“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 100-page 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, Chia-Shun 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 youngest-ever 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 well-liked 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 group-theoretic 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 y-coordinate, and let X denote the x-axis. 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 re-proving 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 complex-valued 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 post-graduate 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**: [[https://faculty.valpo.edu/lpudwell/papers/mm005281.pdf][Digit Reversal Without Apology.pdf]] **Digit Reversal Without Apology** Lara Pudwell Rutgers University Piscataway, NJ 08854 [[mailto:lpudwell@math.rutgers.edu][lpudwell@math.rutgers.edu]] In A Mathematician’s Apology [1] G. H. Hardy states, “8712 and 9801 are the only four-figure 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 short-sighted. 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(ahnh + ah-1nh-1 + • • • + a1 n + a0) = a0nh + a1nh-1 + • • • + ah-1n + ah with n >2, 1 < k < n, 0 < ai < n — 1 for all i, a0 = 0, ah = 0. We shall refer to such an integer a0...ah as an (h + 1)-digit solution for n and write k(ah, ah-1, ..., a1, a0)n = (a0, a1, ..., ah-1, ah)n. For example, 8712 and 9801 are 4-digit solutions in base n = 10 for k = 4 and k = 9 respectively. After characterizing all 2-digit 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 3-digit solution but no 2-digit 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 3-digit solution for n, then deleting the middle digit gives a 2-digit solution for n. Together with Sutcliffe’s work, this proved that there exists a 2-digit solution for n if and only if there exists a 3-digit solution for n. Given the nice correspondence between 2- and 3-digit 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 5-digit 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 one-to-one correspondence. A second extension of Sutcliffe and Kaczynski’s results is to ask, “Is there any value of n for which there is a 5-digit solution but no 4-digit solution?” We will answer this question in the negative; and, furthermore, we will show that there exist 4- and 5-digit solutions for every n >3. *** An attempt at generalization In the case of 3-digit solutions, Kaczynski proved that if n + 1 is prime and k(a, b, c)n = (c, b, a)n is a 3-digit solution for n, then k(a, c)n = (c, a)n is a 2-digit solution. Thus, we consider the following: Question 1. Let k(a, b, c, d, e)n = (e, d, c, b, a)n be a 5-digit solution for n. If n + 1 is prime, then is k(a, b, d, e)n = (e, d, b, a)n a 4-digit solution for n? First, following Kaczynski, let p = n + 1. We have k(an4 + bn3 + cn2 + dn + e) = en4 + dn3 + cn2 + 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[n2 (n2 — 1)a + n2(n + 1)b + fpn2 + n(n + 1)d — (n2 — 1)e] = n2(n2 — 1)e + n2(n + 1)d + fpn2 + n(n + 1)b — (n2 — 1)a. After substituting for p, dividing by n + 1, and rearranging, one sees that k [an3 + (b — a + f)n2 + (d — e)n + e] = en3 + (d — e + f)n2 + (b — a)n + a. Indeed, this is a 4-digit solution for n if f = 0, b — a > 0, and d — e > 0, but not necessarily a 4-digit solution of the form conjectured in Question 1. As in Kaczynski’s proof for 2- and 3-digit 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 5-digit 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 5-digit solutions do, in fact, yield 4-digit 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 5-digit solution but (a, b, d, e)n is not a 4-digit 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 5-digit solutions Although Kaczynski’s proof does not generalize entirely, there exists a family of 5-digit 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 5-digit solution for n if and only if a | (n - a). Proof. We have (n - a)n4 + (n - a - 1)n3 + (n - 1)n2 + (a - 1)n + a an4 + (a - 1)n3 + (n - 1)n2 + (n - a - 1)n + (n - a) (n - a)(n4 + n3 - n - 1) n - a a(n4 + n3 - 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 5-digit solution described in Theorem 1, deleting its middle digit gives a 4-digit 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 5-digit 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 4-digit solution for n. Proof. By Theorem 1, n-a G N. Now (n - a)n3 + (n - a - 1)n2 + (a - 1)n + a an3 + (a - 1)n2 + (n - a - 1)n + (n - a) (n - a)(n3 + n2 - n - 1) n - a a(n3 + n2 - n - 1) a . □ These 4-digit solutions were first described by Klosinski and Smolarski [4] in 1969, but their relationship to 5-digit 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 4-digit solution and a 5-digit 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 5-digit solution but no 4-digit solution. More specifically, we know that there are 4- and 5-digit solutions for every n > 3. Although Kaczynski’s proof does not generalize directly to 4- and 5-digit 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 5-digit 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 5-digit solutions for f = 1 with a particularly nice structure. To date, no other 4- or 5-digit 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][http://www.math.rutgers.edu/~lpudwell/maple.html]]. *** Acknowledgment Thank you to Doron Zeilberger for suggesting this project. *** References [1] G. H. Hardy, A Mathematician’s Apology, Cambridge University Press, New York, NY, 1993. [2] Alan Sutcliffe, “Integers That Are Multipled When Their Digits Are Reversed”, Math. Mag., 39 (1966), 282-287. [3] T. J. Kaczynski, “Note on a Problem of Alan Sutcliffe”, Math. Mag., 41 (1968), 84-86. [4] Leonard F. Klosinski and Dennis C. Smolarski, “On the Reversing of Digits”, Math. Mag., 42 (1969), 208-210. [5] N. J. A. Sloane, Sequence A031877 in “The On-Line Encyclopedia of Integer Sequences”, [[http://www.research.att.com/projects/OEIS?Anum=A031877][http://www.research.att.com/projects/OEIS?Anum=A031877]]. [6] Eric W. Weisstein, “Reversal”, From MathWorld-A Wolfram Web Resource, [[http://mathworld.wolfram.com/Reversal.html][http://mathworld.wolfram.com/Reversal.html]].
lim f(z) = t(x). z -> x z ( vThe 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 t|C - 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 f|S - N = g|S - N. Using these defnitions, Theorems 2 and 3 read as follows. Theorem 2: Let f be a continuous real-valued function in D and let t be a finite-valued 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 half-plane D0: I(z) > 0, and its boundary C0: I(z) = 0, instead of D and C, respectively. Theorem 4 states that if f is a real-valued function of Baire class a > 1 in D0, and t is a finite-valued 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 real-valued Borel-measurable function in D0 and let t be a finite-valued boundary function for f; then t is Borel-measurable. Next, the author proves that for an arbitrary function t on C0, there exists a function f on D0 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 two-dimensional 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
on C such that for every x e C there exists an arc y at x with
lim f (z) =
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, W-dimensional Euclidean space will be denoted by RN, and the Riemann sphere will be denoted by 2. Points in RN will be written in the form {xx , x2 } • • • , xN) rather than (xt , x2 , • • • , xN) (to avoid confusion with open intervals of real numbers in the case N — 2). Whenever we speak of real-valued functions we mean finite-valued 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.
11 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 pr , p2 , p3 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 q2 . Putting
, 72 and 7 together, we obtain an arc T joining pj to p2, with r — {pr, p2} C A. We can assume F is a simple arc, for if r is not simple, and p2 can be joined by some simple arc Fz £ T (see [7]). Let Lx , L2 be the two open arcs of C determined by the pair of points pr , p2 . We may assume, by symmetry, that p3 z Lr . According to [6] (Theorem 11.8, p. 119), D — T has two components Ui and U2 , the boundary of Ui being Lx U P and the boundary of U2 being z2u r.
Let 7' be an arc in A joining q3 to a point q z T A. Putting 73 and 7' together, we obtain an arc 5 joining p3 to q. Starting at p3 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 p3 and r. Clearly, A — {p3} GZ A. We can assume (according to [7]) that A is a simple arc.
Since p3 z Lx , p3 is not in U2 . Since
A - {p3 , r} £ D - T = Ui U2 ,
A — {ps , r} must have a point in Ux . But A — {p3 , r} is connected, so A - {p3, r} C Ui. Hence A is a cross cut of Ur. Let Mr ,M2]oq the two open subarcs of Li with endpoints pt , p3 and p2 , p3 respectively. Let Pi , r2 be the two closed subarcs of T with endpoints pr , r and p2 , r respectively. According to [6] (Theorem 11.8, p. 119), Ur — A has two components Vi and V2, the boundary of Vi being kJ rx kJ A and the boundary of V2 being M2 kJ r2 kJ A.
Since P U A C 2I, Vi^J V2\J U2. Recall that p3 4 U2 . It follows that since p3 z B, B has a point in common with Vi^J V2 . But B is connected, so B £ Vi yj V2. We see that pr $ V2, and therefore that B Vi 4=> (because Pi z B). Hence B £ , so p2 z W . But, since the boundary of Vt is Mx
I\ kJ A, p2 Vi . This contradiction proves the lemma.
Lemma 2. There exists a countable family 8 of open disks such that every open set U Q R2 can be written in the form U — Sn , where Snz § and Sn £ U.
Proof. Let {pn} be a countable dense subset of R2, and let 8 be the family of all open disks of rational radius having some pn 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 Sx z 8 with x z Sx £ Sx £ U. Obviously
u = \J sx.
xt.U
Theorem 1. Let f be a homeomorphism of D onto D, and let C-^ is continuous. Q to C — N.
If U is any open set, write U = Sn , where Sn £ S, Sn CZ U. Suppose x £ (pQ1 (U). Then (x) = n 0
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 (p0 denote the restriction of to C — N, and let G*(r, t) = G(r, t) — N. Since every countable set is an Fff set, G*(r, t) is a G8 set. Observe that
(2) ^([r, +-)) = ^([r, +«.))- N 2 G*(r, f)
2 +-)) - N = ^([Z, +-))•
If t is a fixed rational number, let {rn} be a strictly increasing sequence of rational numbers converging to t. Then, by (2),
C\ 0\[t, + «>)) = ^\[ra ,+«)), 0 can be extended to a real-valued function on C such that for every real u, +00)) is a G8 set and +<»)) is an Fa set. By Theorem IX of the same paper, is of Baire class 1(C, 7?). Since except for xz N, is of honorary Baire class 2(C, 7?). Q.E.D.
Corollary. Let f be a continuous -junction mapping D into RN, and suppose N is a boundary junction jor j. Then 2(C, RN).
Proof. We simply write our functions in terms of their components, say
f — (fl ! f'2 J * ’ • , and (p — (epi , except on the countable set VJi-i Mt • Hence is of honorary Baire class 2(C, RN).
Q.E.D.
Lemma 4. Let g be a continuous junction mapping C into R3. Let q be a point of R3 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 = {yvR3 | \y - g| < e}.
If 0(C) CZ S, let g* : C —> R3 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 < bk], and
k =1= I =>Ik 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 Ifc’s. Say
0’1({0}) ... \JIn.
The endpoints eiak and et6A of Ik are not in 0~1({}), so we can construct, for each k, a continuous function gk : Ik-+ R3 such that
Sk(eiai) = g(eiai), = g^*),
and q is not in the range of gk . Define
0*(x) = 0(a:), if o;£C~ (AUZ2U ... U/n),
0*(x) = gk(x), if x e Ik , 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 2(C, 2).
Proof, Since 2 is a subset of J?3, the corollary to Theorem 2 shows that is of honorary Baire class 2((7, 2?3). Let g be a function of Baire class 1(C, 2?3) which differs from 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 —» R3 such that q does not lie in the range of g* , and for all x £ C,
- g| gn(x) = g*(x). a
It is easy to show that g*n —» g.
We define a function P as follows. If a e R3 — {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 R3 — {} 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 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 real-valued 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 rr-axis 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 = ,
9Ea = ,
91* = 9lco = 912 , If 0 is any class of sets, let 0a denote the class of all countable unions of members of 0, and let 08 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 •
XA + 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 = x n—*a> n—♦<»
Thus gn 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 Nn such that gn | C*-Nn is of Baire class a. Let N = VJZi Nn . Then gn k-# is of Baire class a. But gn Iat-jv —> k-w , so k-^ is of Baire class a + 1.
Now
p (M - N) = & k-k-1((- r]) = Try (m- n), where T s 9la+1 (by IV and V). Clearly P C\ M ~ T C\ M.
We have
L = Po U L, ~ (Go n F) VJ (Gt n E) = (Go U GJ H E,
so L ~ G C\ E where G e g8 . Also
M0=PKM~TC}M = T C\(E - L)
~tc\[e - (g n P)] = [t n (c° - g)] n e.
Since G e g5 , G° - G e , so by VI and VIII, T C\ (C° - G) e 9T+1. Thus
Mo — To n E,
where To e 9l“+1. Now we can examine the properties of P.
P = (GqKE)U (To n E) = (Go U To) n E,
so, again by VI and VIII,
p ~ 7\ n p,
where Tx e 9la+1. Since a countable set is in and the complement of a countable set is in g8 , it is easy to show (using VI and VIII) that
P = T2 H P,
where T2 e 9la+1. Since P C\ Q = is an arbitrary boundary function for /, we can replace /, $>, r by — /, — cp, ~r to find that
^([r, +«>)) e9la+1.
Also,
^((r, +00)) = C° - £snr+1.
By IX, is of Baire class a + 1. Q.E.D.
**** 5. Boundary functions for measurable functions.
Theorem 5. Let f be a real-valued Borel-measurable function in DQ and let
Since every Borel-measurable 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 Lebesgue-measurable function need not be Lebesgue-measurable.
Let u denote Lebesgue measure on R and let /z2 denote Lebesgue measure on R2. 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 real-valued 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 real-valued 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 x0 £ (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 xQ £ (a, e] set f(x0) = A(e). It is easily verified that f is finite-valued 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 In , and H = \J Jm , mN mtM
where {In | n £ N} and {Jm | ms M} are countable families of disjoint open intervals. Let In = (an , bw), and observe that an , bn £ [a, b]. Set
s = (J {g(a„), g(bn)} - {- 00 , + co }.
mN
Notice that S is countable. Set
Kn = (g(a„), g(&n)).
One can easily verify that k =1= n implies Kk P Kn = 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', ge(<7 +') g + ^(J'). Let W = {x + h(x) | x £ E].
Assertion.
w c (E + S) KJ \J \J [(Zn n g-W) 4- (Jn 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 £ In . Since g is increasing,
h(x) = g(x) £ [#«), g(bn)].
If h(x) equals g(an) or #(6n), then h(x) e S, so w = x + h(x) e E + 5. On the other hand, suppose h(x) 4= #(an), (&„). Then h(x) £ Kn . Also, g(x) = h(x) £ Jm for some m. Thus h(x) £ JmC\ Kn and x £ In C\ g~\Jm)} so that
w = x + h(x) £ (In C\ g~\Jm)) + (Jm C\ Kn).
This proves the Assertion.
Since g is increasing, g^J^ is an interval, so both In C\ g~\Jm) and Jm C\ Kn are intervals. Also note that m 4= I implies g~\J„^ g~\Ji) —> . By the Assertion,
M.W g + s) + S E n g~\jmy) + (Jw n #n)] mN mtM
g ne(E + s) + E E [m(/b n n o
mN mzM
= m/U (s + t?)) + E [ E n g~\jmy) + £ »(jm c\ Kn)] szS mN mtM mtM
[[< E + E)]] + E W.) + E mGA. n 2Q]
utS mN mtM
= 0 + ju(G) + E E c\ Kn) nzN mzM
= m(g) + E E m(A. n Kn)
mtM nzN
m(G) + E mW = m( )+ 1EH) < e. mt M
Since e is arbitrary, jne(W) = 0.
Lemma 11. Let L — {(x, a) | x £ E} and M = {{xy b) | x £ R} be two horizontal lines in R2. 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 = \Jkt& s. Then u2(S) = 0.
Proof. Assume without loss of generality that b >a. For any (x, y) £ R2 let ^({x, y)) = x. For any y £ R let ly = {{x, y) | x £ R}. Let
Eo = {z £ E | z is the endpoint of some s £ <£},
and observe that Eo has linear measure 0. For any set A C R2 we of course set tt(A) == {x £ R | (xj y) £ A for some y £ R}.
We define a function h on ir(E0) as follows. If $ £ ^(Eq), then {x, a} zE0 , 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(E0) into tt(F).
Since the segments in £ cannot intersect each other, h must be an increasing function.
Take any yQ with b >yQ > a. Let c = b — y0, d = y0 — a. A simple computation shows that if q s lVo C\ 8, then
tt(q) =
ex + dh(x) c + d
Proof. For each positive integer n let hn be a strictly increasing function on R such that y(hn(R)) = 0, and for every x, ]x — hn(x)j ~ 1/n. Let and En has linear measure 0. For each n, x let sn(x) be the line segment joining (hn(x), 1/n) and (An+1(a;), l/(n + 1)). Since
En is a subset of
hn(x) > hn(x') x > xf => hn+1(x) > hn+1(x'\
we find that x 4= xf implies sH(x) C\ sn(xf) = 0. Since each sn(x) has one endpoint in En and the other in En+1 , Lemma 11 shows that for each n
= 0.
x&R
Hence
M2(0 Usn(a:)) = 0. 'n=sl xtR •
Set
j(z) = £>(($, 0)), if z e sn(x) for some n,
f(z) = 0, if z is not in any sn(x).
j(z) = 0 almost everywhere. Let
y(x) = {{x, 0>} U 0sn(x). 74 = 1
Since the endpoints of sn($) are at (An(z), 1/n) and (hn+1(x)j l/(n + 1)), and since (hn(x), 1/n) —> (x, 0) as n —> oo, it is clear that y(x) is an arc at (x} 0). Obviously
lim j(z) = t we find that is of Baire class a + 1 with R3 as the universal range space. It is then easy to show by means of Satz 2 in Banach’s paper [3] that 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 R3 and let F be one face of K. If f is defined in A, then we say (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 —» R3. 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), 2(x) = =0, if xtE,
=1= 0, if x s F — E.
Therefore ( — ^i)-1({0}) = E, so — x or x and Y in terms of their components; say / = (f i, f2, fa) and , 2, ^3). Since is nonmeasurable, one of its components, say ,- ,
is nonmeasurable. But is a boundary function for the continuous real-valued function f,« , so Theorem 2 and Theorem 4 must be false in three dimensions.
References
[1] F. Bagemihl, Curvilinear cluster sets of arbitrary functions, Proc. Nat, Acad, Sci. U. S. A.> 4 (1955) 379-382.
[2] F. Bagemihl & G. Piranian, Boundary functions for functions defined in a disk, Michigan Math, J., 8 (1961) 201-207.
[3] S. Banach, Uber analytisch darstellbare Operationen in abstrakten Raumen, Fund, Math., 17 (1931) 283-295.
[4] P. T. Church, Ambiguous points of a function homeomorphic inside a sphere, Michigan Math. J., 4 (1957) 155-156.
[5] F. Hausdorff, Uber halbstetige Funktionen und deren Verallgemeinerung, Math. Z., 5 (1919) 292-309.
[6] M. H. A. Newman, Elements of the topology of plane sets of points, Cambridge University Press, 1961.
[7] H. Tietze, Uber stetige Kurven, Jordansche Kurvenbogen und geschlossene Jordansche Kurven, Math. Z., 5 (1919), 284-291.
The University of Michigan
if 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 7s2 (n+l) = ^p2(n) and E(n+1) = E(n), K
and nothing is required to be done.
Case III. Suppose that-F j consists of a single point eQ 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(eo) €£ (0, so that S(eQ, e(eQ), 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 Fn+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 Fn+1 is contained in a complementary interval of F^. Since ^(n) is finite, (8) and (10) show that we can choose 5(Fn+1) £(0, ^y) so that B(Fn+i, 5(Fn+^), 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(Fk) for some k <_ n, so (since Fn+^ either splits with F^ or is contained in a complementary interval of FR) e ^J(Fn+1)*. By (11), B(Fn+1, 5(Fn+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. eQ, e • £E(n). ,
In this case the construction is already finished.
Case IVb. eQ€E(n) and eQ* $E(n).
If e ’ £ F. for some k < n, then F , splits with F. , so that o k — ’ n+1 r k’
eQ’ 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(eo’, e(eo')» 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€ f2(n). By (11), S(eQ’, e(eo’)» 0) and B(Fn+1, 6(Fft+1), a, B) are disjoint. Thus the construction is finished for E(n+1) and(n+1).
/
Case IVc. eQ E(n) and e * € E(n).
This case is essentially the same as Case IVb.
Case IVd. eQ E(n) and eQ'
-1(U) u N. ~"
Then there exists a countable set M ^E such that cpL is of class e-M
CFa (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 -1(B) u N(B)
Let M = VJ N(B). Then M is countable. Let E = E - M and let B€fi °
= o is of class (F (Eq)) . 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
Vb-1(W) = E A ‘p'M = E A VJ • -1(B)
0 0 0 B€ffl(W)
- E A U L(B) 0 Bcacw)
— E U [ -1CB) u N(BJ]
B£ (B(W)
CEO O [CP[[#_ftn3][[‡]]]® U M]
B€Q(W)
= e n U -1CB) ° b e 1CW) = q 1 (W). o
Consequently cp^CW) = Eq 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
En = {x € X : there exists an arc y at x, having one endpoint on Xn, 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 00 00
-1(U) Q Ea|J 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 1(U) .
1. 00
Thus Q m) +
sm) = sm) > 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 R having ipas a boundary function. Let
K = g-1({v € R3 : |v| = | J)
L = g-1({v € R3 : |v| > | })
F = g-1({vGR3 : |v| £y }).
q : R - {0} ■* S
2
be the 0-projection onto S (see page 11), and let f be the composite
2
! function PQ « f^. 1 Then f maps H continuously into S , and Po«
*** 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 Borel-measurable 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 Ui-UI is countable. We can is4J- ie<5-
write
-i(U) Ct Then, if £ >_2, 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 1(ii). Then
-1(W) = T(U)
[[ued(w) T uedcw)]]
[[U€d(W)]]
Thus +\e) is closed under countable
[[uedOD]]
-1 E+1
unions, 5, (E) . Therefore is of Baire class £(E, Y) . ■
i Y such that f(z) = 0 almost everywhere and ipis a boundary
function for f.
Proof. If {y } Yth® family of arcs described in Theorem 10, let X X £ a f(z) =0 if z is in no y
X
f(z) = ag 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 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 Xq and the other on X^, and let SUz . Assume that S is a Borel set.
£ £
If m (S n XQ) 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!].
, and, since |/(p) — p| <21/2— 1 =ctn fw, the part of this line which lies in Ho is contained in s(p, l.fw). We show that hY approaches along this line. By substituting (f(p)—p)y+p for x in the expression for An(,v, y), one obtains
An(x,y) = [(1 - ny) V 0]
(8) 17
rn + 4 + 2(-
?n-
If P^sn, then f(p)^ln, and one can verify directly that (8) vanishes. Ifp>sn, then f(p) rn, and again one can verify directly that (8) vanishes. Thus An(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-(l-y)p)ly, and hence
p = /*(/■(?)) = /*((x-(i -y)p)ly)- .
Therefore (if 0
for some x s tt(£?0)- So
7r(lyo A 8) C
ex + dh{x) . e + d
X £ 7r(Fo)
Now (d/e)h(x) is an increasing function mapping tt(Eq) 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(E0)
has measure 0, so A SY) = 0. But A 8)) = 0 also when yQ 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 yY , y2 s R, let
U(,yi > 2/2) = {{x, y) \ x, y t R, yt < y < y2}.
A set of the form U(yi , y2) 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 b-i).
Since la and lb 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 2e0 + e* < e/(b — a). Let y0 be any member of [a*, b*]. For the time being, yQ will be held fixed.
For each s e <£, let p8 be the endpoint of 8 on lb , let qt be the intersection point of s with lV9 , and let ra be the endpoint of 8 on la .
Choose an open set G £ R such that tt(1Vq A S(n)) C G and g((?) < e0 • Say G = VJ, Ij, where Z, = (af , b^ and the Z, ’s are pairwise disjoint. We may assume that each Z,« contains a point of tt(1V9 S(n)). For each let
Ci = g.l.b. {7r(pa) | s e £(n), 7r(ga) e ZJ,
dj = l.u.b. {^(p,) | s e £(n), 7r(g«) e Z,}, c< = g.l.b. {?r(ra) | 8 e <£(n), ?r(^) e Z,}, d'i = l.u.b. {?r(ra) | 8 s £(n), ir(q8) 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,« c7- , 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 Af 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’^-^pminV’^ + 2^/r
Note that Hf is measurable. Setting ZZ = Hf, 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
M0-(ZZ A ly)) <
__ e__ b — a
We can, of course, assume that
l J €o I • J 7 . €o I |
y e I max \a, ya — , mm S b, y0 + ^(r
An elementary computation, using the geometrical properties of Hj, shows that
n i,)) g (i + 4^--
\ o yQ / o yQ
Therefore
n z„)) g E n w
- G+n ^)e°+2n2
” 2c0 + €0 < t , b — a
so v(ir(H lvY) < e/(b — a) for every y.
We have shown that for each yQ e [a* &*] there exists a horizontal open strip
V(yQ) containing lVo , and there exists a measurable set H(yQ) C y(?/0), such that
S(n) H V(y0) £ H(y0)
and (for every y) ir(H(yQ) lv) is measurable and
y«H(y0) Cy Zy)) <
The various open strips VG/o) (i/0 « [a*, 6*]) clearly cover the compact set {(0, y} I y e fa*> &*]} • Choose a finite subcovering V(y2\ • • • , V(ym). Set
tm— 1 /
m \
U V(y,)]
J“»+l /
n I7(a*, 6*).
H(ym) U \J IW) -
X
Obviously K is measurable, and for each y, ir(K A Z„) is measurable and C\ lv)) < 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 In = [ne, (n + l)e]. Then In = 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
.a1a2a3 • • • an • • • , (binary decimal),
where the decimal does not end in an infinite unbroken string of l’s. Set
f(.a!a2a3 • • • an • • •) = bib2b3 • • • bn • • • , (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 fn :In-> In such that jn is strictly increasing and /z(fn(In)) = 0. Set
h(x) = fn(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 CQ = {{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
** 5. 1966 - On a Boundary Property of Continuous Functions
Original PDF: [[https://archive.org/download/the-mathematical-work-of-ted-kaczynski/5.%201966.%20On%20a%20Boundary%20Property%20of%20Continuous%20Functions.pdf][5. 1966. On a Boundary Property of Continuous Functions.pdf]]
Kaczynski, T.J. 1966. [[https://web.archive.org/web/20190815201940/http:/homepages.rpi.edu/~bulloj/tjk/tjk3.html][On a boundary property of continuous functions.]] Michigan Math. J. 13:313-320.
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 Fa$ [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 x0 e K and r > 0, then
S(r, x0) = {x e K| p(x, x0) < 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 UIe^r I* is open, we can write I* = Un Jn, where {Jn} is a countable family of disjoint open arcs of C. If
X0 € U I - U I* , Ie# Ie#
then for some Io e #, x0 is an endpoint of Io. For some n, Iq c Jn, so that x0 e • But x0 / Jn, so that x0 is an endpoint of Jn. Thus Uj I - Uj I* is contained in the set of all endpoints of the various Jn; 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 cross-cut we shall always mean a cross-cut of D. Suppose y is a cross-cut 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 cross-cuts y with y n D c n. Let
1(0) = U L(y), I0(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 y1 that lies in n. Let y" c n be an arc joining p to p' . The union of y, y', and y11 is an arc o joining a to a'. By [4], there exists a simple arc 01 co that joins a to a^. Clearly, 01 is a cross-cut with 01 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(O1) and Iq(Q2) A ace (Q2)
are not disjoint, then n1 and Oz 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 cross-cut 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 y1n D and y2n 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 u1 nUz, so that U 1 has a point in common with Uz also. Since Ui is connected, this implies that Ui 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 U2 or y2 nD c U1 . By symmetry, we may assume that Yz n D cUi.
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 y4 n D, such that y - { b} c Oz. Then y - { b} and yi 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 Fa $ .
Proof Let {pk}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 I0(U(n, m, k, £)) Cl A(n, m, k, £).
n=l m=l k=l f = 1
Since I0(U(n, m, k, £)) is open, it is of type Fa . It follows that Y is of type Fa<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 € I0(U(n, m[n], k[n], f[n])) A A(n, m[n], k[n], £[n]) (n = 1, 2, 3, •••).
For convenience, set Un = U(n, m[n], k[n], f[n]). By (2) and Lemma 3, Un and Un+i have some point zn in common. For each n, we can choose an arc yn c Un+i with one endpoint at zn and the other at zn+1 . Then yn c Q(n + 1, m[n + 1]). Also,
and therefore
r r
P
Moreover, because p(P(k[ni+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 y1 c y; it follows that y 6 X. Thus Y c X.
Now suppose x e X. Choose an arc y0 at x such that C(f, y0) c Po for some Po e 8. Take any n. Choose k with p(P0, P(k)) < 1/n. Then
poc^(^ w) >
hence C(f, y0) 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 y0 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, y0), contrary to ti । \ ri
the relation C(f, y0) c , P(k) ) . We conclude that there exists a natural number t for which
y0 n s(|,x) o C P(k)) ) •
It follows that y0 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, £) - I0(U(n, m, k, £)) = A(n, m, k, J?) - [l0(U(n, m, k, £)) A A(n, m, k, {)] is countable. It follows easily that
A U A(n, m, k, £) - A U [l0(U(n, m, k, £)) A A(n, m, k, £)] n m,k,£ n m,k,£
is countable. Since
A U [I0(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 Fa$ -set and a countable set, , and hence it is of type Fa $ . ■
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 Fa 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 FQ 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 Fa 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 FQ 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), 299-312.
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), 284-291.
The University of Michigan
** 6. 1967 - PhD thesis at University of Michigan - Boundary Functions
Original PDF: [[https://archive.org/download/the-mathematical-work-of-ted-kaczynski/6.%201967%20-%20PhD%20thesis%20at%20University%20of%20Michigan%20-%20Boundary%20Functions.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 67-17,790
KACZYNSKI, Theodore John, 1942- BOUNDARY FUNCTIONS. -
The University of Michigan, Ph.D„ 1967 Mathematics
University Microfilms, Inc., Ann Arbor, Michigan
BOUNDARY-FUNCTIONS
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 Chia-Shwi Yiih
BOUNDARY FUNCTIONS
By Theodore John Kacijnski
*** Abstract
Let H denote the set of all points in the Euclidean plane having positive y-coordinate, and let X denote the x-axis. 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 complex-valued 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 Borel-measurable function is always Borel-measurable, but we show that a boundary function for a Lebesgue-measurable function need not be Lebesgue-measurable. 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(x0, e, 0)
5. 62
*** INTRODUCTION
**** 1. Preliminary Remarks
Let H denote the upper half-plane, and let X denote its frontier, the x-axis. 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
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 yQ is a point of a metric space Y having metric p, then
S(r, Yo) 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 cross-cut in D whose endpoints divide y into arcs y^ and y£, then D-a has two components, and the frontiers of these components are respectively a u y1 and a u (The term cross-cut 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- ®ive 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 cross-cut of H. If V is the bounded component
of H - y, let L(y) = V f|X. 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 cross-cuts 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 cross-cut 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(fl2) H acc(fl2)
are.not disjoint, then and Q2 are.not disjoint.
Proof. We assume that and fi2 are disjoint and derive a contradic
tion. Let a be a point in both of the two sets (1). Let y^ be a ★ cross-cut 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 y2 /'Mi are disjoint.
Suppose yj H H C v2 and y2 AH G V^. Then, since y^AH $Up has a point in common with V2« But, since U.^ is unbounded, U.^ cannot be contained in V2, so must have a point in common with y2 nH. This contradicts the assumption that y2 A H £Vp so we conclude that either y^ A H^V2 or y2 AH^Vj. Hence, either y^ A H U2 or y2 A H By symmetry, we may assume that
y2 HH$Ur
fi2 does not meet y^, and fi2 does meet U^ (because y2 A H S U-i A ft?) > so fi2 C U^. Since a € acc(fi2), there exists a point b € L(yp such that b € acc(ft2). But then b € fi2£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 {pv}v”-. be a countable dense subset'of Y. Let {Q(n, m)} " K m=l
be a counting of all sets of the form
' {
lim
k-*» Xm(k)
then x
lim k-x»
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^, Eu 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 ns+i equal ng plus the number of members of ,8. Let Fn +^,..., F be all the members Of We must show that conditions fi) ns+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 ng+^ >. r > t >_ 1.
Case I. Assume that n$ > r > t > 1. In this case we already know that either Fr is contained in one of the complementary intervals of Ft or else Fr splits with Ft.
Case II.
Assume that n
. > r > n s+1 —
> t > 1.
There exists v & {1
,s}
such.that Ft Ev» Either Ev and Eg+^ are disjoint or they split.
Case Ila. Assume E^. and Es+j are disjoint. Either F*. =
Hence
Thus we
• ' co
have shown that we can construct sequences {n_.},._p
[[
Lemma 11. Ifandare two special families, then-J-A^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 |([[’/8ic+T]] + 2)[[’/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 be a sequence of elements of f such that every
member of^is equal to some Fn and such that condition (5a) in the
• "*■ 00
definition of a special family is satisfied. Let {Fn)n_^ be a similar sequence for Set
Fk = Fa(k) n Fb(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^^ Fa(k) and F£ — Fa(£) ’ Either Fa(£) 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 Fa(fc) split (in which case F. and F. split).
Case II. b(k) < b(£).
In this case a similar argument shows that either Fo is contained in
b(k) = |([[2!®pJi]] _ [[/SF* *. -]]2) + k
= 4([[»
[[x axis]]
Figure 2. — Trap(J,E ,0)
[[x axis]]
[[Xo]]
[[x axis]]
Figure 4.—S(x0>£ ,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 K2 split, then for any e2, a, 3, B(Kp Ej, a, 3) and B(K2, e2, a, 3)
are disjoint. /
(9) Suppose that £ K, e > e^ > 0, and 0 < 3 < 3-^ < < a < y.
Then
B(Kp Ep (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 xq J(K) . Then, for any e, Ep
B(K, e, a, 3) and S(xq, Ep 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(xQ, n/ 6) n H S B(K, e, a, 3) .
(13) Suppose that e < e’ and 0’ < 0. Then
stx“TT^TAH £S(xq, s’, 0’)-
(14) Suppose xq$K and e, a, 3, 9 are given.. Then there exists 6 > 0 such that for. every h £ 6,
S(xq, n, 0) and B(K, e, a, B) are disjoint.
(15) If xq | x^ and e, 0 are given, then there exists 6 > 0 such that for every n £ 6,
S(xq, e, 6) and S(Xp n, 0) are disjoint.
(16) B(K, e, a, B) n X C K.
(17) S(xo, e, 6) nx = {xo}.
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 Opair of special
and hence
(24)
CO
. g(z) 1 E m=n
1 = 1
2m 2n-1
(z € H - Un) .
Also
if z € U ,, n+1’
then z € U,, U», ,.., U ,, so that 12 n+1
= gn(z) , and
(25)
. i z ns m=n+l 2,u
1
2n
€ Un+1
We assert that
3ir
(26) for each xq € A, g(z) + 0 as z + xq with z € S(xq, li- g—) •
Take any natural number n. Since x C A , = i> either 7 o n+1 v[[#_ftn2][[†]]]z*’n+l’ case, set n = e(n+l, xq). In the.second case, (12) shows that we can choose n > 0 so that
S(xq, ni J1-) (F, 6(n+l, F), an+1, .
Suppose
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) ux^ 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 Xn is unbounded and does not meet a’, X £ A. n
Because x is a two-sided limit point of En, we can choose a point w €. I H En« Let g be an arc at w, having one endpoint on Xn, such that g - {w} C f’^CU). Then g does not meet a’ (because a' - {x, y] ^.f \w) and f \w) f \u) = n +~ 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 - (l-y)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
fo(x, y) =
(yn(n+l) - n)gn(u(x, y)) + ((n+1) - yn(n+]))gn+]L(u(x, y))
■ 1 1 when —=- < y < —. n+1 — 7 — n
Then f is o
continuous
on
we can assume that
is
rn
o
inf x>s
n
2 " “•
defined
h(x)
By the Tietze extension theorem
and continuous on
all
of H. Let
vn
sup xsn) - Ksn)
if s € N n
vn
if s £ N. n T
If x
and y are real numbers, define
x V y = max{x
y>-
set
An(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
An
is continuous in H. Observe that An(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) An(x, y) =
[(1 - ny) V 0] [(l- —i-j- |r_ <• 1 4 2(1 - 1) (s - p) n n 3
- 2h(p) | ) V O] vn<
If p £ s , then h(p) £ &n, and one can verify directly that (31) vanishes. If p > sn, then h(p) £ r , and again one can verify directly that (31) vanishes. Thus An(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 ( -—-Xl-XlE) . Therefore, if 0 < y < 1, p = u(x, y) . So, if ^x, y^ satisfies (30) , n £ 2, and £ y £ ~, then
f0(x, y) = (yn(n+l) - n)gn(p) + ((n+1) - yn(n+l))gn+1(p).
Since the coefficients of gn(p) and gn+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 gnCp) to gn+^Cp)» an^ it follows that fQ(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 An 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) An(x, y) =
[Cl-ny) V 0j[(l - -1- |rn - rm + + 2(1 - 1) (sn- |) V 0]vn.
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
h(a).
x - (l-y)s
*
Since h
is weakly increasing, * * x - (lry)s
: h (h(a))
Let g0 = gl^. H is homeomorphic to R , so by [5,.Lemma 2.9,p. 299],- g can be extended to.a continuous function so
g1 : H -> {v G. R3 : | v | = . •
3
Define f^ : H -> R - {0} by setting <
fjCz) = g(z) if z € L
fx(z) = . gx(z) if z € F.
Then, since F and L are closed, f^ is continuous on H. It is easy to 3 2
verify that
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 real-valued 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 real-valued 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 finite-valued. 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). K-In
Proof. Let N be the set of discontinuities of h. By a well-known 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
Theorem 8. Let Y be a separable arcwise
f : H + Ya function of Baire class g(H
X, and
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 yx
Let
• meets Xn, 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 La (X - K^) . By the definition
y € C (y =1= x) such that y meets yx in
interval having its endpoints at
x and
y-
of L , there exists a’
Let In 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 11 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 € Kn)
n >_ n(t), so
Therefore
t £ M. Now
Hence y
C - B = A.
' In. 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 - Kn)]} n c '
n=l co
- ILau (X -C[ Kn)] nC
CO.
= (L a C) u (C - U K ) = L u
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 Xq 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) - aQ(s))y + aQ(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 < (a1(s) - a (s))y + a (s) + 6 } XU U 3 X v v O
[[Q w.]]
Let
*** SOME UNSOLVED PROBLEMS
1. If A is an arbitrary set of type Fa5 in X, does there necessarily exist a real-valued continuous function f defined in H having A as its set of curvilinear convergence? If
*** REFERENCES
1. F. Bagemihl, Curvilinear cluster .sets of arbitrary functions, Proc. Nat. Acad. Sci. U. S. A., ! (1955) 379-382.
2. F. Bagemihl and G. Piranian, Boundary functions for functions defined in a disk, Michigan Math. J.; 8 (1961) 201-207.
3. F. Bagemihl, G. Piranian, and G. S. Young, Intersections of cluster sets, Bul. Inst. Politehn. Iasi, 5 (9) \1959) 29-34.
[[-]].. ,.
4. s. Banach; Uber analytisch darstellbare Operationen in. abstrakten .Raumen, Fund. Math,, .!I_ (1931) 283-295.
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) 292-309/ . 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) 589-612.
9. T. J. Kaczynski, On. a boundary property of continuous functions, ' Michigan Math. J., 13 (1966) 313-320. .
10. J.E. McMillan, Boundary properties of functions continuous in a disc, Michigan Math. J.,'3 (1966) 299-312. .
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)' '284-291,
* 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: [[https://archive.org/download/the-mathematical-work-of-ted-kaczynski/7.%201968%20-%20Note%20on%20a%20Problem%20of%20Alan%20Sutcliffe.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, • • , ab ao are nonnegative integers, let
(ah, • • • , ai, ao)n denote ahnh + • • • + ain + ao.
Thus if O;;;ai;;;n-1 (i = O, • • •, h), then ah,^^, a1, ao are the digits of the number (ah, • • •, ab 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 (g-l=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(Z2 + r* + l)a + (Z2 + r2 + 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~lva — u~lw. 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 7F*. 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 pk (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 y = z = xyx~l, so x and y commute. Otherwise, x= (z — xyx~l)~'(y — 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, June-July, 1964
y e A(n + 1, m[n+ 1], k[n + 1], f [n 4- 1]) C Un+1 c Q(n 4-1, m[n+ 1]),
2tt i 1
and therefore each point of yn has distance less than from y. Now
^+~F 0 as n -* °°; hence, if we set y = {y} U Un=i yn, then y is an arc with
one endpoint at y.
Since Un and Un+1 have a point in common,
f-1(* S(^> pk[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] — 2n 2n”^
** 8. March 1969 - Boundary Functions for Bounded Harmonic Functions
Original PDF: [[https://archive.org/download/the-mathematical-work-of-ted-kaczynski/8.%20March%201969%20-%20Boundary%20Functions%20for%20Bounded%20Harmonic%20Functions.pdf][8. March 1969 - Boundary Functions for Bounded Harmonic Functions.pdf]]
Kaczynski, T.J. 1969. [[https://web.archive.org/web/20190815201940/http:/homepages.rpi.edu/~bulloj/tjk/tjk6.html][Boundary functions for bounded harmonic functions.]] Trans. Am. Math. Soc. 137:203-209.
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 ambiguous-point 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 1/2 —1. sgN
For each xe X=(— 1, 1), let N(x)={seN: — lf
So the part of the line
(9)
that lies in Ho is contained in s(sm, 1, fir). We show that approaches ^(sm) as z -> sm along this line. Substituting the value of x given by (9) into the expression for An, we obtain
An(x,y) = [(1-ny) V 0]
(10)
■n
If smB, then /mm, then ln
rixi-x2i__________________________________ 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, [y1-y2l
* 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
- [[t-k-ted-kaczynski-the-mathematical-work-of-ted-kac-12.pdf][Cropped Source]]
- [[t-k-ted-kaczynski-the-mathematical-work-of-ted-kac-13.pdf][Full Chapter Source]]
*** 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 geo-board) 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 edge-adjacent 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 briefly went back to playing around with pure math equations in his cabin in Montana. He even submitted one paper to a journal in 1976 called [[https://archive.org/details/the-mathematical-work-of-ted-kaczynski/11.%20Unknown%20Date%20-%20Four-Digit%20Numbers%20that%20Reverse%20Their%20Digits%20When%20Multiplied/][Four-Digit Numbers that Reverse Their Digits When Multiplied]] which he claimed proved an earlier posited solution:[124]
[124] [[https://www.thetedkarchive.com/library/the-unabom-task-force-a-review-and-compilation-of-the-writings-of-ted-kaczynski][A Review and Compilation of the Writings of Ted Kaczynski]]
... Some time ago - (Last Nov. or Dec.) I submitted a mathematical paper for publication, and I am rather ashamed of this. Not because of any idea that the paper will advance technical progress — I feel confident that it will never have any practical applications, direct or indirect — but because it represents, to a certain degree, a personal surrender to one of the escape mechanisms which keep people distracted so that they can forget the purposelessness, subordination, and indignity of life in a technological society ...
Here's how Ted explained the paper in relation to his other work:[125]
(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.
SPV Laboratories:[126]
[125] Conversations with the producer of this great documentary on Ted's math career: [[https://www.youtube.com/watch?v=wD4xrnzKN1Y][Ted Kaczynski's PhD Thesis]]
His perspective of intellectual tension with Piranian is interesting and I maybe sensed it reading their papers. Piranian and collaborators were in the complex analysis space, it seemed like they were developing language for talking about boundary functions from their work on cluster sets, and all of it seemed geared toward complex analysis applications. Ted's papers treated boundary functions like its own distinct discipline and I don't think he mentions cluster sets once. So it's funny that George seemed to urge him to prove this statement (a) about "functions admitting disjoint arcs", a concept originating from cluster sets, while Ted proved this statement (b) first which is primarily a boundary function concept:
Statement (b) that continuous half plane functions admit honorary Baire class 2 functions (functions differing from baire class 1 on at most a countable subset) shows up in his thesis and it seems like he proved a version of that fairly early in 1964 based on that letter.
As far as statement (a) I was not too far in the weeds of cluster sets but tbh it sounds like something that should have been proved already. Wouldn't take my word for it over his though.
Finally, here's a glimpse into Ted's headspace when writing it, from a journal entry at the time:[127]
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.
[126] [[https://www.thetedkarchive.com/library/ted-kaczynski-truth-versus-lies-original-draft][Truth versus Lies (Original Draft)]]
[127] [[https://www.thetedkarchive.com/library/ted-kaczynski-journal-1-of-4-from-series-vii-1984-1986][Journal #1 of 4 from Series VII (1984-1986)]]
Would the paper he wrote in his cabin get published today? Or would all the formulas and explanations have already been well covered by other papers?
SPV Laboratories:[130]
[130] Conversations with the producer of this great documentary on Ted's math career: [[https://www.youtube.com/watch?v=wD4xrnzKN1Y][Ted Kaczynski's PhD Thesis]]
Can't speak to whether or not his stuff would get published today... if my paper gets published then that would be in his favor. The things complex analysts were interested in during the mid 20th century are not so much in vogue anymore, in fact it seems like they fell out of style quite fast based on what I've seen in the literature. So most things Ted would have been working on post-Berkeley would have been non-trivial but also probably not discovered sooner by anyone due to a lack of interest. Not implying anything negative about his work, just that certain inquiries in math come and go.
** 13. Four-Digit Numbers that Reverse Their Digits When Multiplied
Original PDF: [[https://archive.org/download/the-mathematical-work-of-ted-kaczynski/11.%20Unknown%20Date%20-%20Four-Digit%20Numbers%20that%20Reverse%20Their%20Digits%20When%20Multiplied.pdf][11. Unknown Date - Four-Digit Numbers that Reverse Their Digits When Multiplied.pdf]]
FOUR-DIGIT NUMBERS THAT REVERSE THEIR
DIGITS WHEN MULTIPLIED
T. J. KACZYNSKI
If n
** 14. Handwritten Math equations and procedures
Original PDF: [[https://archive.org/download/the-mathematical-work-of-ted-kaczynski/12.%20Unknown%20date%20-%20Handwritten%20Math%20equations%20and%20procedures.pdf][12. Unknown date - Handwritten Math equations and procedures.pdf]]
* Ted's Work from Prison
He would sometimes write hard questions for the teachers of kids who would write to him in prison:[128]
[128] [[https://www.thetedkarchive.com/library/ted-kaczynski-math-tutor][Ted Kaczynski, Math Tutor]]
Now I'm going to play a really nasty trick on your teacher. I'm going to give you a problem to give her, and if she doesn't get it right, you be sure to give her an F.
SPV Laboratories:[129]
[129] Conversations with the producer of this great documentary on Ted's math career: [[https://www.youtube.com/watch?v=wD4xrnzKN1Y][Ted Kaczynski's PhD Thesis]]
One of these letters had some conjectures based on a question he posed in his thesis. He claims to have had proofs for three of these. I am able to prove two; I plan on publishing my results soon.
Finally, here is a link to that paper:
[[https://arxiv.org/abs/2311.16210][An uncountable union of line segments with null two-dimensional measure by Parker Kuklinski]]