, Theorem 2], which tells us that there exist continuous realvalued functions and f_{2} defined in H such that, for each x e A_{9} has angular limit Re (0(x)) at x and f_{2} has angular limit Im (^(x)) at x. For each x e R_{9} set
gn(x) = fdx, + if_{2}(x, iy
\ n) \ n/
(IV)Now we proceed to the proof of statement (B). Let 0 be a function of Baire class 1(A_{9} K^{2}) and let £ be a (possibly empty) countable subset of A such that <£(x) = ^(x) for each x e A —E. Let N be an infinite countable set with E^N^X. Let w be a realvalued function defined on N such that w(s) > 0 for each se N and
2 M<s) < 2^{1/2} —1. sgN
For each xe X=(— 1, 1), let N(x)={seN: — l<s<x}. Define f on (—1,1) by setting
f(x) = x+ 2 HO
seN(x)
Then/is a bounded, strictly increasing function on (— 1, 1), and /(x)—x < 2^{1/2}—1. By (I), there exists a continuous, weakly increasing function f*: R+ R such that f*(f(^{x}))=^{x} for each x e ( 1, 1).
Let
H_{o} = {<x,y>_{6j}R^{2}:O<y
For fixed <x, y) e H_{o},
_{u}__{f}Jx(ly)u\
\ y /
is a strictly increasing continuous function of u that approaches +oo as u > +oo and — oo as u > — oo. Consequently there exists precisely one number u(x, y) that satisfies the equation
(6) <&.y)  o.
I assert that u(x, y) is a continuous function on H_{o}. We show this by using (II). Suppose that <x, y>e H_{o}, u_{0} e E u {oo, +oo}, {<x_{n}, y_{n}>}£/f_{0}, On, y_{n}> > <x, y>, and u(x_{n}, y_{n}) > w_{0} If u_{0} = +°o, then, as n >oo,
x_{n}(ly_{n})u(x_{n}, y_{n}) _{; x} y_{n} '
and so
U<X_{n}, A)/*(^{Xn}~^{(1}~^^{XXn}’^{K)}) +«>,
\ zn /
which contradicts (6). So i/_{0}^ +°o, and a similar argument shows that m_{o}0 — oo. Thus, by (6),
0= lim [u(x_{n}, K)/*(^{Xn}~° ~^{y}”^{)u(x}” n>co L \ yn /J

Consequently u_{0} = u(x, y). By (II), u is continuous.
From (III), there exists a sequence {g_{n}} of continuous complexvalued functions defined on R such that g_{n}(x) > 0(x) as n > oo for each x e A. For n 2, define
h_{0}(x, y) = (yn(n +1) n)g_{n}(u(x, y)) + ((n +1)yn(n + l))g_{n+}j(u(x, y))
when 1/(72 41) ^y 1 /«. Then h_{0} is continuous on H_{o}. Let {s_{n}}n=i be all the elements of N, where w/zn implies s_{n}^s_{m}. Let
r_{n} = inf /(x), X>S_{n} In = SUP /(X) = f(s_{n}), x < s_{n}
Zn = if s_{n}eE,
z„ = 0 if s_{n} $ E.
Notice that r_{n}—/_{n}>0. If x and y are real numbers, define xvy=max{x, y} and x Ay=min {x, y}. For <x, y>6 H_{o}, set
A„(x,y) = [(1 ny) v O]I(1
r_{n}+l_{n}2s_{n}+2
v o z_{n}.
L \ 'n h
Then A_{n} is continuous in H_{o}. Observe that A_{n}(x, y)=0 when y^l/n. Using this fact, it is easy to show that, if we set
oo
hl = h_{0}+ 2 △n, n = 1
then is defined and continuous on H_{o}.
Let p be any point of A. The line (7) passes through <p, 0>, and, since /(p) — p <2^{1/2}— 1 =ctn fw, the part of this line which lies in H_{o} is contained in s(p, l.fw). We show that h_{Y} approaches along this line. By substituting (f(p)—p)y+p for x in the expression for A_{n}(,v, y), one obtains
A_{n}(x,y) = [(1  ny) V 0]
(8) 17
^{r}n + 4 + 2(
?n
If P^s_{n}, then f(p)^l_{n}, and one can verify directly that (8) vanishes. Ifp>s_{n}, then f(p) r_{n}, and again one can verify directly that (8) vanishes. Thus A_{n}(x, y) vanishes along that part of the line (7) which lies in H.
Solving (7) for f(p), we find that, along the given line,
/(p) = (x(ly)p)ly, and hence
p = /(/■(?)) = /((x(i y)p)ly) .
Therefore (if 0<y^) p=u(x, y). Hence, if <x, y>satisfies (7), n^2, and l/(n+l) 1/n, then
h_{0}(x, y) = (yn(n +1)  n)g„(p) + ((»+!)yn(n + l))g_{n+}j(p),
so that A_{0}(x, y) lies on the line segment joining g_{n}(p) to g_{n}+i(p) It follows that /z_{0}(x, y) approaches 0(p) as <x, y>approaches p along the line (7). Since each A_{n} vanishes on the part of this line lying in H, h_{r}(x, y) also approaches ^(p) along this line.
Let s_{m} be any point of E. The definition of f shows that
/(x)x 2 ^{w}(^{5})
seN
for all x, and from this it easily follows that
KJml 2 l^{/m}“^{J}ml = 2
sgN sgN
Hence
So the part of the line
(9)
that lies in H_{o} is contained in s(s_{m}, 1, fir). We show that approaches ^(s_{m}) as z > s_{m} along this line. Substituting the value of x given by (9) into the expression for A_{n}, we obtain
A_{n}(x,y) = [(1ny) V 0]
(10)
■n
If s_{m}<s_{B}, then /_{m}<r_{m}£/_{n}<r_{n}, and one can verify that (10) vanishes. If s_{n}<s_{m}, then l_{n}<r_{n}£l_{m}<r_{m}, and again one can verify that (10) vanishes. Thus, for n/m, A_{n}(x, y)=0 when <x, y>lies on the line (9) and in H.
If we take n=m in (10), we obtain
A_{m}(x,y) = [(1wy) v 0]z_{m}.
Therefore A_{m}(x,y) approaches z_{m}=^(s_{m})^(s_{m}) along the given line.
Take any <x, y>e H_{o} satisfying (9), and take any a and b satisfying
(11) a < s_{m} < b.
Then f(a) ^l_{m}< (r_{m}+/_{m}) < r_{m} £f(b), so that
(f(a)s_{m})y+s_{m} < x < (f(b)—s_{m})y+s_{m};
from which we deduce that
/(a) < (x(l y)s_{m})ly < f(b).
Since f* is weakly increasing,
a = £f*((x(ly)_{Sm})ly) £f*(f(b» = b.
Because a and b were taken to be any two numbers satisfying (11), we conclude that
s_{m} = f*(.(x(ly)s_{m})/y),
whence it follows that u(x,y)=s_{m}. Thus
h_{0}(x, y) = (yn(n+1)  ri)g_{n}(s_{m})+((«+!)  yn(n + l))g„_{+}i(s_{m})
when l/(n+l)^y^ 1/n. Consequently h_{0}(x,y) approaches ^(s_{m}) along the line (9); so Ai(x, y) approaches along the given line.
We have shown that, for each x e A, there exists a line segment at x, lying in s(x, such that A_{1}(z)^<^(x) as z>x along the line segment. We do not know that h_{Y} is bounded, but this is easily patched up. Choose a real number B such that, for all x e A,
—B < Re^(x) < B, —B < Im<£(x) < B, and set
h(z) = ([(ReA_{1}(z)) v (B)] A B)+i([(Imft_{1}(z)) v (B)] A B).
If we extend h to a bounded continuous function defined in H, and then restrict h to Q, we have the desired function.
(C) Let d{t) be a weakly increasing, positive, realvalued function defined for 0< t 1. Then there exists a continuous, complexvalued function k defined in Q, with &(z) ^2^{1/2} for all z e Q, such that for each a e (0, 1] and for each arc
y £ { <x, y>: — 1 x 1, 0 < y g a},
{diameter y) 2: d(a) implies {diameter k{y)) 2.
Proof. Let p{x)=%d{t) dt (0<x^l). Then p is positive, continuous, and strictly increasing, and p{x)^\d{x). Let ae(0, 1] be given. Since p{xY^{r} is uniformly continuous on each compact subset of (0,1], there exists ee(0,1] such that
(a Xi 1 and x_{x}x_{2} < e)
implies
IX*1)~^{1}~P(*2)~^{1} 1
Let e{a) be the supremum of all such e. Then e{a) is an increasing function of a, and
Qu x_{x} S 1 and x_{x}x_{2} < e(a))
implies
Set ?(x)=fo e(t) dt. Then q is positive, continuous, and strictly increasing, and tf(x) = «(*)• Let m{x)=min {p(x), q{x)}. For <x, y>e Q, define
kfy) = sin {2n/ym{y)), k_{2}{x, y) = sin {4nxlp{y)),
k{x,y) = i(y) + i_{2}(x,y).
Now suppose that a e (0, 1] is given, and suppose that y£{ <x,y>: l^x^l, 0<y^a} is an arc with (diameter y)^d(a). Choose z_{1} = <x_{1}, y_{x}> and z_{2}=<x_{2}, y_{2}> in y so that lz_{x}—z_{2} S d{a). Assume without loss of generality that y_{2}^yi. We can choose a' so that 0<a'^y_{1}^a'^a. Since m{a’)^ d{a')^$ d{a), and since \z_{1}—z_{2}\^d{a), we must have either
(12)  yiy_{2} m{a')
or
(13) biJal < m{a’) and x!x_{2}  d{a').
First assume that (12) holds. Here m{y_{2})^m{y_{1})^.m{a'), so
2?r/y_{1}m(y_{1}) 2jrly_{2}m{y_{2}),
and we have 2tt 2tt 27r(j_{x}ffl(^i)y_{2}m(y_{2})) yzm(y_{2}) ypntyj > 2n(y_{1}m(y_{2})y_{2}m(y_{2})) _{=} 2n(y_{1}y_{2}) ~ yiy2m(y_{1})m(y_{2}) yxy_{2}w(yx) _{>} 27r^(fl^{/}) _{>} 2tt _{>}
^{= </sup> J1J2 <sup>}
Thus, as <x, y) moves along y from <x_{x}, y^ to <x_{2}> J2X ^{we} see that 2irlym(y) varies over an interval of length at least 2tt, and hence k_{Y}{y) varies over the whole of the interval [—1,1]. Therefore (diameter k(y))^2.
Now assume that (13) holds. Then
4ttX! 4ttX_{2} p(yi)~ p(y 2)
4t7
Xx *2 p(yi) p(yi)
x_{2} x_{2}
P&2) p{yi)
rixix_{2}i__________________________________ 1_____ 1 1
L /»Oi) p(yz) p(yi) J
[i<)_ 11 1
Lx«) Xh) p(yi) J
S 477 11 
1 1
p(y2) p(yi)
Now, [y_{1}y_{2}l<m(a')^q(a')^e(a’), so [p(y_{2})~^{1}p(yi)~^{1}[^j Therefore 47rx_{1}/p(y_{1})4irx_{2}/p(y_{2}) 2tt, and we see that as <x, y>varies along y from <x_{x}, y_{x}> to <x_{2>} y_{2}>, the quantity 4irx/p(y) varies over an interval of length at least 2tt, so that k_{2}(x, y) takes on every value in the interval [—1, 1], Thus (diameter k(y))^2.
(D) Let A^X be a set of type F_{a{}, and let</> be a bounded function of honorary Baire class 2(A, R^{2}). Then there exists a bounded continuous complexvalued function f defined in Q such that A is the set of curvilinear convergence of f and</> is a boundary function for f.
Proof. Let g be the function of (A) and let h be the function of (B). For t e (0, 1], let
d_{t}(t) = sup {8 e (0,1] : t,y_{2}Z t, <x_{15} y_{x}> e Q, <x_{2}, y_{2}> e Q, and
<xi,yi><x_{2},y_{2}>< 8) implies A(x_{1},y_{1})/i(x_{2},y_{2}) t},
d_{2}(t) = sup {8 e (0,1] : (jx t, y_{2} t, <Xx, jx>e Q, <x_{2}, y_{2}> e Q, and
<Xx, yx><x_{2}, y_{2}> < 8) implies g(x_{1;} y_{1})g(x_{2}, y_{2}) S t}, d(t) = min <4(10,10
Let k be the function of (C) for this d(t), and set f(z)=h(z)+g(z)k(z) (z e Q). We show that f is the desired function.
Suppose x e A. Then there exists an arc y at x, lying in s(x, 1, such that h approaches</> (x) along y. But g(z) approaches 0 through s(x, 1, $77) and k is bounded, so g(z)k(z) approaches 0 along y. Hence f(z) approaches</> (x) along y. Thus is a boundary function for /, and A is a subset of the set of curvilinear convergence off. It only remains to show that if x is a point of the set of curvilinear convergence of/, then x e A. To show this, let y be an arc at x along which f approaches a limit. We may assume without loss of generality that y has an end point in { <x, 1 >: — 1 ^x^ 1}. By the properties of g, it will be enough to show that g approaches zero along y. Assume that g does not approach zero along y. Then there exists e e (0, 1] and there exists a sequence {z_{n}} such that z_{n}ey — {x}, z_{n}+x as «>oo, and g(z_{n}) e for all n. Write z_{n} = (x_{n}, y_{n}). Choose N so that n^N implies y_{n}<
For the time being, let n be a fixed integer greater than or equal to N. Set a = 4y_{n}/3. Let y be the component of y n Cl [S(J(tz), z_{n})] that contains z_{n}. (Recall that S(d(a), z_{n})={z : \zz_{n}\ <d(a)}.) Then
d(a) diameter / 2 d(a),
and, since d(a) la,
7 £ {<x, y) : la g y £ a}.
By the choice of k, there exist points p and q in y with  £(/»)—&(<7) S2. We have lpq{ £2 d(a)< dfila), so, by the definition of J/t),
A(p)A(?) la < e.
Similarly,
\g(p)g(Zn)\ la < e,
g(?)S(O la <
Thus
I/(P)/(9) \g(p)k(p)~g(.q)k(q)\\h(p)h(q)\
Z 2e2^{ll2}le2^{ll2}lel_{e} > e.
Note that \pz_{n}\^d(a)^la=ly_{n}, and similarly ^z_{n}^y_{n}.
We have now shown that, for each n^N, there exist points p_{n}, q_{n}ey with Ia>~z„ ily_{n}, knZnl Hy_{n}, and /(p_{n})>«■ But then p_{n}>x and q_{n}>x as n > oo, so f does not approach a limit along y. This is a contradiction. We conclude that g(z) > 0 along y, and hence that x e A.
(E)Let A^C be a set of type F_{ad}, and let <f>be a bounded function of honorary Baire class ^2(A, B^{2}). Then there exists a bounded continuous complexvalued function f defined in D such that A is the set of curvilinear convergence of f and</> is a boundary function for f
Proof. If A = 0, this is trivial. If A / 0, then we can assume, by making a suitable rotation of the disk, that <1, 0>g A. Let G = DS(&, <, 0» and let L = C —{<1,0>}. Because Q u X is homeomorphic with G u £, we see from (D) that there exists a bounded continuous complexvalued function f_{Y} defined in G such that
(i) A n L is the set of all points x e L such that/i approaches a limit along some arc at x, and
(ii) the restriction of</> to L is a boundary function for f_{r}.
Since G is closed relative to Z>, we can extend f_{±} to a bounded continuous function f defined in D in such a way that /has <£«1, 0» as a radial limit at <1, 0>. This f will have all the desired properties.
(F) Let S^{2} denote the Riemann sphere, let A^C be a set of type F_{o6}, and let</> be a function of honorary Baire class ^2(A, S^{2}). Then there exists a continuous function f: D+ S^{2} such that A is the set of curvilinear convergence of f and</> is a boundary function for f
(1)We suppose that
S'^{2} = {<x, y, z) g K^{3} : x^{2}+y^{2} + z^{2} = 1}. We let
U = {<x, y, z)eS^{2} : < ^{2 1}
V = <x, y, z) e S^{2} : 1 z <
Zu = y, z> e s^{2} : < z £ 1,
Z_{v} =  <x, y, z~>e S^{2} : 1 z < jkj"
We define mappings 0^: Z_{v} > U and <J>_{y}: Z_{v} > V by setting
<Mx, y, z) = <x(4z^{2} 1), y(4z^{2}  1), z(4z^{2}  3)> «x, y, z) g Zu)
and
<D_{y}(x, y, z) = <x(4z^{2} 1), y(4z^{2} — 1), z(4z^{2}  3)> «x, y, z) g Z_{y}).
Then 0^ is a onetoone continuous function from Zu onto U. Since Z_{a} and U are each homeomorphic to the unit disk D, it follows from [7, Corollary 1, p. 122] that <>[/ is a homeomorphism of Z_{v} onto U. Similarly, O_{y} is a homeomorphism of Z_{y} onto V,
We define a continuous function O: S^{2} > S^{2} by setting
O(x, y, z) = <D_{u}(x, y, z), < z 1,
0>(x, y, z) = <x, y, z}, z
<D(x, y, z) = <D_{v}(x, y, z), 1 z <
Notice that for each p e S^{2}, the inverse image set ^{<}h~^{1}({p}) contains at most three points.
(II) Most of the results of Hausdorff [2] on realvalued Baire functions can easily be shown to hold also for functions taking values in A". We shall make free use of these results in this more general form.
(III)Now we proceed to the proof of (F). Let N be a countable subset of A such that the restriction of 0 to A—N is of Baire class \(A — N, S^{2}), and let A=^ —N. It will be convenient to let F_{a}(Ai) denote the class of all subsets of A_{t} that are of type F„ relative to A_{lt} and G_{t}{A^) the class of all subsets of A_{x} that are of type G_{6} relative to A. Since U and V are open subsets of S^{2} and U u V=S^{2}, we see that A_{±} n j^U) e F^A^, A,/^{1}^) e G^AJ, and A_{1}/~^{1}(F)^A_{1} n 0^{1}(G). An elegant theorem of Sierpinski [8] now enables us to choose a set X e F„(A_{1}) n G_{d}(A) such that
A0^{1}(F) S K £ Ai n ^(U).
Let L = AiK. Then L e F^AJ n G_{f}(Ai). Moreover, </>(K)q U and 0(L)£ V. Let/>! = <!, 0,0>, and define 0: A >S^{2} — (pj as follows. Set
^(x) = <D^(0(x)), ^(x) =
xeK, xeL.
If xeN, we let 0(x) be any element of Z_{V}KJZ_{V} for which d>(0(x))=0(x). This choice of 0(x) is always possible, because d>(Z_{y} u Z_{y})2 U'J V=S^{2}. Let </>o be the restriction of 0 to A^KkjL. I assert that 0_{O} is of Baire class
1(A» 5^{2}{p_{1}}). Since S^{2}  {pi} is homeomorphic to R^{2}, it will suffice to show that 0_{O}~ \G) e F^Ai) for every open set G£S'^{2}{p_{1}}. But
0o \G) = A n i/r\G) = [Kn^G)]u[Ln ^>~\G)]
= [Kn fWulZu n G))] u [L n <f>\^Z_{v} c\ G))] e F^),
so 0o is of Baire class 1(A, S^{2}—{pj). Now, A_{±}A —N</em> is of type <em>G<sub>6</sub></em> relative to <em>A,</em> so (again using the fact that <em>S<sup>2</sup>—</em>{pj is homeomorphic to <em>R<sup>2</sup>)</em> we can extend 0o to a function 0<sub>L</sub> of Baire class <em>^UA, S<sup>2</sup>—</em>{pj). The existence of 0! shows that 0 is of honorary Baire class <em>£2(A, S<sup>2</sup>—</em>{pj). The range of 0 is contained in <em>Z<sub>v</sub> KJ Z<sub>v</sub>,</em> so that the values of 0 are bounded away from p<sub>x</sub>. Thus, if we still think of <em>S<sup>2</sup> </em> {pi} as corresponding to the plane <em>R<sup>2</sup>,</em>0 corresponds to a bounded function. By (E), there exists a continuous function <em>f<sub>1</sub>: D</em> 5<sup>2</sup>{pj} such that the values of are bounded away from p<sub>n</sub> <em>A</em> is the set of curvilinear convergence of /<sub>1(</sub> and 0 is a boundary function for <em>f<sub>r</sub>.</em> Let <em>f</em> denote the composite function o Then <em>f</em> is continuous and <!>° 0=0 is a boundary function for <em>f.</em> It only remains to show that if x is a point of the set of curvilinear convergence of <em>f,</em> then x e <em>A.</em> Let y be an arc at x along which <em>f</em> approaches a limit, and let <em>C(f<sub>lt</sub></em> y) denote the cluster set of/i along y. Assume that x 0 <em>A.</em> Then does not approach a limit along y, so C(/i, y) contains infinitely many points. Now, O maps at most three points to any one given point, so O(C(/<sub>X</sub>, y)) contains infinitely many points. But O(C(/i, y)) is the cluster set of/
<I>° along y, and hence f does not approach a limit along y, contrary to our assumption. We conclude that x e A after all. This completes the proof of the theorem.
The following questions remain open.
Problem 1. If A is an arbitrary set of type F_{a6} in C, does there necessarily exist a continuous realvalued function in D having A as its set of curvilinear convergence!
Problem 2. If A^C is a set of type F_{o6}, and if</> is a function of honorary Baire class ^2(A, R), does there necessarily exist a continuous realvalued function in D having A as its set of curvilinear convergence and</> as a boundary function ?
Appendix. Some theorems concerning functions of Baire class 1 which take values on the Riemann sphere can be obtained by the technique used to prove (F). We use the notation set up in the proof of (F).
Theorem (a). Let M be a metric space, and let</> : M > S^{2} be a function such that ^(G) is an F_{a} set for every open set G^S^{2}. Then (f> is of Baire class l(Af, S^{2}).
Proof. Since U and V are open and U u F=S^{2}, it follows that the set
is F_{a}, the set M—is G_{6}, and By the theorem of
Sierpinski [8], there exists a set K that is simultaneously F_{a} and G_{d} such that
Let L = M—K. Then L is simultaneously F_{a} and G_{6}, and
W U, c: V.
Define 0: M S^{2}—{pj (where ^ = <1,0, 0» by setting
^(x) = 0  i(^(x)), x g K, 0(x) = <D_{y} ^(x)), x g L.
If G is an open subset of S^{2} — {/h}, then
= [KhWUZij G))l u [£ n n G))],
so 0^{1}(G) is an F_{ff} set. Since S^{2}—{p_{1}} is homeomorphic to the plane, it follows that there exists a sequence {ip_{n}} of continuous functions, each mapping M into S^{2}—{p_{1}}, such that pointwise on M. But then O(0_{n}(x))0(^(x))=^(x) for each fixed x g M, so</> is of Baire class 1(M, S^{2}).
A special case of Theorem (b) was proved (in effect) in [6, proof of Theorem 6] by means of a rather messy lemma (Lemma 3). Theorem (a) provides a proof that is both more general and more esthetically satisfactory.
Theorem (b). Let M be a metric space, and let <£: M >S^{2} be a function. Then <f>is of honorary Baire class ^2(M, S^{2}) if, and only if, there exists a countable set N^M such that, for every closed set F^ S^{2},</> ~^{X}(F) — N is a G_{6} set.
Proof. The implication in one direction is trivial. Now assume that N is countable and that — N is a G_{d} set for every closed set F^S^{2}. Let</> _{0} be the restriction of</> to M—N. Since S^{2} is a subset of R^{3}, <£_{0} is of Baire class ^1(M—N, R^{3}). Because M—N is a G_{6} set,</> _{0} can be extended to a function of Baire class
1(M, A^{3}). Now, <^(x) e S^{2} except for only countably many x, so there exists some point q in the open ball enclosed by S^{2} such that q is not in the range of fa. Define a mapping P: R^{3}{q}+ S^{2} as follows. If a g R^{3}{q}, let L be the ray with end point at q which passes through a, and let P(a) be the intersection point of L with S^{2}, Then P is continuous and P(a) = a for each aeS^{2}. Let ^Po<^<sub>1</sub>. If <em>G^S<sup>2</sup></em> is open, then 0<sup>1</sup>(G) = ^r<sup>1</sup>(P<sup>1</sup>(G)), so that 0<sup>1</sup>((7) is an <em>F<sub>a</sub></em> set. Thus, by Theorem (a), <em>i/j</em> is of Baire class 1(M, <em>S<sup>2</sup>).</em> Moreover, if <em>x $ N,</em> then ^<sub>1</sub>(x)
^_{0}W = <£(x)g S^{2}, so that 0(x)P(^(x))
^(x). Therefore</> is of honorary Baire class ^2(M, S^{2}).
An alternative proof of Theorem (b) could be given by combining Theorem (a) with the following result.
Theorem (c). Let M be a metric space, E a G_{6} set in M, <f>a function of Baire class 1(£, S^{2}). Then</> can be extended to a function of Baire class l(Af, S^{2}).
To prove this, use the technique appearing in the proof of Theorem (a).
Finally, we note that a theorem proved by Bagemihl and McMillan for realvalued functions [1, Theorem 2] can be transferred to the Riemann sphere by means of our technique.
References
1. F. Bagemihl and J. E. McMillan, Characterization of the sets of angular and global convergence, and of the sets of angular and global limits, of functions in a halfplane, Fund. Math. 59 (1966), 177187.
2. F. Hausdorff, Uber halbstetige Funktionen und deren Verallgemeinerung, Math. Z. 5 (1919), 292309.
3. T. J. Kaczynski, Boundary functions for functions defined in a disk, J. Math. Meeh. 14 (1965), 589612.
4.  , On a boundary property of continuous functions, Michigan Math. J. 13 (1966),
313320.
5.  , Boundary functions, Doctoral dissertation, University of Michigan, Ann Arbor,
Mich., 1967.
6. J. E. McMillan, Boundary properties of functions continuous in a disc, Michigan Math. J. 13 (1966), 299312.
7. M. H. A. Newman, Elements of the topology of plane sets of points, 2nd ed., Cambridge Univ. Press, New York, 1964.
8. W. Sierpinski, Sur une propriete des ensembles ambigus, Fund. Math. 6 (1924), 15.
9. L. E. Snyder, Biarc boundary functions, Proc. Amer. Math. Soc. 18 (1967), 808811.
University of Michigan, Ann Arbor, Michigan
University of California, Berkeley, California
9. July 1969  Boundary functions and sets of curvilinear convergence for continuous functions
Original PDF: 9. July 1969  Boundary functions and sets of curvilinear convergence for continuous functions.pdf
Kaczynski, T.J. 1969. Boundary functions and sets of curvilinear convergence for continuous functions. Trans. Am. Math. Soc. 141:107125.
MR0243078 Kaczynski, T. J. Boundary functions and sets of curvilinear convergence for continuous functions. Trans. Amer. Math. Soc. 141 1969 107.125. (Reviewer: J. E. McMillan) 30.62
Explanation by John D. Bullough
The author completes the investigation, initiated by Bagemihl and Piranian, of boundary functions of continuous complexvalued functions defined in the open unit disk D. the set of curvilinear convergence A of such a function f is defined to be the set of those e^{iT} at which f has a finite or infinite limit along some open Jordan arc lying in the disk and having one endpoint at e^{iT}. A boundary function of f is a function t defined on A such that each t(e^{iT}) is one of these limit values. The author proved that t differs from some function of the first Baire class at at most countably many points, and McMillan proved that A is of type F(sd). By means of an intricate construction, the author proves that for any set A on the unit circle of type F(sd), and for any function t defined on A that differs from some function of the first Baire class at at most countably many points, there exists a continuous complexvalued function f defined in D having A as its set of curvilinear convergence and having t as its boundary function. The author points out the the problem remains open for realvalued functions.
Article by Ted
10. Nov 1969  The Set of Curvilinear Convergence…
Original PDF: 10. Nov 1969  The Set of Curvilinear Convergence….pdf
Kaczynski, T.J. 1969. The set of curvilinear convergence of a continuous function defined in the interior of a cube. Proc. Am. Math. Soc. 23:323327.
MR0248339 Kaczynski, T. J. The set of curvilinear convergence of a continuous function defined in the interior of a cube. Proc. Amer. Math. Soc. 23 1969 323.327. (Reviewer: J. E. McMillan) 30.62
Explanation by John D. Bullough
The set of points of the unit circle at which a continuous complexvalued function in the open unit disk has limits along curves (asymptotic values) is of type F(sd) and, in general, has no other properties. The author shows that for continuous complexvalued functions defined in a cube, this set of "curvilinear convergence" does not even need to be a Borel set. He asks whether such an example can be given for realvalued functions.
Article by Ted
THE SET OF CURVILINEAR CONVERGENCE OF A CONTINUOUS FUNCTION DEFINED IN THE INTERIOR OF A CUBE
T. J. KACZYNSKI
Let Q be an open connected set in a finitedimensional Euclidean space, and let f be a function mapping Q into another finitedimensional Euclidean space. We define the set of curvilinear convergence of f to be
{pE:boundary of Q: there exists a simple arc 7 with one endpoint at p such that 7 — {p} CQ and/(fl) converges to a finite limit as v—+p along 7}.
J. E. McMillan [6] has shown that if Q is an open disk in the plane and if f is continuous in Q, then the set of curvilinear convergence of f is of type F_{o}$. In this paper we prove that there exists a bounded continuous complexvalued function /, defined in the interior of a threedimensional cube, such that the set of curvilinear convergence of f is not a Borel set. Thus McMillan’s theorem does not generalize to three dimensions. However, the following question remains open.
Problem. Does there exist a continuous realvalued function f, defined in the interior of a threedimensional cube, such that the set of curvilinear convergence off is not a Borel set?
Let
R be the set of real numbers
R^{n} — wdimensional Euclidean space
Q= l(x,y)Gi?^{2}:0<ySland
■K={(x,y, z)Gi?^{3}:0<y^l, l^x^l.and l^z^l}
<2° = interior of Q
K° = interior of K.
Let Q again represent an open connected subset of R^{n}. If/: Q—>R^{m} is a function, we shall say that a£R^{m} is an asymptotic value of / iff there exists a continuous function v: [0,1)—such that dist(fl(/), R^{n} — Q)—>0 and/(fl(£))—>& as t—>1. (Note that a limit approached by / along a path which tends to 00 may or may not be an asymptotic value by our definition.) We say that a is a point asymptotic value of / (at p) iff v can be chosen so that, as t—*1, v(t) approaches
Received by the editors April 3, 1969.
323
a point —Q. Because of the result of [8], the set of curvilinear convergence of f is
{pE:R^{n} — has a point asymptotic value at p}.
Lemma. There exists a continuous complexvalued function s defined in
{(x,y) G R^{2}: y> 0},
with  six, y)  1 for all x and y, such that s has the following property.
Let E be the set of all asymptotic values of s that are real and lie in the interval (— 1, 1). Then E is equal to the set of all point asymptotic values of s that are real and lie in (— 1, 1), and E is not a Borel set.
Proof. Let A be an analytic subset of R that is not a Borel set. (This exists [7, p. 254].) We see from the paper of Kierst [4] that there exists a holomorphic function h defined in {z: z is a complex number and  z\ < 11 such that h omits the three values —i, i, oo and A\J { — i, i] is the set of all (finite) asymptotic values of h. The function h is then normal [5, p. 53], so, as pointed out by McMillan [6, p. 311], it follows from Theorem 1 of [2] that A\J {—i, i\ is just the set of all (finite) point asymptotic values of h. We now obtain the desired function by setting
s(x, y)
Z?((l — y)e^{ix})
(0<y^ 1),
1 +  h(fl  y)e^{ix}) 
/z(0)
(y 1).
$(x, y) =  i r
1+ h(O) 
Remark. Since the theorem we want to prove has nothing to do with meromorphic functions, it is unfortunate that the proof of the lemma depends on the theory of meromorphic functions. This can be avoided. The lemma can be proved by using [7, Theorem 113, p. 216], [1, Theorem 2, p. 179], and the methods of [3], but this involves a messy construction, so we omit the details.
Theorem. There exists a bounded continuous complexvalued function f defined in K° such that the set of curvilinear convergence off is not a Borel set.
Proof. Let 5 and E be as described in the lemma, and set g(x, y) = s(x/y, y) for (x, y)GQ The reader can verify that E equals the set of all real point asymptotic values of g at the point (0, 0) which lie in the interval ( — 1, 1). For each /G(0, 1], define
tZo(O = sup{a G (0, 1]: ((x, y) G Q, (%', y^{f}) G Q, y t, y' t, and  (x, y) — (x', y')  < 5) implies  g(x, y) — g(x', y')  t}, d(t) = min{d_{o}(JO, 1^
By statement (C) of [3], there exists a continuous complexvalued function k defined in Q, with fe(x, y)  ^2^{1/2} for all (x, y)GQ, such that for each oG(0, 1] and for each arc
7 C {(x, y): — 1 x 1 and 0 < y g a}, (diameter 7) ^d(a) implies (diameter £(7)) ^2.
Let f be the function with domain K° defined by /(x, y, 2) = (g(x, y) —2)&(x, y). We note that the following inequality holds for any three points (x, y, 2), (x', y', 2'), (x", y", 2") in K°:
I/O', y’, z') f(x",y",z")\
=  (g« y')  z')k(x', v')  (g(x, y)  z)k(x', y') + (g(x, y) ~ ^k(x', y')  (g(x, y)  z)k(x", y") + (g(x,y)  z)k(x”,y")  (,g(x'',y")  z")k(x",y")  I g(x, y)  z I I k(x', y')  k(x", y") I
(1)   k(x', y)  g« y')  z'  g(x, y) + z I
  k(x", y") 11 g(x, y) — z — g(x", y") + z" 
I g(x, y)  z 11 k(*', y'}  k(x", y”) I
 2 I g(x',y ')  g(x, y) I  2 I g(x, y)  g(x", y") 
 2  z — z'  — 2  z" — z  .
LetZ= {(0, 0, 2): — 1 <2 < 1}, and let T be the set of curvilinear
convergence of f. We wish to show that rP\L= {(0, 0, 2):2G^}.
Suppose &G£ Then there is an arc 7 with one endpoint at (0, 0) such that 7— {(0, 0)} CQ° and g approaches b along 7. Let
7' = {(*, y, b): (x, y) G 7}.
Then g(x, y)—2—>0 as (x, y, 2)—*(0, 0, b) along 7'. Thus, since k is bounded,/(x, y, 2)—>0 along 7', so (0, 0, &)GLHL.
Now let us assume, conversely, that (0, 0, &)GrO£ and deduce that &G£ Let 7' be an arc with one endpoint at (0, 0, b) such that 7'— {(0, 0, b)} QK° and / approaches a limit along 7'. Let
7 = {(x, y) G R^{2}' (x, y, 2) G 7' for some 2}.
Then 7 is a (not necessarily simple) arc with one endpoint at (0, 0) and 7— {(0, 0)} CQ°. I assert that g(x, y) —2 approaches 0 along 7'.
Assume this is false. Then there exists c > 0 and there exists a sequence of points {(x_{n}, y_{n}, z_{n})} *_{=1} in 7' — {(0, 0, b)} such that
(x_{n}, y_{n}, Zn) (0, 0, b) as n 00
and  g(x_{n}, y_{n}) —z_{n}\ for all n. Let 8>0 be chosen so that whenever (w, v, w)£y', (x, y, z)Ey', and v, y^b, then w—z\ <e. Let N be chosen so that n N implies y_{n} < min {3e/32, 33/4, 3/4}.
For the present, let n be a fixed integer greater than N. Set a = ^y_{n}/3. There exists an arc 7* contained in
7 C\ {(x, y) G R^{2}*.  (x, y) — (x„, y_{n})  d(a)\
joining (x_{n}, y_{n}) to a point on the circle of radius d(a) about (x_{n}, y_{n}). Clearly (diameter 7*) (a), so (diameter k(y*))^2. Choose points
(x» , yn), (xX, y”) in y* with \k(xj, yn)k(x’_{n}’, y") ^2. Choose Zn , z’n so that (x_{n}', y_{n}', Zn) and (x„, y„', z„) are in 7'. It is easy to check that Ja^y/ <8 and a^y"<8, so
(2)  z_{n} — Zn  < je and  z_{n}" — z_{n}  < e.
Moreover, since  (x_{n}', y_{n}') — (x_{n}, y_{n}) ^d(a) g %d^a), we have
g(Xn, <em>yn)  g(%n , yn)  </em> <  e; 
and similarly
g(.r„", yn')  g(x_{n}, y_{n}) I <  e. 
Combining these inequalities with (1) and (2), we get
 f (x_{n} J yn y Z_{n} ) f (x_{n} , y_{n} , z_{n} )  >
 g(%n, yn) Z_{n}   ^(x_{n} , y_{n} ) k(x_{n} , y_{n} )  £ = 2fi € — €.
But y_{n}', yn ^y_{n}/3, so (x_{n}', y_{n}', z„)—»(0, 0, b) and (x", y_{n}", 2/') —>(0, 0, b) as n—»oo ; hence f cannot approach a limit along 7', which is a contradiction. We conclude that g(x, y)— z—>0 as (x, y, z) ^(0, 0, b) along 7'.
It follows immediately that g(x, y)^b along 7, so b^E. We have now shown that
rnz = {(o,0,z):zE £}.
Thus TC\L is not a Borel set. Hence T is not a Borel set; for if it were, then TPiL would also be a Borel set.
References
1. F. Bagemihl and J. E. McMillan, Characterization of the sets of angular and global convergence, and of the sets of angular and global limits, of functions in a halfplane, Fund. Math. 59 (1966), 177187.
2. F. Bagemihl and W. Seidel, Koebe arcs and Fatou points of normal functions, Comment. Math. Helv. 36 (1961), 918.
3. T. J. Kaczynski, Boundary functions and sets of curvilinear convergence for continuous functions, Trans. Amer. Math. Soc. (to appear).
4. S. Kierst, Sur Vensemble des valeurs asymptotiques d'une fonction meromorphe dans le cercleunite, Fund. Math. 27 (1936), 226233.
5. O. Lehto and K. I. Virtanen, Boundary behavior and normal meromorphic functions, Acta Math. 97 (1957), 4765.
6. J. E. McMillan, Boundary properties of functions continuous in a disc, Michigan Math. J. 13 (1966), 299312.
7. W. Sierpinski, General topology, Univ, of Toronto Press, Toronto, Canada, 1952. S. R. Foguel, The ergodic theory of Markov processes, Van Nostrand, New York.
8. H. Tietze, Uber stetige Kurven, Jordansche Kurvenbogen und geschlossene Jordansche Kurven, Math. Z. 5 (1919), 284291.
University of California, Berkeley
Ted's Work from his Parents Home in Illinois
11. Problem 786
January, 1971
https://doi.org/10.2307%2F2688865
By T. J. Kaczynski, Lombard, Illinois.
Suppose we have a supply of matches of unit length. Let there be given a square sheet of cardboard, n units on a side. Let the sheet be divided by lines into n2 little squares. The problem is to place matches on the cardboard in such a way that: a) each match covers a side of one of the little squares, and b) each of the little squares has exactly two of its sides covered by matches. (Matches are not allowed to be placed on the edge of the cardboard.) For what values of n does the problem have a solution?
12. A Match Stick Problem
November–December 1971
https://doi.org/10.2307%2F2688646
Problem 786. [January, 1971] Proposed by T. J. Kaczynski, Lombard, Illinois.
Suppose we have a supply of matches of unit length. Let there be given a square sheet of cardboard, n units on a side. Let the sheet be divided by lines into n2 little squares. The problem is to place matches on the cardboard in such a way that: a) each match covers a side of one of the little squares, and b) each of the little squares has exactly two of its sides covered by matches. (Matches are not allowed to be placed on the edge of the cardboard.) For what values of n does the problem have a solution?
I. Solution by Richard A. Gibbs, Hiram Scott College, Nebraska.
A necessary and sufficient condition that a solution exist is that n be even.
Sufficiency is easy. If n = 2k, consider the cardboard as consisting of k2 2X2 squares. Simply place a match on each of the four segments adjacent to the center point of each 2X2 square.
For necessity, assume a solution exists for an nXn sheet of cardboard. To each unit square correspond the point at its center. Connect two points if their corresponding squares share a match. By the hypotheses, every point will be joined to exactly two others. Therefore, according to a basic result of Graph Theory, the resulting graph will be a collection of disjoint cycles. Each cycle will enclose a polygonal region whose sides are either horizontal or vertical line segments. Consequently, since the length of each segment is an integer, the area of each polygonal region will be an integer. By Pick's theorem (a beautiful result familiar to anyone who has played with a geoboard) the area of the 2th polygonal region is
A = %Pi + li  1
where there are Pi points on the perimeter and li points in the interior of the 2th polygonal region. Since each area is an integer, each Pi is even. As each point is on exactly one perimeter, the sum of the Pi is the total number of points, n2. Hence n is even.
II. Solution by Richard L. Breisch, Pennsylvania State University.
A generalization of the stated problem will be demonstrated. Let the cardboard be an m X n rectangle. The problem of covering the cardboard in the stated manner has a solution if and only if m and 72^2, and m and n are not both odd.
An alternative representation of the problem will be used to demonstrate this. Consider the mXn array of the center points of the little squares. If two edgeadjacent squares have a match on their mutual edge, connect the centers of these squares with a line segment. Since each little square has exactly two of its sides covered by matches, in the alternative representation, there are exactly two line segments from each point in the array. Hence each connected set of line segments forms a polygon, and the mXn array is covered by a collection of polygons. Each polygon must have an even number of horizontal segments and an even number of vertical segments. Since there are m n segments, m and n cannot both be odd integers.
Suppose m is even. Then the mXn array can be covered with m/2 rectangular polygons each of which has dimensions 1 segment by n segments. The arrangement of matches in the original representation is easily derived from this representation.
Also solved by Dan Bean, Dave Harris and E. F. Schmeichel (Jointly), College of Wooster, Ohio; Thomas A. Brown, Santa Monica, California; Melvin H. Davis, New York University; Roger Engle and Necdet Ucoluk (jointly), Clarion State College Pennsylvania; Michael Goldberg, Washington, D.C.; M. G. Greening, University of New South Wales, Australia; Heiko Harborth, Braunschweig, Germany; Herbert R. Leifer, Pittsburgh, Pennsylvania; Joseph V. Michalowicz, Catholic University of America; George A. Novacky, Jr., University of Pittsburgh; J. W. Pfaendtner, University of Michigan; Sally Ringland, Shippenville, Pennsylvania; Rina Rubenfeld, New York City Community College; E. P. Starke, Plainfield, New Jersey; and the proposer.
Ted's Work as a Montana Hermit
Never published new ground?
Ted went back to playing around with pure math equations briefly in his cabin in Montana. He even wrote the kind of math paper you would submit to a journal, but never sent it anywhere.
Here's how Ted explained the paper in relation to his other work:{2}
(Ca) FL #80, letter from me to my parents, Spring, 1964, p. 1: “It’s a good thing I didn't follow Piranian’s suggestions about how to attack the problem, or I never would have solved it!”
Piranian urged me to prove (a) that every continuous function in the disk admits a family of disjoint arcs, and to deduce from this (b) that every boundary function for a continuous function can be made into a function of the first Baire class by changing its values on at most a countable set. (The terminology is explained in F. Bagemihl and G. Piranian, “Boundary Functions for Functions Defined in a Disk,” Michigan Mathematical Journal, 8 (1961), pp. 201–207.)
I maintained that it would be much easier to prove (b) by examining inverse–image sets, and I even suggested that (b) might then be used to prove (a). And that’s how it turned out. I did prove (b) within three months or so by using inverse–image sets. The proof of (a) was vastly more difficult. I didn’t succeed in proving (a) until two decades later, and I had to use (b) in order to do it. The proof of (a) has not been published.
And here's a glimpse into Ted's headspace when writing it, from a journal entry at the time:{3}
Ever since seeing how the Trout Creek area has been ruined I feel so much grief whenever I am sitting quietly, or when I am walking slowly through the woods just looking and listening, that I have to keep occupied almost all the time in order to escape this grief. That was my favorite spot. Whoever has read my notes knows very well what the other causes have been. Where can I go not to enjoy in peace nature and the wilderness life? — which are the best things I have ever known. Even in the officially designated “wilderness” there must be the continued noise of airplanes, especially the jets, since I know that planes are permitted to fly over the Bob Marshal and Scapegoat wildernesses. Are there fewer planes there than here. Maybe, maybe. Perhaps one of these days I’ll go and find out. But so many times I’ve gone looking for a place where I can escape completely from industrial society, and always . . . [three dots in the original] well, I’m very discouraged. So, I’ve been playing around with mathematics a good deal lately. It’s a rather contemptible game, but while I’m involved in it, it enables me to escape from my grief.
13. FourDigit Numbers that Reverse Their Digits When Multiplied
Original PDF: 11. Unknown Date  FourDigit Numbers that Reverse Their Digits When Multiplied.pdf
FOURDIGIT NUMBERS THAT REVERSE THEIR
DIGITS WHEN MULTIPLIED
T. J. KACZYNSKI
If n <i2 is an integer and ag, . .. * a^ are integers satisfying 0 n for i « 0,1, • • • ,h ,
then we let ... , ap ag)_{n} denote the number
^{a}j^{aJ} • Whenever we write a symbol of the form (a_{h}, ... , a., ag)_{n} * it is to be understood that
0 a_£ n for i — Oy1f.««fh so that ^{a}h® • • • > ^{a}l ag are the digits of the number (a_{h}, ... , *p ag)_{n} ^{in} base n notation.
If k is an integer and 1 k < n , we say that (a^, ... , ap ^{a}o)_{n} i® reversible for n, k if and only if a^ / 0 and kCa^, ... , ap a$)_{n} a (ag, a^_{t} ... , ^{a}h)_{n} • Reversible numbers have been studied in [1], [2j, [3]. The purpose of thia paper is to construct a rather involved family of 4digit reversible numbers that illustrates the complexity of the reversible number problem. We use the abbreviation RN for "reversible number".
Sutcliffe [jj showed that there exists a 4digit RN for any base n >3 • Let d be any divisor of n
(possibly n itself) with d "3 , and set t = n/d and k = d1 . Then
k(t, t1, nt1, nt)_{n} = (nt, nt1, t1, t)_{n} .
(This family of ^{in Let us}
refer to a RN of this type as a Sutcliffe RN. Note that the Sutcliffe reversible number (t, t1, nt1, nt)_{n} is equal to (n+1)(t1, n1, n~t;_{n} .
At least two other types of 4digit RNs may exist for certain values of n. ,
If (a,b,c)_{Q} is a Jdigit RN for n, k , and if a+b £ n1 and b+c < n1 , then (n*1)(a,b,c)_{n} is a 4digit RN for n, k • (For instance, 4X (2,5,9) =
(9,5»2)_{17} ; multiplying by 18 yields 4X(2,7,14,9)_{17} ^{a} (9,14,7,2)•)
If (a,b,c)_{n} is any solution of the system of conditions
k(a,b,c)_{n} = (c1 ,b+1 ,a)_{Q} ,
(1)
a+b £ n2 , b+c >■ n , a 0 ,
then (n+1)(a,b,c)_{Q} is a 4digit RN for n, k , as can
be verified by computation. We note that 4^/7 Jw /•
from a solution of (1) can never be a Sutcjlffe RN for n, k , because if t = n/(k+1) then (t1, n1, nt)_{Q} cannot satisfy (1).
One family of solutions of (1) can be obtained by taking any integers u  1 and k 3 and setting n = u(k^{2}1)+k , a = (k1)u , b a (u(k+1) + 1)(k2) , c = (uk+1)(k1) . Observe that the corresponding 4digit RN is (n+1)(a,b,c)_{n} = (k1 ,k3,k1)_{n}(u,uk+1)_{n} , and that (u,uk+1)_{n} is a 2df g’RN for n, k .
Sutcliffe [J] showed that there exists a 2digit RN in base n notation if and only if n+1 is not prime. It was shown in [ 1 ] that there exists a 3digit RN for n if and only if n+1 is not prime. This directs our attention to 4digit RNs in the case where n+1 is prime.
Does (1) ever have a solution when n+1 is prime? The answer is yes. With n+1 = 59 we have 19X(2,41,52)53 = (51,42,2)^ , which yields 19 *(2,44,35,52)^ = (52,35,44,2)53 .
Do there exist infinitely many such examples? The answer is again yes. Let s be any nonnegative integer, take k « 19 , n = 5&+360s , a = 2+17a , b = 41+260s , c = 52+323s , and we have a solution of (1). By Dirichlet’s Theorem, there are infinitely many positive integers s for which n+1 « 59*360s is prime.
However, all these solutions are in a sense
isomorphic; w© do not regard them as essentially different. What we really want to show is this:
There exist infinitely many positive integers k having the property that there exist integers n, a, b, c for which n+1 is prime and the system of conditions (1) is satisfied.
This is our main result. To prove it, set
f(x) = 41067x^{2}  UOhx ♦ 9 g(x) = 10179x^{2}  222x ♦ 1 .
The discriminant of g(x) is 8%8 » 2^1071 , not a square, so g(x) has no linear facWr with rational coefficients. Therefore f(x) and g(x) have no nonconstant common factor with rational coefficients. Consequently there exist polynomials p(x) and q(x) , with rational coefficients, such that p(x)f(x) •> q(x)g(x) * 1 . Let d>0 be the product of the denominators of all the fractions that appear as coefficients of p(x) and q(x) , and let pfx) » dp(x) and Q(x) = dq(x) • Then P(x) and Q(x) have integer coefficients and P(x)f(x) ♦ Q(x)g(x) = d .
Let k be any number of the form k « 1l7yd2 , where y is a positive integer. Let D = yd and let v be the greatest common divisor of f(D) and g(D) . Then v divides D a yP(D)f(D) ♦ yQ(D)g(D) . Since v divides g(D) it follows that v divides 1 . Thus f(D) and g(D) are relatively prime.
By Dirichlet*s Theorem, we can choose a positive integer t for which f(D)t * g(D) is prime. Set
n = f(D)t ♦ g(D)  1 = (4iO67D^{2}UO4D*9)t*1O179D^{2}222D ,
u = 1JD , r = 2(u1) , m = 117Dtt*29D = (9ul)t+29D ,
U = 3u1 , R = 3r+1 a 6u5 = 78D5 , M = 9m*1 ,
w = 9rm*Jm*r .
We compute
k = 117D2 a 9u2 = 3U+1 , n a MU*1 , MR = 3w*1 .
Modulo 9u1 we have the following congruences:
nR+w a (MU*1)(6u5)*9rm*3m+r
« (27mu+3u9m)(6u5)♦18mu15m*2u2
= (3»3u9m)(6u5)2ia15m>2u2
a 18u^{2}13u36mu+17m2 = 2u*13m3
= 26l>377D3 = 351D3 « 3(117D1) = 3(9u1)
= 0 (mod 9u1) .
Thus nR+w is divisible by 9u1 • Choose an integer c so that (k+1)c = (9u1)c a nR+w . Set S = knR(k^{2}1)c1 . Because (n+1)R = (MU+2)R  1 (raod 3) , we see that k1 = 3U divides MIl[(n+1 )Rl]. Thus
SnR+1 = (kn^{2}1)R(k^{2}1)nc(n1)
E(n^{2}1)R(n1) = MU[(n+1)R1 j 5 0 (mod k1) .
Choose an integer b so that (k1)b = SnR+1 . Sot a  kcRn . We then have
(2) kc = Rn+a
(3) kb+R = Sn+b+1
(4) ka+S = c1
We must show that certain inequalities are satisfied. Clearly 2 k <i n , c >2 , 2 R k1 . Thus (k^{2}1)c = 3U(k+1)c = 3U(nR+w) < 3UnR+UMR < 3UnR+nR _{3} knR < kn(kl) < (k^{2}1)n .So 2 < c < n .
Observe that R1+U < 3U 2(R1)*U . Adding 3U(k*1)c = 3V(nR+w) to thia inequality gives
< J 3U(nR+w)4R1+U < 3U(k+1)c^^^3U(nR+w)+2(R1)+U , (k1)nR*R1+MRU <(k^{2}1)c+k1 4 (k1)nR*2R2+MRU , (k1)nR>nR1 < (k^{2}1)c*k1 (k1)nR+nR+R2 ,
1 <(k^{2}1) cknR+k+1 <R ,
kR<.S < k1
Thus 2 < S < k1 (from which we see that b >0 ) and
(5) S+R > k+1 .
Also, (k1)b = SnR+1 Sn (k2)n^ ffe^so that b<T < ■^'^{n} = n1 , and b+1 < n .
p
Note that (k+1) C n , so that (k+1)c = nR+w > 2 (k+1) and c1 > k >S • Thus ka = c1S > 0 , so that a 0 .
From (J) and (4), we find k(a+b)+R+S = Sn+b+c <^ (S+2)nt~kn • Therefore a+b < n . Suppose a+b = n1 • Then from (4) and the definition of b we have (k1)(n1) = (k1)(a+b) = S(n1)+caR . Consequently n1 divides caR . But c >ka by (4), so n1 >caR >(k1)aR>0 . This contradiction shows that a+b n2 .
From (3) and (5) we see that (k1)(b+c) s SnR+1+(k+1)c2c = (S+R)n+w+1R2c (k+1)n+w+1R2c >
(k1)n+w+1R . But 3R MR = 3w+1, so that R < w+1 . Therefore b+c >n •
Equations (2), (3)» (4), together with the inequalities we have just proved, show that (a,b,c)_{n} satisfies (1). ©
In the foregoing argument there is no need to restrict ourselves to the case where n+1 is prime, so the construction also yields many 4digit RNs for composite values of n+1 .
We hooe to publish at a later date a more general treatment of reversible numbers, in which we shall prove (among other tilings) that if n*1 is prime, then every 4digit RN for n is either a Sutcliffe RN, or of the form (n+DfajbjC) , where (a,b,c)_{n} is a solution of (1).
REFERENCES
1. T. J. Kaczynski, Note on a problem of Alan Sutcliffe, Math. Mag. 41 (1968), 8486.
2. L. F. Klosinski and D. C. Smolarski, On the reversing of digits, Math. Mag. 42 (1969), 208210.
3. A. Sutcliffe, Integers that are multiplied when their digits are reversed, Math. Mag. 39 (1966), 282287.
14. Handwritten Math equations and procedures
Original PDF: 12. Unknown date  Handwritten Math equations and procedures.pdf
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[1] F. Bagemihl, Curvilinear cluster sets of arbitrary functions, Proc. Nat, Acad, Sci. U. S. A.> 4 (1955) 379382.
[1] F. Bagemihl, Curvilinear cluster sets of arbitrary functions, Proc. Nat, Acad, Sci. U. S. A.> 4 (1955) 379382.
[1] F. Bagemihl, Curvilinear cluster sets of arbitrary functions, Proc. Nat, Acad, Sci. U. S. A.> 4 (1955) 379382.
[1] F. Bagemihl, Curvilinear cluster sets of arbitrary functions, Proc. Nat, Acad, Sci. U. S. A.> 4 (1955) 379382.
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[2] F. Bagemihl & G. Piranian, Boundary functions for functions defined in a disk, Michigan Math, J., 8 (1961) 201207.
[2] F. Bagemihl & G. Piranian, Boundary functions for functions defined in a disk, Michigan Math, J., 8 (1961) 201207.
[2] F. Bagemihl & G. Piranian, Boundary functions for functions defined in a disk, Michigan Math, J., 8 (1961) 201207.
[2] F. Bagemihl & G. Piranian, Boundary functions for functions defined in a disk, Michigan Math, J., 8 (1961) 201207.
[2] F. Bagemihl & G. Piranian, Boundary functions for functions defined in a disk, Michigan Math, J., 8 (1961) 201207.
[2] F. Bagemihl & G. Piranian, Boundary functions for functions defined in a disk, Michigan Math, J., 8 (1961) 201207.
[2] F. Bagemihl & G. Piranian, Boundary functions for functions defined in a disk, Michigan Math, J., 8 (1961) 201207.
[2] F. Bagemihl & G. Piranian, Boundary functions for functions defined in a disk, Michigan Math, J., 8 (1961) 201207.
[2] F. Bagemihl & G. Piranian, Boundary functions for functions defined in a disk, Michigan Math, J., 8 (1961) 201207.
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[3] S. Banach, Uber analytisch darstellbare Operationen in abstrakten Raumen, Fund, Math., 17 (1931) 283295.
[3] S. Banach, Uber analytisch darstellbare Operationen in abstrakten Raumen, Fund, Math., 17 (1931) 283295.
[3] S. Banach, Uber analytisch darstellbare Operationen in abstrakten Raumen, Fund, Math., 17 (1931) 283295.
[3] S. Banach, Uber analytisch darstellbare Operationen in abstrakten Raumen, Fund, Math., 17 (1931) 283295.
[3] S. Banach, Uber analytisch darstellbare Operationen in abstrakten Raumen, Fund, Math., 17 (1931) 283295.
[3] S. Banach, Uber analytisch darstellbare Operationen in abstrakten Raumen, Fund, Math., 17 (1931) 283295.
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[4] P. T. Church, Ambiguous points of a function homeomorphic inside a sphere, Michigan Math. J., 4 (1957) 155156.
[4] P. T. Church, Ambiguous points of a function homeomorphic inside a sphere, Michigan Math. J., 4 (1957) 155156.
[4] P. T. Church, Ambiguous points of a function homeomorphic inside a sphere, Michigan Math. J., 4 (1957) 155156.
[4] P. T. Church, Ambiguous points of a function homeomorphic inside a sphere, Michigan Math. J., 4 (1957) 155156.
[4] P. T. Church, Ambiguous points of a function homeomorphic inside a sphere, Michigan Math. J., 4 (1957) 155156.
[4] P. T. Church, Ambiguous points of a function homeomorphic inside a sphere, Michigan Math. J., 4 (1957) 155156.
[4] P. T. Church, Ambiguous points of a function homeomorphic inside a sphere, Michigan Math. J., 4 (1957) 155156.
[4] P. T. Church, Ambiguous points of a function homeomorphic inside a sphere, Michigan Math. J., 4 (1957) 155156.
[4] P. T. Church, Ambiguous points of a function homeomorphic inside a sphere, Michigan Math. J., 4 (1957) 155156.
[4] P. T. Church, Ambiguous points of a function homeomorphic inside a sphere, Michigan Math. J., 4 (1957) 155156.
[4] P. T. Church, Ambiguous points of a function homeomorphic inside a sphere, Michigan Math. J., 4 (1957) 155156.
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[6] M. H. A. Newman, Elements of the topology of plane sets of points, Cambridge University Press, 1961.
[6] M. H. A. Newman, Elements of the topology of plane sets of points, Cambridge University Press, 1961.
[6] M. H. A. Newman, Elements of the topology of plane sets of points, Cambridge University Press, 1961.
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[6] M. H. A. Newman, Elements of the topology of plane sets of points, Cambridge University Press, 1961.
[6] M. H. A. Newman, Elements of the topology of plane sets of points, Cambridge University Press, 1961.
[6] M. H. A. Newman, Elements of the topology of plane sets of points, Cambridge University Press, 1961.
[6] M. H. A. Newman, Elements of the topology of plane sets of points, Cambridge University Press, 1961.
[6] M. H. A. Newman, Elements of the topology of plane sets of points, Cambridge University Press, 1961.
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{1} Better known for other work.