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\title{The Mathematical Work of Ted Kaczynski}
\date{Mar 16, 2020}
\author{Ted Kaczynski}
\subtitle{}
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\chapter{Original PDFs}
\begin{itemize}
\item\relax
\href{https://archive.org/download/the-mathematical-work-of-ted-kaczynski/1.\%20June\%201964\%20-\%20Another\%20Proof\%20of\%20Wedderburn\%E2\%80\%99s\%20Theorem.pdf}{1. June 1964 - Another Proof of Wedderburn’s Theorem.pdf}
\item\relax
\href{https://archive.org/download/the-mathematical-work-of-ted-kaczynski/2.\%201964\%20Distributivity\%20and\%20\%28\%E2\%88\%921\%29x\%20\%3D\%20\%E2\%88\%92x\%20\%28Advanced\%20Problem\%205210\%29.pdf}{2. 1964 Distributivity and (−1)x = −x (Advanced Problem 5210).pdf}
\item\relax
\href{https://archive.org/download/the-mathematical-work-of-ted-kaczynski/3.\%20Distributivity\%20and\%20\%28\%E2\%88\%921\%29x\%20\%3D\%20\%E2\%88\%92x\%20\%28Advanced\%20Problem\%205210\%2C\%20with\%20Solution\%20by\%20Bilyeu\%2C\%20R.G.\%29.pdf}{3. Distributivity and (−1)x = −x (Advanced Problem 5210, with Solution by Bilyeu, R.G.).pdf}
\item\relax
\href{https://archive.org/download/the-mathematical-work-of-ted-kaczynski/4.\%201965\%20Boundary\%20Functions\%20for\%20Functions\%20Defined\%20in\%20a\%20Disk.pdf}{4. 1965 Boundary Functions for Functions Defined in a Disk.pdf}
\item\relax
\href{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}
\item\relax
\href{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}
\item\relax
\href{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}
\item\relax
\href{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}
\item\relax
\href{https://archive.org/download/the-mathematical-work-of-ted-kaczynski/9.\%20July\%201969\%20-\%20Boundary\%20functions\%20and\%20sets\%20of\%20curvilinear\%20convergence\%20for\%20continuous\%20functions.pdf}{9. July 1969 - Boundary functions and sets of curvilinear convergence for continuous functions.pdf}
\item\relax
\href{https://archive.org/download/the-mathematical-work-of-ted-kaczynski/10.\%20Nov\%201969\%20-\%20The\%20Set\%20of\%20Curvilinear\%20Convergence\%E2\%80\%A6.pdf}{10. Nov 1969 - The Set of Curvilinear Convergence\dots{}.pdf}
\item\relax
11. Problem 787
\item\relax
\href{t-k-ted-kaczynski-the-mathematical-work-of-ted-kac-12.pdf}{12. A Match Stick Problem (Problem 787, with Solutions}
\item\relax
\href{https://archive.org/download/the-mathematical-work-of-ted-kaczynski/11.\%20Unknown\%20Date\%20-\%20Four-Digit\%20Numbers\%20that\%20Reverse\%20Their\%20Digits\%20When\%20Multiplied.pdf}{13. Unknown Date - Four-Digit Numbers that Reverse Their Digits When Multiplied.pdf}
\item\relax
\href{https://archive.org/download/the-mathematical-work-of-ted-kaczynski/12.\%20Unknown\%20date\%20-\%20Handwritten\%20Math\%20equations\%20and\%20procedures.pdf}{14. Unknown date - Handwritten Math equations and procedures.pdf}
\end{itemize}
\chapter{Introduction by Jørgen Veisdal}
\textbf{The Mathematics of Ted Kaczynski}
\textbf{Disclaimer:} As should be fairly evident, this essay is in no way meant to glorify Ted Kaczynski. Rather, it was written with two goals in mind: 1. To orient to reality some of the myths of Kaczynski's "genius" and 2. To illustrate yet another example of a mathematician whose abstract endeavours ultimately defeated him.
Before terrorist Theodore John Kaczynski (1942-) began sending mail-bombs to faculty members at various American universities, he had a promising career in mathematics. In particular, between 1964–69, he published a total of six single-authored research papers in renowned mathematical journals, including \emph{The American Mathematical Monthly} and \emph{Proceedings of the American Mathematical Society.}
The young Kaczynski did work in analysis, specifically geometric function theory in the narrow subfield of boundary values of continuous functions. The purpose of this article is to give an introduction to this work.
\section{Education (1958–67)}
Kaczynski grew up in Illinois, where he attended Sherman Elementary School and Evergreen Park Central Junior High school. At the age of 10 years old, his IQ was evaluated to be 167, and so he skipped the sixth grade (Chicago Tribune, 2017), an event later described as pivotal to his development (Chase, 2004):
\begin{quote}
“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.”
\end{quote}
\section{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 \emph{“shy”}, \emph{“quiet”} and \emph{“a loner”} who \emph{“never talked to anyone“} (Song, 2012):
\begin{quote}
“He would just rush through the suite, go into his room, and slam the door [\dots{}] When we would go into his room there would be piles of books and uneaten sandwiches that would make the place smell”
\end{quote}
His personality notwithstanding, Kaczynski’s talent was however still recognized among his Harvard peers, one of which in 2012 stated:
\begin{quote}
“It’s just an opinion — but Ted was brilliant [\dots{}]. He could have become one of the greatest mathematicians in the country”
\end{quote}
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).
\section{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:
\begin{quote}
“[My] memories of the University of Michigan are \textbf{not} pleasant [\dots{}] 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”
\end{quote}
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 \emph{“}\emph{Boundary Functions}\emph{”} 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
\begin{quote}
“The best I have ever directed”
\end{quote}
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
\begin{quote}
“I would guess that maybe 10 or 12 men in the country understood or appreciated it”
\end{quote}
Another, Peter Duren, stated
\begin{quote}
“He was really an unusual student”
\end{quote}
Kaczynski at UCB in 1967 (Photo: Wikimedia Commons)
\section{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 \emph{“straight from the textbook and refused to answer questions”}.
He resigned on June 30th, 1969 without explanation.
\section{Work (1964–69)}
\section{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:
\begin{itemize}
\item\relax
\textbf{Kaczynski, T.J. (1964).} “Another proof of Wedderburn’s theorem”. \emph{The American Mathematical Monthly} 71(6), pp. 652–653.
\end{itemize}
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.
\section{Boundary Functions}
Kaczynski’s Ph.D. dissertation concerned boundary values of continuous functions and was entitled, simply
\begin{itemize}
\item\relax
\textbf{Kaczynski, T.J. (1967).} \emph{Boundary Functions}. Ann Arbor: University of Michigan.
\end{itemize}
\begin{quote}
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.
\end{quote}
\begin{quote}
\textbf{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)
\end{quote}
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:
\begin{quote}
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.
\end{quote}
Kaczynski would publish five journal papers related to the work from his dissertation between 1965–69:
\begin{itemize}
\item\relax
\textbf{Kaczynski, T.J. (1965)}. “Boundary functions for functions defined in a disk”. \emph{Journal of Mathematics and Mechanics.} 14(4), pp. 589–612.
\item\relax
\textbf{Kaczynski, T.J. (1966)}. “On a boundary property of continuous functions”. \emph{Michigan Math. J.} 13, pp. 313–320.
\item\relax
\textbf{Kaczynski, T.J. (1969)}. “The set of curvilinear convergence of a continuous function defined in the interior of a cube”. \emph{Proceedings of the American Mathematical Society} 23(2), pp. 323–327.
\item\relax
\textbf{Kaczynski, T.J. (1969)}. “Boundary functions and sets of curvilinear convergence for continuous functions”. \emph{Transactions of the American Mathematical Society.} 141, pp. 107–125.
\item\relax
\textbf{Kaczynski, T.J. (1969)}. “Boundary functions for bounded harmonic functions”. \emph{Transactions of the American Mathematical Society.} 137, pp. 203–209.
\end{itemize}
\section{The Distributivity Problem}
The only other trace of Kaczynski in a mathematical journal is two notes in the American Monthly in 1964 and 65:
\begin{itemize}
\item\relax
\textbf{Kaczynski, T.J. (1964).} “Distributivity and (−1)x = −x (Advanced Problem 5210)”. \emph{The American Mathematical Monthly.} 71(6), pp. 689.
\item\relax
\textbf{Kaczynski, T.J. (1965)}. “Distributivity and (−1)x = −x (Advanced Problem 5210, with Solution by Bilyeu, R.G.)”. \emph{The American Mathematical Monthly} 72(6), pp. 677–678.
\end{itemize}
In the first note, Kaczynski proposes the following problem, concerning group theory:
\begin{quote}
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\dots{}+1 (n times). Prove that, for all x ∈ K, (-1)x = -x.
\end{quote}
In the second note, the solution to the problem is — somewhat dismissively — provided by R. G. Bilyeu:
\begin{quote}
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.
\end{quote}
\section{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):
\begin{quote}
[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.
\end{quote}
In another 1996 article, in the Los Angeles Times article, Professor Donald Rung similarly expressed:
\begin{quote}
“The field that Kaczynski worked in doesn’t really exist today [\dots{}]. 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.”
\end{quote}
\chapter{An Advanced Explanation of His Breakthrough by Lara Pudwell}
\textbf{Original PDF}: \href{https://faculty.valpo.edu/lpudwell/papers/mm005281.pdf}{Digit Reversal Without Apology.pdf}
\textbf{Digit Reversal Without Apology}
Lara Pudwell Rutgers University Piscataway, NJ 08854 \href{mailto:lpudwell@math.rutgers.edu}{lpudwell@math.rutgers.edu}
In A Mathematician’s Apology\footnote{F. Bagemihl, Curvilinear cluster sets of arbitrary functions, \emph{Proc. Nat, Acad, Sci. U. S. A.>} 4 (1955) 379-382.} 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\footnote{F. Bagemihl \& G. Piranian, Boundary functions for functions defined in a disk, \emph{Michigan Math, J.,} 8 (1961) 201-207.} 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
\emph{k(ahn\textsuperscript{h}} + a\textsubscript{h-1}n\textsuperscript{h-}1 + • • • + \emph{a}\emph{1 n} + a\textsubscript{0}) = a\textsubscript{0}n\textsuperscript{h} + a1n\textsuperscript{h-1} + • • • + \emph{a}\emph{\textsubscript{h-1}n} + ah
with n \emph{>2, 1 < k < n, 0 < ai} \emph{< n —} 1 for all \emph{i,} a\textsubscript{0} = 0, a\textsubscript{h} = 0. We shall refer to such an integer \emph{a}0\emph{\dots{}a}h as an (\emph{h} + 1)-digit solution for \emph{n} and write \emph{k}(\emph{a}\textsubscript{h}\emph{, a}\textsubscript{h-1}\emph{, \dots{}, a}\textsubscript{1}\emph{, a}\textsubscript{0})\textsubscript{n} = (\emph{a}\textsubscript{0}\emph{, a}\textsubscript{1}\emph{, \dots{}, a}\textsubscript{h-1}\emph{, a}\textsubscript{h})\textsubscript{n}. For example, 8712 and 9801 are 4-digit solutions in base \emph{n} = 10 for \emph{k} = 4 and \emph{k} = 9 respectively. After characterizing all 2-digit solutions for fixed \emph{n} and generating parametric solutions for higher digit solutions, Sutcliffe left the following open question: Is there any base \emph{n} for which there is a 3-digit solution but no 2-digit solution?
Two years later T. J. Kaczynski\footnoteB{Better known for other work.}\footnote{S. Banach, Uber analytisch darstellbare Operationen in abstrakten Raumen, \emph{Fund, Math.,} 17 (1931) 283-295.} answered Sutcliffe’s question in the negative. His elegant proof showed that if there exists a 3-digit solution for \emph{n}, then deleting the middle digit gives a 2-digit solution for \emph{n}. Together with Sutcliffe’s work, this proved that there exists a 2-digit solution for \emph{n} if and only if there exists a 3-digit solution for \emph{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 \emph{>3.}
\section{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)\textsubscript{n} = (e, d, b, a)\textsubscript{n} a 4-digit solution for n?
First, following Kaczynski, let p = n + 1. We have
k(an\textsuperscript{4} + bn\textsuperscript{3} + cn\textsuperscript{2} + dn + e) = en\textsuperscript{4} + dn\textsuperscript{3} + cn\textsuperscript{2} + bn + a. (1)
Reducing this equation modulo p, we obtain
\emph{k(a — b} + c — d + e) = e — d + c — b + \emph{a} = \emph{a — b} + c — d + e mod p.
Thus, (k — 1)(a — b + c — d + e) = 0 mod \emph{p,} and
p \textbar{} (k — 1)(a — b+ c — d + e). (2)
Ifp \textbar{} (k—1), then k—1 > p, which is impossible because k < n. Therefore, p \textbar{} (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 \emph{a — b} + c — d + e = \emph{fp}\emph{,} where f G \{—1,0,1,2\}. Substituting c = -a + b + d - e + fp into equation 1 gives:
k[n\textsuperscript{2} (n\textsuperscript{2} — 1)a + n\textsuperscript{2}(n + 1)b + fpn\textsuperscript{2} + n(n + 1)d — (n\textsuperscript{2} — 1)e]
= n\textsuperscript{2}(n\textsuperscript{2} — 1)e + n\textsuperscript{2}(n + 1)d + fpn\textsuperscript{2} + n(n + 1)b — (n\textsuperscript{2} — 1)a.
After substituting for p, dividing by n + 1, and rearranging, one sees that k [an\textsuperscript{3} + (b — a + f)n\textsuperscript{2} + (d — e)n + e] = en\textsuperscript{3} + (d — e + f)n\textsuperscript{2} + (b — a)n + a. Indeed, this is a 4-digit solution for n if f = 0, b — \emph{a >} 0, and d — e \emph{>} 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.
\section{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.
\section{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 \emph{>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 \textbar{} (n - a).
Proof. We have
\emph{(n} \emph{- a)n}\emph{\textsuperscript{4}} \emph{+ (n} \emph{- a} \emph{- 1)n}\emph{\textsuperscript{3}} \emph{+ (n} \emph{- 1)n}\emph{\textsuperscript{2}} \emph{+ (a} \emph{- 1)n + a} an\textsuperscript{4} + (a - 1)n\textsuperscript{3} + (n - 1)n\textsuperscript{2} + (n - a - 1)n + (n - a)
\emph{(n} \emph{- a)(n}\emph{\textsuperscript{4}} \emph{+ n}\emph{\textsuperscript{3}} \emph{- n} \emph{- 1)} n - a a(n\textsuperscript{4} + n\textsuperscript{3} - n - 1) a \textsuperscript{,}
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)\textsubscript{n} = (n - a,n - a - 1, a - 1, a)\textsubscript{n}
is a 4-digit solution for n.
\emph{Proof.} By Theorem 1, n-a G N. Now
\emph{(n} \emph{- a)n}\emph{\textsuperscript{3}} \emph{+ (n} \emph{- a} \emph{- 1)n}\emph{\textsuperscript{2}} \emph{+ (a} \emph{- 1)n + a}
an\textsuperscript{3} + (a - 1)n\textsuperscript{2} + (n - a - 1)n + (n - a)
\emph{(n} \emph{- a)(n}\emph{\textsuperscript{3}} \emph{+ n}\emph{\textsuperscript{2}} \emph{- n} \emph{- 1)} n - a
a(n\textsuperscript{3} + n\textsuperscript{2} - n - 1) a \textsuperscript{.}
□
These 4-digit solutions were first described by Klosinski and Smolarski\footnote{P. T. Church, Ambiguous points of a function homeomorphic inside a sphere, \emph{Michigan Math. J.,} 4 (1957) 155-156.} 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. \emph{There is a} 4\emph{-digit solution and a} 5\emph{-digit solution for every} n \emph{>} 3.
\emph{Proof.} Let \emph{a} = 1 in the statements of Theorem 1 and Theorem 2 above. □
\section{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 \href{http://www.math.rutgers.edu/\%7Elpudwell/maple.html}{http:\Slash{}\Slash{}www.math.rutgers.edu\Slash{}\textasciitilde{}lpudwell\Slash{}maple.html}.
\section{Acknowledgment}
Thank you to Doron Zeilberger for suggesting this project.
\section{References}
\bigskip
\part{Ted's Work as a Michigan PhD Student}
\chapter{1. June 1964 - Another Proof of Wedderburn’s Theorem}
\textbf{Original PDF}: \href{https://archive.org/download/the-mathematical-work-of-ted-kaczynski/1.\%20June\%201964\%20-\%20Another\%20Proof\%20of\%20Wedderburn\%E2\%80\%99s\%20Theorem.pdf}{1. June 1964 - Another Proof of Wedderburn’s Theorem.pdf}
Kaczynski, T.J. 1964. Another proof of Wedderburn's theorem. \emph{Am. Math. Month.} 71:652-653.
\textbf{Another Proof of Wedderburn's Theorem}
\textbf{Author(s): T. J. Kaczynski}
\textbf{Source: \emph{The American Mathematical Monthly,} Vol. 71, No. 6 (Jun. - Jul., 1964), pp. 652-653}
\textbf{Published by: Taylor \& Francis, Ltd. on behalf of the Mathematical Association of America}
\textbf{Stable URL:} \href{https://www.jstor.org/stable/2312328}{\textbf{https:\Slash{}\Slash{}www.jstor.org\Slash{}stable\Slash{}2312328}}
\bigskip
The author holds a NSF Coop fellowship. The preparation of this manuscript was partly supported by the NSF (G 24335).
\textbf{References}
1. E. A. Coddington and Norman Levinson, Theory of ordinary differential equations, McGraw-Hill, New York, 1955.
2. Witold Hurewicz, Lectures on ordinary differential equations, Wiley and MIT Press, New York, 1958.
\textbf{ANOTHER PROOF OF WEDDERBURN’S THEOREM}
\textbf{T. J.} Kaczynski, Evergreen Park, Illinois
In 1905 Wedderburn proved that every finite skew field is commutative. At least seven proofs of this theorem (not counting the present one) are known. See\footnote{F. Bagemihl, Curvilinear cluster sets of arbitrary functions, \emph{Proc. Nat, Acad, Sci. U. S. A.>} 4 (1955) 379-382.},\footnote{F. Bagemihl \& G. Piranian, Boundary functions for functions defined in a disk, \emph{Michigan Math, J.,} 8 (1961) 201-207.},\footnote{F. Hausdorff, Uber halbstetige Funktionen und deren Verallgemeinerung, \emph{Math. Z.,} 5 (1919) 292-309.} (Part Two, p. 206 and Exercise 4 on p. 219),\footnote{M. H. A. Newman, \emph{Elements of the topology of plane sets of points,} Cambridge University Press, 1961.} (two proofs), and\footnote{H. Tietze, Uber stetige Kurven, Jordansche Kurvenbogen und geschlossene Jordansche Kurven, \emph{Math. Z.,} 5 (1919), 284-291.}. Unlike these proofs, the proof to be given here is group-theoretic, in the sense that the only non-group-theoretic concepts employed are of an elementary nature.
Lemma. \emph{Let q be a prime. Then the congruence} Z\textsuperscript{2}+r\textsuperscript{2}= — 1 (mod \emph{q) has a solution t, r with t\^{}O} (mod \emph{q).}
\emph{Proof.} If —1 is a quadratic residue, take r = 0 and choose \emph{t} appropriately. Assume — 1 is a nonresidue. Then any nonresidue can be written in the form \emph{— s\textsuperscript{2}} (mod \emph{q)} with s\^{}O. If \emph{t\textsuperscript{2}+r\textsuperscript{2}} is ever a nonresidue for some \emph{t,} r, set \emph{t\textsuperscript{2}+r\textsuperscript{2}} s— s\textsuperscript{2}, and we have (\Slash{}5\textasciitilde{}\textsuperscript{1})\textsuperscript{2} + (r5”\textsuperscript{1})\textsuperscript{2} = — 1. (Throughout this note, x\textsuperscript{-1} denotes that integer for which xx\textsuperscript{-1} = l (mod \emph{q).)} On the other hand, if \emph{t\textsuperscript{2}+r\textsuperscript{2}} is always a residue, then the sum of any two residues is a residue, so —l=g—1 = 1 + 14- • • • + 1 is a residue, contradicting our assumption.
\emph{Proof of the theorem.} Let \emph{F} be our finite skew field, E* its multiplicative group. Let 5 be any Sylow subgroup of F*, of order, say, \emph{p\textsuperscript{a}.} Choose an element \emph{g} of order \emph{p} in the center of 5. If some \emph{h\^{}S} generates a subgroup of order \emph{p} different from that generated by g, then \emph{g} and \emph{h} generate a commutative field containing more than \emph{p} roots of the equation x\textsuperscript{p}=l, an impossibility. Thus 5 contains only one subgroup of order \emph{p} and hence is either a cyclic or a generalized quaternion group (\footnote{S. Banach, Uber analytisch darstellbare Operationen in abstrakten Raumen, \emph{Fund, Math.,} 17 (1931) 283-295.} p. 189).
If S is a generalized quaternion group, then 5 contains a quaternion subgroup generated by two elements \emph{a} and \&, both of order 4, where \emph{ba — a\textasciitilde{}\^{}b.} Now \emph{a\textsuperscript{2}} generates a commutative field in which the only roots of the equation \emph{x\textsuperscript{2} —} 1 or (x+l)(x—1) =0 are ±1, so since (a\textsuperscript{2})\textsuperscript{2} = l, we have
(1) \emph{a\textsuperscript{2} =} - 1.
Hence \emph{a\^{} — a\textsuperscript{2} — —a\textsubscript{t}} so
(2) \emph{ba = — ab.}
This content downloaded from 82.46.120.253 on Thu, 21 Feb 2019 17:17:06 UTC All use subject to \href{https://about.jstor.org/terms}{https:\Slash{}\Slash{}about.jstor.org\Slash{}terms}
Similarly,
(3) 5\textsuperscript{2} = - 1.
Taking \emph{q —} characteristic of \emph{F} (\#-l = 0), choose \emph{t} and \emph{r} as specified in the lemma. Using relations (1), (2), (3), we have
\emph{(\Slash{} + ra} + 5)(r\textsuperscript{2} + 1 + \emph{rta + tb)} = r(\Slash{}\textsuperscript{2} + r\textsuperscript{2} + l)a + (\Slash{}\textsuperscript{2} + r\textsuperscript{2} + 1)5 = 0.
One of the factors on the left must be 0, so for some numbers \emph{u, v, w, u} 0 (mod g), we have w+\^{}a+\^{}5 = 0, or \emph{b= -u\^{}wa-u\^{}w.} So \emph{b} commutes with \emph{a,} a contradiction. We conclude that 5 is not a generalized quaternion group, so 5 is cyclic.
Thus every Sylow subgroup of F* is cyclic, and F* is solvable (\footnote{P. T. Church, Ambiguous points of a function homeomorphic inside a sphere, \emph{Michigan Math. J.,} 4 (1957) 155-156.}, pp. 181— 182). Let \emph{Z} be the center of F* and assume Z\^{}F*. Then \emph{F*\Slash{}Z} is solvable, and its Sylow subgroups are cyclic. Let \emph{A\Slash{}Z} (with ZC\^{}4) be a minimal normal subgroup of \emph{F*\Slash{}Z. A\Slash{}Z} is an elementary abelian group of order (F prime), so since the Sylow subgroups of F*\emph{\Slash{}Z} are cyclic, \emph{A\Slash{}Z} is cyclic. Any group which is cyclic modulo its center is abelian, so \emph{A} is abelian. Let \emph{x} be any element of F*, \emph{y} any element of \emph{A.} Since \emph{A} is normal, \emph{xyx\textasciitilde{}\textsuperscript{x}\^{}A,} and (l+x)y = z(l+x) for some \emph{zf\^{}A.} An easy manipulation shows that \emph{y — z — zx — xy = (z — xyx\textasciitilde{}\textsuperscript{l})x.}
If \emph{y — z = z —} xyx\textsuperscript{-1} = 0, then \emph{y = z = xyx\textasciitilde{}\textsuperscript{1},} so \emph{x} and \emph{y} commute. Otherwise, x= \emph{(z — xyx\textasciitilde{}\textsuperscript{1}')\textasciitilde{}\textsuperscript{1}(y — z).} But \emph{A} is abelian, and 2, y, xyx\textsuperscript{-1}(E\^{}4, so \emph{x} commutes with \emph{y.} Thus we have proven that \emph{A} is contained in the center of F*, a contradiction.
\textbf{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.
\section{A NOTE ON PRODUCT SYSTEMS OF SETS OF NATURAL NUMBERS}
T. G. McLaughlin, University of California at Los Angeles
In this note, we apply a slight twist to a trick exploited about twelve years ago by J. C. E. Dekker (\emph{[2} ]), our purpose being to expose a couple of elementary facts about nonempty, countable "product systems” of infinite sets of natural numbers which are, at the same time, "finite symmetric difference systems.” We proceed in terms of the following definitions.
Definition. \emph{By a product system of subsets of N (N the natural numbers), we mean a collection of subsets of N which contains, along with any two of its members, their intersection.}
\bigskip
\chapter{2. 1964 - Distributivity and (−1)x = −x (Advanced Problem 5210)}
\textbf{Original PDF}: \href{https://archive.org/download/the-mathematical-work-of-ted-kaczynski/2.\%201964\%20Distributivity\%20and\%20\%28\%E2\%88\%921\%29x\%20\%3D\%20\%E2\%88\%92x\%20\%28Advanced\%20Problem\%205210\%29.pdf}{2. 1964 Distributivity and (−1)x = −x (Advanced Problem 5210).pdf}
\textbf{Kaczynski, T.J. (1964).} “Distributivity and (−1)x = −x (Advanced Problem 5210)”. \emph{The American Mathematical Monthly.} 71(6), pp. 689.
\textbf{ADVANCED PROBLEMS}
\emph{All solutions of Advanced Problems should be sent to} J. \emph{Barlaz, Rutgers} - \emph{The State University, New Brunswick, N.J. Solutions of Advanced Problems in this issue should be submitted on separate, signed sheets and should be mailed before December 31, 1964 .}
5210. \emph{Proposed by T.} J. \emph{Kaczynski, Evergreen Park, Illinois}
Let \emph{K} be an algebraic system with two binary operations (one written additively, the other multiplicatively), satisfying:
1. \emph{K} is an abelian group under addition,
2. \emph{K} - \{O\} is a group under multiplication, and
3. \emph{x(y} + \emph{z)} == \emph{xy} + \emph{xz} for all \emph{x,y,z} EK.
Suppose that for some \emph{n,} 0 \texttt{= 1 + 1 + ... + 1 ( n times). Prove that, for all x eK, (-l)x =} \emph{-x.}
\chapter{3. 1964 - Distributivity and (−1)x = −x (Advanced Problem 5210, with Solution by Bilyeu, R.G.)}
\textbf{Original PDF}: \href{https://archive.org/download/the-mathematical-work-of-ted-kaczynski/3.\%20Distributivity\%20and\%20\%28\%E2\%88\%921\%29x\%20\%3D\%20\%E2\%88\%92x\%20\%28Advanced\%20Problem\%205210\%2C\%20with\%20Solution\%20by\%20Bilyeu\%2C\%20R.G.\%29.pdf}{3. Distributivity and (−1)x = −x (Advanced Problem 5210, with Solution by Bilyeu, R.G.).pdf}
\textbf{Kaczynski, T.J. (1965)}. “Distributivity and (−1)x = −x (Advanced Problem 5210, with Solution by Bilyeu, R.G.)”. \emph{The American Mathematical Monthly} 72(6), pp. 677–678.
Distributivity and ( -1 \emph{)x} == \emph{-x}
5210 [1964, 689]. \emph{Proposed by T.} J. \emph{Kaczynski, Evergreen Park, Illinois}
Let \emph{K} be an algebraic system with two binary operations (one written additively, the other multiplicatively), satisfying:
1. \emph{K} is an abelian group under addition,
2. \emph{K} - \{O\} is a group under multiplication, and
3. \emph{x(y} + \emph{z)} == \emph{xy} + \emph{xz} for all \emph{x,y,z} EK.
Suppose that for some \emph{n,} 0 \texttt{= 1 + 1 + ... + 1 ( n times). Prove that, for all x eK, (-l)x =} \emph{-x.}
\emph{Solution by R. G. Bilyeu, North Texas State University.} The last part of the hypothesis is unnecessary. If \emph{z} denotes -1, then \emph{z} + \emph{z} + \emph{z z} \texttt{= z (1 + 1 + z) =} \emph{z,} so \emph{z(?)} \texttt{= 1. Now z(x + zx) =} \emph{zx} + \emph{x} \texttt{= x + zx, so either x + zx =} 0 or \emph{z} \texttt{= 1. In either case z(?) =} \emph{-x.}
Also solved by Carol Avelsgaard, Richard Bourgin, Robert Bowen, Joel Brawley, Jr., F. P. Callahan, M. M. Chawla (India), R. A. Cunninghame-Green (England), M. J. DeLeon, M. Edelstein, N. J. Fine, Harvey Friedman, Anton Glaser, M. G. Greening (Australia), A. G. Heinicke, Sidney Heller, G. A. Heuer, Stephen Hoffman, K. G. Johnson, A. J. Karson, Max Klicker, Kwangil Koh, C. C. Lindner, C. R. MacCluer, H. F. Mattson, C. J. Maxson, R. V. Moddy, Jose Morgado (Brazil), W. L. Owen, Jr., P. R. Parthasarathy (India), Harsh Pittie, Kenneth Rogers, Toru Saito (Japan), Camilio Schmidt, Leonard Shapiro, Frank A. Smith, George Van Zwalenberg, W. C. Waterhouse, Kenneth Yanosko, and the proposer.
\bigskip
\chapter{4. 1965 - Boundary Functions for Functions Defined in a Disk}
\textbf{Original PDF}: \href{https://archive.org/download/the-mathematical-work-of-ted-kaczynski/4.\%201965\%20Boundary\%20Functions\%20for\%20Functions\%20Defined\%20in\%20a\%20Disk.pdf}{4. 1965 Boundary Functions for Functions Defined in a Disk.pdf}
Kaczynski, T.J. 1965. \href{https://web.archive.org/web/20190815201940/http:/homepages.rpi.edu/\%7Ebulloj/tjk/tjk2.html}{Boundary functions for functions defined in a disk.} \emph{J. Math. and Mech.} 14(4):589-612.
MR0176080 Kaczynski, T. J. Boundary functions for function defined in a disk. J. Math. Mech. 14 1965 589.612. (Reviewer: C. Tanaka) 30.62
\section{Explanation by John D. Bullough}
Let D denote the unit disk \textbar{}z\textbar{} < 1, C its boundary, and let f(z) be any function that is defined in D and takes its values in some metric space S. Then a boundary function for f is a function t on C such that for every x ( C there exists an arc v at x with
\begin{quote}
lim f(z) = t(x).
z -> x
z ( v
\end{quote}
The author proves several theorems on boundary functions in the following four cases: (1) f(z) a homeomorphism of D onto D, (2) f(z) a continuous function, (3) f(z) a Baire function and (4) f(z) a measurable function. These theorems include answers to two questions raised by Bagemihl and Piranian.
Theorem 1 states that if f(z) is a homeomorphism of D onto D, then there exists a countable set N such that t\textbar{}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\textbar{}S - N = g\textbar{}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 D\textsuperscript{0}: I(z) > 0, and its boundary C\textsuperscript{0}: 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 D\textsuperscript{0}, 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 D\textsuperscript{0} 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 C\textsuperscript{0}, there exists a function f on D\textsuperscript{0} such that f(z) = 0 almost everywhere and t is a boundary function for f. The paper concludes with some remarks concerning extensions of these theorems into three dimensions.
\section{Article by Ted}
\emph{Boundary Functions for Functions Defined in a DisB}
T. J. KACZYNSKI
Communicated \emph{by} F. Bagemihl
\subsection{1. Introduction}
Throughout this paper \emph{D} will denote the open unit disk (in two-dimensional Euclidean space) and \emph{C} will denote its boundary, the unit circle. Bagemihl and Piranian\footnote{F. Bagemihl \& G. Piranian, Boundary functions for functions defined in a disk, \emph{Michigan Math, J.,} 8 (1961) 201-207.} have introduced the following definition.
\emph{Definition.} If \emph{x} e C, an \emph{arc at x is \&} simple arc \emph{y} having one endpoint at \emph{x} such that \emph{y — \{x\}} \emph{C} \emph{D.} Let \Slash{} be any function that is defined in \emph{D} and takes its values in some metric space S. Then a \emph{boundary junction} for f is a function \emph{
. Then} acc (A) \emph{and B have at most two points in common.} \emph{Proof.} Assume that \emph{p\textsubscript{r} , p\textsubscript{2} , p\textsubscript{3}} are three distinct points of acc (A) A \emph{B} and derive a contradiction. Let 7* be an arc joining \emph{pi} to a point \emph{qi z A,} with \emph{\{Pi\} A (i} = 1, 2, 3). Let \emph{y} be an arc in \emph{A} joining and \emph{q\textsubscript{2}} . Putting , 7\textsubscript{2} and 7 together, we obtain an arc T joining pj to \emph{p\textsubscript{2},} with r — \emph{\{p\textsubscript{r}, p\textsubscript{2}\}} \emph{C} \emph{A. We} can assume F is a simple arc, for if r is not simple, and \emph{p\textsubscript{2}} can be joined by some simple arc F\textsuperscript{z} £ T (see\footnote{H. Tietze, Uber stetige Kurven, Jordansche Kurvenbogen und geschlossene Jordansche Kurven, \emph{Math. Z.,} 5 (1919), 284-291.}). Let \emph{L\textsubscript{x} , L\textsubscript{2}} be the two open arcs of \emph{C} determined by the pair of points \emph{p\textsubscript{r} , p\textsubscript{2}} . We may assume, by symmetry, that \emph{p\textsubscript{3} z L\textsubscript{r} .} According to\footnote{M. H. A. Newman, \emph{Elements of the topology of plane sets of points,} Cambridge University Press, 1961.} (Theorem 11.8, p. 119), \emph{D —} T has two components \emph{U}i and \emph{U\textsubscript{2} ,} the boundary of \emph{Ui} being \emph{L\textsubscript{x}} U P and the boundary of \emph{U\textsubscript{2}} being z\textsubscript{2}u r. Let 7' be an arc in \emph{A} joining \emph{q\textsubscript{3} to a} point \emph{q z} T \emph{A.} Putting 73 and 7' together, we obtain an arc 5 joining \emph{p\textsubscript{3}} to \emph{q.} Starting at \emph{p\textsubscript{3}} and proceeding along 5, let \emph{r} be the first point of T that we reach. Let A be the subarc of 5 with endpoints at \emph{p\textsubscript{3}} and \emph{r.} Clearly, A — \{p\textsubscript{3}\} \emph{GZ} \emph{A.} We can assume (according to\footnote{H. Tietze, Uber stetige Kurven, Jordansche Kurvenbogen und geschlossene Jordansche Kurven, \emph{Math. Z.,} 5 (1919), 284-291.}) that A is a simple arc. Since \emph{p\textsubscript{3} z L\textsubscript{x} , p\textsubscript{3}} is not in \emph{U\textsubscript{2} .} Since A - \emph{\{p\textsubscript{3}} , r\} £ \emph{D} - T = \emph{Ui U\textsubscript{2} ,} A — \emph{\{ps , r\}} must have a point in \emph{U\textsubscript{x} .} But A — \emph{\{p\textsubscript{3} ,} r\} is connected, so A - \emph{\{p\textsubscript{3},} r\} \emph{C} \emph{Ui.} Hence A is a cross cut of \emph{U\textsubscript{r}}. Let \emph{M\textsubscript{r} ,M\textsubscript{2}]oq} the two open subarcs of \emph{Li} with endpoints \emph{p\textsubscript{t} , p\textsubscript{3}} and \emph{p\textsubscript{2} , p\textsubscript{3}} respectively. Let Pi , r\textsubscript{2} be the two closed subarcs of T with endpoints \emph{p\textsubscript{r} , r} and \emph{p\textsubscript{2} , r} respectively. According to\footnote{M. H. A. Newman, \emph{Elements of the topology of plane sets of points,} Cambridge University Press, 1961.} (Theorem 11.8, p. 119), \emph{U\textsubscript{r}} — A has two components \emph{Vi} and \emph{V\textsubscript{2},} the boundary of \emph{Vi} being \emph{kJ} r\textsubscript{x} kJ A and the boundary of \emph{V\textsubscript{2}} being \emph{M\textsubscript{2} kJ} r\textsubscript{2} kJ A. Since P U A C 2I, \emph{Vi\^{}J V\textsubscript{2}\textbackslash{}J U\textsubscript{2}.} Recall that \emph{p\textsubscript{3} 4 U\textsubscript{2}} . It follows that since \emph{p\textsubscript{3} z B, B} has a point in common with \emph{Vi\^{}J V\textsubscript{2} .} But \emph{B} is connected, so \emph{B £ Vi yj V\textsubscript{2}}. We see that \emph{p\textsubscript{r} \$ V\textsubscript{2}}, and therefore that \emph{B Vi} 4=<\Slash{}> (because \emph{Pi z B).} Hence \emph{B £} , so \emph{p\textsubscript{2} z} W . But, since the boundary of \emph{V\textsubscript{t}} is \emph{M\textsubscript{x}} I\textbackslash{} \emph{kJ} A, \emph{p\textsubscript{2} Vi} . This contradiction proves the lemma. \emph{Lemma 2. There exists a countable family} 8 \emph{of open disks such that every open set U Q R\textsuperscript{2} can be written in the form U — S\textsubscript{n} , where S\textsubscript{n}z § and S\textsubscript{n} £ U.} \emph{Proof.} Let \emph{\{p\textsubscript{n}\}} be a countable dense subset of \emph{R\textsuperscript{2},} and let 8 be the family of all open disks of rational radius having some \emph{p\textsubscript{n}} as center. 8 is clearly countable. If \emph{U} is an open set it is easy to show that for each \emph{x z U} there exists an \emph{S\textsubscript{x} z} 8 with \emph{x z S\textsubscript{x} £ S\textsubscript{x} £ U.} Obviously \emph{u = \textbackslash{}J s\textsubscript{x}.} xt.U \emph{Theorem 1. Let f be a homeomorphism of} \emph{D onto} \emph{D, and let
x} and f (z\textsubscript{n}) —> w\}. Let \emph{E = \{x £ C} \textbar{} there exist arcs y\textsubscript{x}, \emph{y\textsubscript{2}} at \emph{x} such that C(\Slash{}, \emph{yj IP C(J,} y\textsubscript{2}) =<\Slash{}> \}• A theorem of Bagemihl\footnote{F. Bagemihl, Curvilinear cluster sets of arbitrary functions, \emph{Proc. Nat, Acad, Sci. U. S. A.>} 4 (1955) 379-382.} states that \emph{E} is countable. Let \emph{N = E \textbackslash{}J F\textsubscript{s} .} \textbf{\emph{St}} s \emph{N} is countable. Let \emph{
0 W) C \emph{\textbackslash{}J} acc \emph{(f-\textbackslash{}S\textsubscript{n}} n Z>)) - \emph{N.} \textbf{\emph{n}}
On the other hand, suppose \emph{x} £ acc \emph{IP} D)) for some \emph{n,} and \emph{x 4 N.} Choose an arc \emph{y} in \emph{fXSn (P D)} with one endpoint at \emph{x.} Clearly,
C(f, \emph{y)} C \emph{S\textsubscript{n} (P D CS\textsubscript{n}CU.}
Since \emph{x 4 E,}
?.(\emph{) = p(}) \textsubscript{£} \emph{C(j, y)} c \emph{u,}
so \emph{x t} \^{}’(G). Thus
\emph{\textbackslash{}J} acc \emph{(T\textsuperscript{l}(S\textsubscript{n} r\textbackslash{}D))-NC \^{}\textbackslash{}U), n}
\textbf{SO}
\emph{ ,
it is clear that \emph{g} is of Baire class 1(C, \emph{R\textsuperscript{N}),} and that \emph{g} agrees with \emph{ 7?\textsuperscript{3} \emph{such that q does not lie in the range of g*, and for all x v} C,
\emph{\textbackslash{}g(x) - q\textbackslash{}} e => \emph{g(x) = g*(x).}
\emph{Proof.} Let
S = \emph{\{yvR\textsuperscript{3} \textbar{} \textbackslash{}y} - g\textbar{} < e\}.
If 0(C) \emph{CZ} \emph{S,} let \emph{g*} : C —> \emph{R\textsuperscript{3}} be any continuous function whose range does not include \emph{q.} Otherwise, 0\textasciitilde{}\textsuperscript{1}(\Slash{}S) is a proper open subset of \emph{C} and hence can be written in the form
\emph{g\textasciitilde{}XS)} = UA, \textbf{\emph{k}}
where
\emph{Ik =} \{e’‘ \textbar{} \emph{at < t < b\textsubscript{k}],} and
\emph{k} \texttt{1} \emph{I =>I\textsubscript{k} Ii =<\Slash{}> .}
Since 0\textasciitilde{}\textsuperscript{1}(\{0\}) is a closed (and therefore compact) subset of 0\textasciitilde{}\textsuperscript{1}(>S), 0\textasciitilde{}\textsuperscript{1}(\{\^{}\}) is covered by a finite number of I\textsubscript{fc}’s. Say
0’\textsuperscript{1}(\{0\}) \emph{\dots{} \textbackslash{}JI\textsubscript{n}.}
The endpoints \emph{e\textsuperscript{iak}} and e\textsuperscript{t6A} of \emph{I\textsubscript{k}} are not in 0\textasciitilde{}\textsuperscript{1}(\{\}), so we can construct, for each \emph{k,} a continuous function \emph{g\textsubscript{k} : I\textsubscript{k}-+ R\textsuperscript{3}} such that
\emph{Sk(e\textsuperscript{iai}) = g(e\textsuperscript{iai}), = g\^{}*),}
and \emph{q} is not in the range of \emph{g\textsubscript{k}} . Define
0*(x) = 0(a:), if o;\textsubscript{£}C\textasciitilde{} (AUZ\textsubscript{2}U \dots{} U\Slash{}\textsubscript{n}),
0*(x) = \emph{g\textsubscript{k}(x),} if \emph{x} e \emph{I\textsubscript{k} , k} = 1, • • • , \emph{n.}
It is easy to show that g* has the desired properties.
\emph{Theorem 3. Let f be a continuous function mapping D into the Riemann sphere} 2, \emph{and let 8 —> \emph{R} is of \emph{Baire class} 0 if and only if it is continuous. For any ordinal number \emph{a} > 0, \emph{f} is of \emph{Baire class a} if and only if \Slash{} is the pointwise limit of a sequence of functions each of Baire class less than \emph{a.}
Let denote the class of all sets M C S such that
\emph{M} = r\textbackslash{}(r, +«>)),
for some real rand some function \Slash{} of Baire class \emph{a} on >S. Let 912 denote the class of all sets \emph{N} C \& such that
\emph{N} = r\textbackslash{}[r, +-)),
for some real r and some function \Slash{} of Baire class \emph{a} on \emph{8.} It is easily shown that 9TCj = and 9l£ = .
Let
9 — 9c«> = g«,
gj = ,
9E\textsuperscript{a} = ,
91* = 9lco = 912 , If 0 is any class of sets, let 0\textsubscript{a} denote the class of all countable unions of members of 0, and let 0\textsubscript{8} denote the class of all countable intersections of members of 0. Each of the following facts is either explicitly stated in\footnote{F. Hausdorff, Uber halbstetige Funktionen und deren Verallgemeinerung, \emph{Math. Z.,} 5 (1919) 292-309.}, or can be easily deduced from statements found in\footnote{F. Hausdorff, Uber halbstetige Funktionen und deren Verallgemeinerung, \emph{Math. Z.,} 5 (1919) 292-309.}, or is obtained by a routine transfinite induction argument.
I. 9TC2 = \emph{(\textbackslash{}J} 9lX , 9i? = (U 9TtX •
X, +«))e\^{} .}
Then \Slash{} is of Baire class \emph{a.}
\emph{Definition.} If \emph{A} and \emph{B} are two sets, we will call \emph{A} and \emph{B equivalent,} and write \emph{A \textasciitilde{} B,} if and only if \emph{A — B} and \emph{B — A} are both countable. It is easily verified that \textasciitilde{} is an equivalence relation.
\emph{Lemma 5. If A \textasciitilde{} E, then 8 — A \textasciitilde{} 8 — E for any set 8. If A\textsubscript{n} \textasciitilde{} E\textsubscript{n} (for all n in some countable set} V), \emph{then}
\emph{)jA\textsubscript{n}\textasciitilde{}\textbackslash{}jE\textsubscript{n} and \textasciitilde{} •}
ntN ntN ntN ntN
The proof of this lemma is routine.
\emph{Definition.} An interval of real numbers will be called \emph{nondegenerate} if it contains more than one point.
\emph{Lemma 6. Any union of nondegenerate intervals is equivalent to} \emph{an open set.}
\emph{Proof.} Let 4 be a family of nondegenerate intervals and let \emph{H =} For any \emph{x} and \emph{y} let
ZO, \emph{y) = [x, y\},} if \emph{x} g \emph{y,}
and let
\emph{i(\%>y) = [y,} \^{}1,
Define a relation (R on \emph{H} by
\emph{x(Sly <=>I(x, y)} \emph{CZ} \emph{H,}
\emph{(x, yzH).}
It is easy to show that (R is an equivalence relation on \emph{H.} In view of the fact that a set \emph{A} of real numbers is an interval if and only if
\emph{x, y} e \emph{A =>I(x, y)} \emph{C} \emph{A,}
it is obvious that each equivalence class is an interval. For each \emph{x £ H,} there exists an \emph{I £ \$} with \emph{x £ I.} Every member of \emph{I} is equivalent to \emph{x.} Thus each equivalence class contains more than one point, and hence is a nondegenerate interval. Let \{J\textsubscript{a}\}be the family of equivalence classes. Any disjoint family of nondegenerate intervals is countable, so there are only countably many \emph{J}\textsubscript{a}’s. Let \emph{E} be the set of all endpoints of the various \emph{J}\textsubscript{a}’s. Then \emph{E} is countable and
\emph{H = \textbackslash{}J J\textsubscript{a}\textasciitilde{}} IJ \emph{J\textsubscript{a} - E = \textbackslash{}Jj*,} \textbf{\emph{a a a}}
where <7* is the interior of \emph{J\textsubscript{a}} . This proves the lemma.
\emph{Lemma 7. Let h be an increasing real-valued -function on a nonempty set E} \emph{CZ} \emph{R, Suppose that \textbackslash{}x — h(x)\textbackslash{}} 1 \emph{for every x} s \emph{E. Then h can be extended}
\emph{to an increasing real-valued function h\textsubscript{x} on R.}
\emph{Proof.} Let \emph{e} \textasciitilde{} g.l.b. \emph{E} (e may be — oo). For each \emph{x\textsubscript{Q}} e (e, + °°) set
\emph{hi(x .} For each \emph{xtE,} choose an arc \emph{y\textsubscript{x}} at \emph{x} such that lim f\emph{(z) = x}}
and
(a) \Slash{}(?*) £ (— \textsuperscript{00} , r + e), if \emph{x} s \emph{P}
(b) f(\textsubscript{7}») £ (J - \textsubscript{€}, +«>), if \emph{xtQ.}
(This is accomplished by cutting the arc off sufficiently close to a;.) We remark that if \emph{x e P} and \emph{y £ Q,} then \emph{y\textsubscript{x} C\textbackslash{} y\textsubscript{y} =<\Slash{}> .}
We will say that \emph{y\textsubscript{x}} meets \emph{y\textsubscript{y}} in \emph{A\textsuperscript{Q}\textsubscript{n}} provided that \emph{y\textsubscript{x}} and \emph{y\textsubscript{y}} have subarcs y' and y' respectively such that \emph{x} £ y' \emph{GZ} \emph{A°, y} £ y' \emph{C} \emph{A°,} and y' P \emph{y\textsubscript{y}} \texttt{4}<\Slash{}> . Let
L\textsubscript{o} = \emph{\{x £ P} \textbar{} (Vn)(3?\Slash{} 4= \emph{x)(y\textsubscript{x}} meets \emph{y\textsubscript{y}} in \emph{A°\textsubscript{n})\},}
\emph{L\textsubscript{r} = \{x £ Q} I (Vn)(3?\Slash{} 4= \emph{x)(y\textsubscript{x}} meets \emph{y\textsubscript{y}} in A®)\},
\emph{M\textsubscript{o} = £ P \textbar{} (ln)(y\textsubscript{x}} meets no \emph{y\textsubscript{v} (y} 4= \emph{x)} in A®)\},
\emph{\{x £ Q \textbar{} (?ri)(y\textsubscript{x}} meets no \emph{y\textsubscript{y} (y} 4= \emph{x)} in A®)\},
\emph{L} = Lo V Lt ,
\emph{M = M\textsubscript{0}\textbackslash{}J M\textsubscript{r}.}
Observe that \emph{L\textsubscript{o} , L\textsubscript{r} , M\textsubscript{o} , Mt} are pairwise disjoint, and that \emph{P = L\textsubscript{o} M\textsubscript{o}} and \emph{Q = L\textsubscript{x}} U \emph{M\textsubscript{r} .}
For each \emph{x £ M,} let \emph{n\textsubscript{x}} be an integer such that \emph{y\textsubscript{x}} meets no \emph{y\textsubscript{y}} (with \emph{y} 4= \emph{x)} in A®* . Notice that \emph{n n\textsubscript{x}} implies \emph{y\textsubscript{x}} meets no \emph{y\textsubscript{y}} in A® . Let
\emph{K\textsubscript{n} = \{x £ E \textbar{} y\textsubscript{x}} meets C®, and if \emph{x £ M, n\textsubscript{x} n\}.}
Clearly \emph{E} = VJn-i \emph{K\textsubscript{n} .} Moreover, \emph{K\textsubscript{n}} \emph{C} \emph{K \textsubscript{n+1}} for each \emph{n.}
Take any fixed integer \emph{n.} For each \emph{x} £ L\textsubscript{o} we can find a \emph{y} 4= \emph{x} such that \emph{y\textsubscript{x}} meets \emph{y\textsubscript{v}} in A® . Let \emph{I\textsuperscript{n}\textsubscript{x}} be the nondegenerate closed interval between \emph{x} and \emph{y.} We shall show that \emph{I\textsuperscript{n}\textsubscript{x} C L\textsubscript{o} U (C® - K\textsubscript{n}\textbackslash{}} If \emph{t £ I\textsuperscript{n}\textsubscript{x}} , either \emph{t £ C\textsuperscript{Q} - K\textsubscript{n}} or \emph{t £ K\textsubscript{n} .} Suppose \emph{t £ K\textsubscript{n}} . Then \emph{y\textsubscript{t}} meets C®, and (if \emph{t £ M) n\textsubscript{t}} n. It is clear from Figure 1 that \emph{y\textsubscript{t}} must meet either \emph{y\textsubscript{x}} or \emph{y\textsubscript{y}} in A® . (This can be rigorized by means of Theorem 11.8 on p. 119 in\footnote{M. H. A. Newman, \emph{Elements of the topology of plane sets of points,} Cambridge University Press, 1961.}.)
Consequently, \emph{t \% M.} Now \emph{x iL\textsubscript{0} Q P, so} since \emph{y\textsubscript{x}} intersects \emph{y\textsubscript{y} , y \$ Q.} So \emph{y} e \emph{E — Q} = P. Similarly, since \emph{y\textsubscript{t}} meets \emph{y\textsubscript{x}} or \emph{y\textsubscript{y}, t} e \emph{E — Q = P.} Thus \emph{t} e \emph{P — M = Lq} . We have shown that \emph{t} e \emph{I\textsuperscript{n}\textsubscript{x}} implies that \emph{t} e \emph{C° — K\textsubscript{n}} or \emph{t} e \emph{L\textsubscript{q} ,} so \emph{I\textsuperscript{n}\textsubscript{x} £ Lq} V (C° — \emph{K\textsubscript{n}).} It follows that (for each ri)
\emph{L\textsubscript{0}Q(\textbackslash{}Jr\^{}r\textbackslash{}E q[l\textsubscript{0}\textbackslash{}j (c° -} xj] n \emph{e. XzLq}
Let \emph{W„ = kJxet, I\textsuperscript{n}\textsubscript{x} .} By Lemma 6, \emph{W\textsubscript{n}} is equivalent to an open set. \emph{l\textsubscript{o}q(c\textbackslash{}w\}c\textbackslash{}e} 'n==l \Slash{}
C M \emph{[L\textsubscript{o} kJ (C° -} K\textsubscript{n})]\} H \emph{E = \textbackslash{}l\textsubscript{0} v f\} (.0°} - K„)\} n \emph{E}
\emph{= \{L\textsubscript{o} n E]} H (C° - K„) \emph{n e\textbackslash{} = L\textsubscript{o}} V \emph{= L\textsubscript{a} .}
Therefore \emph{L\textsubscript{Q} = W\textsubscript{n}) A E.} Since each \emph{W\textsubscript{n}} is equivalent to an open set there exists a \emph{G\textsubscript{Q}} e g\textsubscript{8} such that
L\textsubscript{o} — Go n \emph{E.}
Similar reasoning shows there exists a G\textsubscript{x} e \emph{Q\textsubscript{§}} such that
\emph{E.}
Next we study the properties of \emph{M\textsubscript{Q} ,} It is convenient to define a function 7T : \emph{R\textsuperscript{2} —» R} by \emph{ir(\{x, y)) =} If \emph{x} e \emph{M C\textbackslash{} K\textsubscript{n}} , then, starting at \emph{x} and proceeding along \emph{y\textsubscript{x}} , let \emph{(?\textsubscript{n}(x)} be the first point of \emph{C\textsuperscript{Q}\textsubscript{n}} reached. Set \emph{h\^{}(x) = ir(a\textsubscript{n}(x))} (for \emph{x} e \emph{M C\textbackslash{} Kn).}
is an increasing function on \emph{M C\textbackslash{} K\textsubscript{n}} ; for if \emph{Xi, x\textsubscript{2} £ M K\textsubscript{n}} and \emph{x\textsubscript{x} < x\textsubscript{2},} then, since cannot meet \emph{y\textsubscript{Xi}} in \emph{A\textsuperscript{Q}\textsubscript{n},} it is evident (see Figure 2) that \emph{tt\^{}Xxj)) <} \^{}( \textsubscript{\}}}
\emph{P} £ T\textsubscript{2} C C° - \emph{Q.}
Remembering the definitions of \emph{P} and \emph{Q\textsubscript{\}}} and observing the fact that C° — \^{}\textsuperscript{-1}(k +\textsuperscript{00})) = \^{}\textsuperscript{-1}((” \textsuperscript{00}, 0)> \textsuperscript{we can} summarize the results of the first part of the proof as follows.
\emph{For each pair r, t of real numbers with r < t, there exists a set T(r, t)} e 3l“\textsuperscript{+1} \emph{such that}
\^{}((\textasciitilde{} r]) C \emph{T(r, t)} C a,, \emph{t».}
Given any real r, let \{Z\textsubscript{n}\} be a strictly decreasing sequence of real numbers converging to \emph{r.} Then
<’((—\textsuperscript{00}, »•]) = °°, \Slash{}„)). n = l
So
<*((- »,r]) £ H \emph{T(r,} \Slash{}„) C \emph{f\textbackslash{}} <\textsuperscript{1}((- «>, i\textsubscript{n})) = \textsubscript{v}\textasciitilde{}\textbackslash{}(- «, r]), n“l \emph{n—}1
and hence
\emph{\^{}((\_\textsubscript{ro},\textsubscript{r]}) = C\textbackslash{}T\{r,Q.} n\textasciitilde{}l
By VIII,
<\textsuperscript{1}((\textasciitilde{}°o,r]) e9T\textsuperscript{+1}.
Since \Slash{} is an arbitrary function of Baire class \emph{a} in Z)° and \emph{ , \emph{r} by — \Slash{}, \emph{— cp, \textasciitilde{}r} to find that
\^{}([r, +«>)) e9l\textsuperscript{a+1}.
Also,
\^{}((r, +00)) = \emph{C° -} \textsubscript{£}snr\textsuperscript{+1}.
By IX, \emph{ a} then \emph{e £ E.} For \emph{x\textsubscript{Q}} £ (a, \emph{e]} set f(x\textsubscript{0}) = A(e). It is easily verified that \emph{f} is finite-valued and increasing, and is an extension of \emph{h.}
\emph{Lemma 10. Let E} \emph{C} \emph{R be a set of measure} 0 \emph{and let h be an increasing function on E. Suppose h(E) has measure} 0. \emph{Then \{x + h(x) \textbar{} x £ E\} has measure} 0.
\emph{Proof.} Extend \emph{h} to an increasing function \emph{g} on an open interval \emph{I = (a,} b) 2 \emph{E.} Set \emph{g(a) =} — 00 and \emph{g(b)} = + 00. Take any e > 0. Choose an open set \emph{G} such that \emph{I} and \emph{utG) <} e\Slash{}2. Choose an open set \emph{H} \emph{Z)} \emph{h(E)} with \Slash{}z(ZZ) < e\Slash{}2.
Say
\emph{G} = U \emph{I\textsubscript{n} ,} and \emph{H = \textbackslash{}J J\textsubscript{m} , mN mtM}
where \emph{\{I\textsubscript{n} \textbar{} n £ N\}} and \emph{\{J\textsubscript{m} \textbar{} ms M\}} are countable families of disjoint open intervals. Let \emph{I\textsubscript{n} = (a\textsubscript{n}} , b\textsubscript{w}), and observe that \emph{a\textsubscript{n} , b\textsubscript{n}} £ [a, b]. Set
\emph{s = (J \{g(a„), g(b\textsubscript{n})}\} - \{- 00 , + co \}.
\emph{mN}
Notice that \emph{S} is countable. Set
\emph{K\textsubscript{n}} = (g(a„), g(\&n)).
One can easily verify that \emph{k} \texttt{1} \emph{n} implies \emph{K\textsubscript{k}} P \emph{K\textsubscript{n}} = 0.
If \emph{A} and \emph{B} are two subsets of 2?, let
\emph{A + B} = \{a + b \emph{\textbackslash{} as, A,b sB\}.}
It is easy to show that for any two intervals \emph{J} and <7', g\textsubscript{e}(<7 +<\Slash{}') g + \^{}(J'). Let W = \emph{\{x + h(x) \textbar{} x £ E].}
\emph{Assertion.}
\emph{w} c \emph{(E + S) KJ \textbackslash{}J \textbackslash{}J} [(Z\textsubscript{n} n \emph{g-W)} 4- \emph{(J\textsubscript{n} r\textbackslash{} K.)]. mN imM}
To prove this, let \emph{w} be an arbitrary point of \emph{W.} Write \emph{w = x + h(x)\textsubscript{f}} where \emph{x sE.} For some n, \emph{x £ I\textsubscript{n} .} Since \emph{g} is increasing,
\emph{h(x) = g(x) £ [\#«), g(b\textsubscript{n})].}
If \emph{h(x)} equals \emph{g(a\textsubscript{n})} or \#(6\textsubscript{n}), then \emph{h(x)} e \emph{S,} so \emph{w = x + h(x)} e \emph{E} + 5. On the other hand, suppose \emph{h}(x) 4= \#(a\textsubscript{n}), (\&„). Then \emph{h(x) £ K\textsubscript{n} .} Also, \emph{g(x) = h(x) £ J\textsubscript{m}} for some \emph{m.} Thus \emph{h(x) £ J\textsubscript{m}C\textbackslash{} K\textsubscript{n}} and \emph{x £ I\textsubscript{n} C\textbackslash{} g\textasciitilde{}\textbackslash{}J\textsubscript{m})\textsubscript{\}}} so that
w = \emph{x + h(x)} £ (I\textsubscript{n} \emph{C\textbackslash{} g\textasciitilde{}\textbackslash{}J\textsubscript{m})) + (J\textsubscript{m} C\textbackslash{} K\textsubscript{n}).}
This proves the Assertion.
Since \emph{g} is increasing, \emph{g\^{}J\^{}} is an interval, so both \emph{I\textsubscript{n} C\textbackslash{} g\textasciitilde{}\textbackslash{}J\textsubscript{m})} and \emph{J\textsubscript{m} C\textbackslash{} K\textsubscript{n}} are intervals. Also note that m 4= \emph{I} implies \emph{g\textasciitilde{}\textbackslash{}J„\^{} g\textasciitilde{}\textbackslash{}Ji) —}<\Slash{}> . By the Assertion,
M.W g \emph{+ s)} + S E n \emph{g\textasciitilde{}\textbackslash{}j\textsubscript{m}y)} + (J\textsubscript{w} n \#\textsubscript{n})] \textbf{\emph{mN mtM}}
g \emph{n\textsubscript{e}(E + s)} + E E [m(\Slash{}\textsubscript{b} n n o
\textbf{\emph{mN mzM}}
= m\Slash{}U (s + t?)) + E [ E n \emph{g\textasciitilde{}\textbackslash{}j\textsubscript{m}y) + £ »(j\textsubscript{m} c\textbackslash{} K\textsubscript{n})]} \textbf{\emph{szS mN mtM mtM}}
\href{\%3C\%20E\%20\%3Cem\%3E\%2B\%20E\%29\%3C/em\%3E}{\texttt{< E + E)}} + E \emph{W.)} + E mGA. n 2Q]
\textbf{\emph{utS mN mtM}}
= 0 + ju(G) + E E \emph{c\textbackslash{}} \textbf{\emph{K\textsubscript{n}) nzN mzM}}
= m(g) + E E m(A. n \emph{K\textsubscript{n})}
mtM nzN
m(G) + E mW = m( )\textbf{\emph{+ 1EH) <}} e. \textbf{\emph{mt M}}
Since e is arbitrary, \emph{jn\textsubscript{e}(W)} = 0.
\emph{Lemma 11. Let L — \{(x, a) \textbar{} x £ E\} and M = \{\{x\textsubscript{y} b) \textbar{} x £ R\} be two horizontal lines in R\textsuperscript{2}. Let E be a set of (linear) measure} 0 \emph{in L and let F be a set of (linear) measure} 0 \emph{in M. Let £>be a set of closed line segments such that}
(a) , \$2 £ <£, \emph{S2 ■—}Si \emph{F\textbackslash{} S2 \textsuperscript{=}}
(b) \$£ <£=>\emph{one endpoint of s lies in E and the other endpoint lies in F.}
\emph{Let S = \textbackslash{}J\textsubscript{kt\&} s. Then u\textsuperscript{2}(S) =} 0.
\emph{Proof.} Assume without loss of generality that \emph{b >a.} For any \emph{(x, y) £ R\textsuperscript{2}} let \emph{\^{}(\{x, y)) = x.} For any \emph{y £ R} let \emph{l\textsubscript{y} = \{\{x, y) \textbar{} x £ R\}.} Let
\emph{E\textsubscript{o} = \{z £ E \textbar{} z} is the endpoint of some s £ <£\},
and observe that \emph{E\textsubscript{o}} has linear measure 0. For any set \emph{A} C \emph{R\textsuperscript{2}} we of course set tt(A) == \emph{\{x £ R \textbar{} (xj y) £ A} for some \emph{y £ R\}.}
We define a function \emph{h} on \emph{ir(E\textsubscript{0})} as follows. If \$ £ \emph{\^{}(Eq),} then \emph{\{x, a\} zE\textsubscript{0} ,} so we can choose a (unique) segment s £ <£ with one endpoint at (x, \emph{a).} If the other endpoint of s is p, we set \emph{h(x)} = 7r(p). Clearly \emph{h} maps \emph{ir(E\textsubscript{0})} into \emph{tt(F).}
Since the segments in £ cannot intersect each other, \emph{h} must be an increasing function.
Take any \emph{y\textsubscript{Q}} with \emph{b >y\textsubscript{Q} > a.} Let \emph{c = b — y\textsubscript{0}, d = y\textsubscript{0} — a.} A simple computation shows that if \emph{q} s \emph{l\textsubscript{Vo} C\textbackslash{} 8,} then
tt(q) =
\emph{\emph{ex} + dh(x) c + d}
\bigskip
for some \emph{x} s tt(£?\textsubscript{0})- So
7r(l\textsubscript{yo} A 8) C
\emph{\emph{ex + dh\{x)} . e + d}
\emph{X} £ 7r(Fo)
\bigskip
Now \emph{(d\Slash{}e)h(x)} is an increasing function mapping \emph{tt(E\textsubscript{q})} into (d\Slash{}c)7r(F), so by Lemma 10
\emph{x + h(x)}
\emph{X £ Tt(Eq)}
\bigskip
has measure 0. Hence
\emph{X £ Tt(Eq)}
\emph{\emph{ex + dh(x)} . e + d}
\emph{X £ Tt(E\textsubscript{0})}
\bigskip
has measure 0, so A \emph{SY)} = 0. But A 8)) = 0 also when \emph{y\textsubscript{Q} 4} (a, b), so A 8)) = 0 for every \emph{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 \emph{y\textsubscript{Y} , y\textsubscript{2} s R,} let
\emph{U(,yi >} 2\Slash{}2) = \emph{\{\{x, y) \textbackslash{} x, y t R, y\textsubscript{t} < y < y\textsubscript{2}\}.}
A set of the form \emph{U(yi , y\textsubscript{2})} will be referred to as a \emph{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 \emph{\{\{Xj b) \textbar{} x} e (—\emph{n, ri)\}.} Let
\emph{S(n)} = [ U \textsubscript{S}]H + i b-i).
Since \emph{l\textsubscript{a}} and \emph{l\textsubscript{b}} have (plane) measure 0, and since
sc I.U4U Q \emph{S(n),} \textbf{\emph{n = l}}
it is sufficient to show that each \emph{S(ri)} has measure 0.
Let \emph{n} be a fixed positive integer. Set a* = \emph{a} + 1\Slash{}n and b* = b — \emph{1\Slash{}n.} Take any \emph{e} > 0. Choose €\textsubscript{0} so that 2e\textsubscript{0} + \emph{e*} < e\Slash{}(b — \emph{a).} Let \emph{y\textsubscript{0}} be any member of [a*, b*]. For the time being, \emph{y\textsubscript{Q}} will be held fixed.
For each s e <£, let \emph{p\textsubscript{8}} be the endpoint of 8 on \emph{l\textsubscript{b} ,} let \emph{q\textsubscript{t}} be the intersection point of s with \emph{l\textsubscript{V9} ,} and let r\textsubscript{a} be the endpoint of 8 on \emph{l\textsubscript{a} .}
Choose an open set \emph{G £ R} such that \emph{tt(1\textsubscript{Vq}} A \emph{S(n))} \emph{C} \emph{G} and g((?) < e\textsubscript{0} • Say \emph{G} = VJ, \emph{Ij}, where Z, = (a\textsubscript{f} , \emph{b\^{}} and the Z, ’s are pairwise disjoint. We may assume that each Z,« contains a point of \emph{tt(1\textsubscript{V9} S(n)).} For each let
\emph{Ci = g.l.b.} \{7r(p\textsubscript{a}) \textbar{} s e £(n), 7r(g\textsubscript{a}) e ZJ,
\emph{dj = l.u.b.} \{\^{}(p,) \textbar{} s e £(n), 7r(g«) e Z,\}, c< = \emph{g.l.b.} \{?r(r\textsubscript{a}) \textbar{} 8 e <£(n), ?r(\^{}) e Z,\}, \emph{d'i = l.u.b.} \{?r(r\textsubscript{a}) \textbar{} 8 s £(n), \emph{ir(q\textsubscript{8})} e ZJ.
Note that c, g d,- and c< g d< . Since the segments in £ cannot intersect each other, it is easily seen that the intervals (c,- , d,) are all pairwise disjoint. It is also clear (from the definition of £(n)) that each (c,-, d,) is a subset of (\textasciitilde{} n, ri). Hence, if we set a,- = d,« \emph{c\textsubscript{7}- ,} we have 22,- a,- g \emph{2n.}
For each \emph{j,} let \emph{s(j)} be the line segment joining the two points (c< , \emph{a),} , b), and let \emph{t(j)} be the line se\emph{gment j}oining the two points (d< , a), (d,- , \emph{b).} Let \emph{A\textsubscript{f}} be the closed subset of \emph{U(a,} b) which is enclosed by the two line segments 80), \^{}0). Let \emph{Hj} denote the intersection of A,- with the horizontal open strip
\emph{V} — LI max
€o I \emph{• J 7} I \textsuperscript{€}0 I I \textsuperscript{o}’\^{}-\^{}p\textsuperscript{min}V’\^{} \textsuperscript{+} 2\^{}\Slash{}r
Note that \emph{H\textsubscript{f}} is measurable. Setting ZZ = \emph{H\textsubscript{f}, it} is clear from the definition of the A\Slash{}s that
\emph{S(ri) C\textbackslash{}V QH.}
Take any \emph{y} s \emph{R.} We wish to show that
\textsubscript{M}0-(ZZ A l\textsubscript{y})) <
\_\_ e\_\_ \emph{b — a}
We can, of course, assume that
\emph{l J} \textbf{\textsuperscript{€}o I • J 7 . €o I \textbar{}}
\emph{y e} I max \emph{\textbackslash{}a, y\textsubscript{a} —} , mm S \emph{b, y\textsubscript{0} + \^{}(r}
An elementary computation, using the geometrical properties of \emph{Hj,} shows that
n i,)) g (i + \emph{4\^{}--}
\emph{\textbackslash{} o y\textsubscript{Q} \Slash{} o y\textsubscript{Q}}
Therefore
n z„)) g E n \emph{w}
- G+\textsuperscript{n} \^{})\textsuperscript{e}°+\textsuperscript{2n2}
” 2c\textsubscript{0} + €\textsubscript{0} < t , \emph{b — a}
so \emph{v(ir(H l\textsubscript{v}Y) < e\Slash{}(b — a)} for every \emph{y.}
We have shown that \emph{for each y\textsubscript{Q}} e [a* \&*] \emph{there exists a horizontal open strip}
\emph{V(y\textsubscript{Q}) containing l\textsubscript{Vo} , and there exists a measurable set H(y\textsubscript{Q})} \emph{C} y(?\Slash{}\textsubscript{0}), \emph{such that}
\emph{S(n) H V(y\textsubscript{0}) £ H(y\textsubscript{0})}
\emph{and (for every y) ir(H(y\textsubscript{Q}) l\textsubscript{v}) is measurable and}
\emph{y«H(y\textsubscript{0}) Cy} Z\textsubscript{y})) <
The various open strips VG\Slash{}o) (i\Slash{}\textsubscript{0} « [a*, 6*]) clearly cover the compact set \{(0, \emph{y\}} I \emph{y} \textsuperscript{e} f\textsuperscript{a}\emph{> \&}]\} • Choose a finite subcovering \emph{V(y\textsubscript{2}\textbackslash{} • • • , V(y\textsubscript{m}).} Set
\emph{tm—} 1 \Slash{}
m \textbackslash{}
U \emph{V(y,)]}
J“»+l \Slash{}
n I7(a*, 6*).
\emph{H(y\textsubscript{m}) U \textbackslash{}J} IW) -
X
Obviously \emph{K} is measurable, and for each \emph{y, ir(K} A Z„) is measurable and \emph{C\textbackslash{} l\textsubscript{v})) < e\Slash{}(b - a).} Moreover, \emph{S(n)} C \emph{K.} We have
\emph{\emph{\textbackslash{}K) = £ \^{}(K C\textbackslash{}} 1,)) \emph{dy} g £ \emph{dy} = (6} - a*) \emph{< e.}
Since \emph{e} is arbitrary, this shows that
g.l.b. \{\Slash{}(X) \textbar{} \emph{K} measurable, \emph{S(n)} C \emph{K\}} = 0.
Therefore \emph{S(n)} has measure 0.
\emph{Lemma 12, For every} \emph{e} > 0 \emph{there exists a strictly increasing function h on R such that h(R) has measure} 0, \emph{and for every x,} I\# — A(x)\textbar{} e
\emph{Proof.} For each (not necessarily positive) integer n, let \emph{I\textsubscript{n}} = [ne, (n + l)e]. Then \emph{I\textsubscript{n} = R.} There exists a strictly increasing function \emph{f} : [0, 1] [0, 1]
such that m(\Slash{}([0, 1])) = 0. For example, such a function may be defined as follows. Any number in [0, 1) may be written in the form
\emph{.a\textsubscript{1}a\textsubscript{2}a\textsubscript{3} • • • a\textsubscript{n} • • • ,} (binary decimal),
where the decimal does not end in an infinite unbroken string of l’s. Set
\emph{f(.a!a\textsubscript{2}a\textsubscript{3} • • • a\textsubscript{n} • • •) = bib\textsubscript{2}b\textsubscript{3} • • • b\textsubscript{n} • • • ,} (ternary decimal), where \emph{bi} = 0 if = 0 and 6,- = 2 if a,- = 1. Set f(l) = 1. \emph{f} maps [0, 1] into the Cantor set, so m(\Slash{}([0, 1])) = 0. It is easily shown that \Slash{} is strictly increasing.
For each n, it is easy to obtain from \Slash{} a function \emph{f\textsubscript{n} :I\textsubscript{n}-> I\textsubscript{n}} such that \emph{j\textsubscript{n}} is strictly increasing and \Slash{}z(fn(In)) = 0. Set
\emph{h(x)} = f\textsubscript{n}(x) for \emph{x} e (ne, (n + l)e].
There is no difficulty in proving that \emph{h} has the desired properties.
\emph{Theorem 6. Let be an arbitrary junction on C\textsuperscript{Q} = \{\{x,} 0) \textbar{} \emph{x} e \emph{R}\}. Then \emph{there exists a junction j on D° — \{\{x, y) \textbar{} y} > 0\} \emph{such that j(z)} = 0 \emph{almost everywhere and h\textsubscript{n}(x') x > x\textsuperscript{f} => h\textsubscript{n+1}(x) > h\textsubscript{n+1}(x'\textbackslash{}}
we find that \emph{x} 4= \emph{x\textsuperscript{f}} implies \emph{s\textsubscript{H}(x) C\textbackslash{} s\textsubscript{n}(x\textsuperscript{f})} = 0. Since each \emph{s\textsubscript{n}(x)} has one endpoint in \emph{E\textsubscript{n}} and the other in \emph{E\textsubscript{n+1} ,} Lemma 11 shows that for each \emph{n}
= 0.
x\&R
Hence
M\textsuperscript{2}(0 Us\textsubscript{n}(a:)) = 0. 'n=\textsuperscript{s}l \textbf{\emph{xtR •}}
Set
\emph{j(z)} = £>((\$, 0)), if \emph{z} e \emph{s\textsubscript{n}(x)} for some n,
\emph{f(z)} = 0, if \emph{z} is not in any \emph{s\textsubscript{n}(x).}
\emph{j(z)} = 0 almost everywhere. Let
\emph{y(x)} = \{\{x, 0>\} U 0s\textsubscript{n}(x). 74 = 1
Since the endpoints of s\textsubscript{n}(\$) are at (A\textsubscript{n}(z), \emph{1\Slash{}n)} and \emph{(h\textsubscript{n+1}(x)j l\Slash{}(n} + 1)), and since \emph{(h\textsubscript{n}(x), 1\Slash{}n)} —> (x, 0) as n —> oo, it is clear that \emph{y(x)} is an arc at \emph{(x\textsubscript{\}}} 0). Obviously
lim \emph{j(z) = ® vzy}
\emph{Lemma 13. Suppose that every point of F is an ambiguous point of the function f : K —» R\textsuperscript{3}. Then f has a nonmeasurable boundary function.}
\emph{Proof.} Let \emph{E} be a nonmeasurable subset of \emph{F.} Since each point of \emph{F} is an ambiguous point we can choose, for each \emph{x} e \emph{F,} two distinct points \emph{\^{}(x), X zey
J. E. McMillan proved that if f is a continuous function mapping D into the Riemann sphere, then the set of curvilinear convergence of f is of type F\textsubscript{a}\$ [2, Theorem 5]. In this paper we shall provide a simpler proof of this theorem than McMillan’s, and we shall give a generalization and point out some of its corollaries.
\emph{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 \emph{p.} If x\textsubscript{0} e K and r > 0, then
S(r, x\textsubscript{0}) = \{x e K\textbar{} p(x, x\textsubscript{0}) < r\} .
An arc of C will be called \emph{nondegenerate} if and only if it contains more than one point.
LEMMA 1. \emph{Let bea family of nondegenerate closed arcs of} C. \emph{Then} Uie\^{} I " Uje\Slash{}Z I* \emph{countable.}
\emph{Proof.} Since U\textsubscript{Ie}\^{}r I* is open, we can write I* = U\textsubscript{n} J\textsubscript{n}, where \{J\textsubscript{n}\} is a countable family of disjoint open arcs of C. If
X\textsubscript{0} € U I - U I* , Ie\# Ie\#
then for some I\textsubscript{o} e \#, x\textsubscript{0} is an endpoint of I\textsubscript{o}. For some n, Iq c J\textsubscript{n}, so that x\textsubscript{0} e • But x\textsubscript{0} \Slash{} J\textsubscript{n}, so that x\textsubscript{0} is an endpoint of J\textsubscript{n}. Thus Uj I - Uj I* is contained in the set of all endpoints of the various J\textsubscript{n}; this proves the lemma. ■
In what follows we shall repeatedly use Theorem 11.8 on page 119 in\footnote{S. Banach, Uber analytisch darstellbare Operationen in abstrakten Raumen, \emph{Fund, Math.,} 17 (1931) 283-295.} 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.
\bigskip
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), I\textsubscript{0}(n) = U L(y)* .
yE r yE r
Let ace (n) denote the set of all points on C that are accessible by arcs in n.
The following lemma is weaker than it could be, but there is no point in proving more than we need.
LEMMA 2. \emph{The set} ace (n) - I0(n) \emph{is countable.}
\emph{Proof.} By Lemma 1, I(n) - I0(n) is countable; therefore it will suffice to show that ace (n) - I(n) is countable. If ace (n) has fewer than two points, we are done. Suppose, on the other hand, that ace (n) has two or more points. If a E ace (n), then there exists a' E ace (0) with a' =f. a. Let y, y' be arcs at a, a' , respectively, with
ynncn , y'n n cn.
Let p be the endpoint of y that lies in n, p\^{} the endpoint of y\textsuperscript{1} that lies in n. Let y" c n be an arc joining p to p' . The union of y, y', and y\textsuperscript{11} is an arc o joining a to a'. By\footnote{P. T. Church, Ambiguous points of a function homeomorphic inside a sphere, \emph{Michigan Math. J.,} 4 (1957) 155-156.}, there exists a simple arc 0\textsuperscript{1} co that joins a to a\^{}. Clearly, 0\textsuperscript{1} is a cross-cut with 0\textsuperscript{1} n D c n and a, a\^{} e L(0 '). Thus a e I(n), and so ace (n) c 1(0). •
LEMMA 3. \emph{Suppose} 01 \emph{and} Oz \emph{are domains contained in} D - \{O\}. If
(1) IqCOj) A acc(O\textsubscript{1}) \emph{and} Iq(Q\textsubscript{2}) A ace (Q\textsubscript{2})
\emph{are not disjoint, then} n1 \emph{and O\textsubscript{z} are not disjoint.}
\emph{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 y\textsubscript{1}n D and y\textsubscript{2}n D are disjoint.
Suppose yi nD cVz and \emph{yz A} D c Vj. Then, since yi n D c Ui, Ui has a point in common with Vz. But O E u\textsubscript{1} nUz, so that U \textsubscript{1} has a point in common with Uz also. Since U\textsubscript{i} is connected, this implies that U\textsubscript{i} has a common point with Yz n D, which contradicts the assumption that Yz nD cVp Therefore yi nD \^{} V2 or Yz n D \^{} Vi • We conclude that either y 1 nD c U\textsubscript{2} or y\textsubscript{2} nD c U1 . By symmetry, we may assume that Yz n D cU\textsubscript{i}.
It is possible to choose a point b E L(yi) * that is accessible by an arc in Oz, because a is in the closure of ace (Oz). Let y be a simple arc joining b to a point of y\textsubscript{4} n D, such that y - \{ b\} c Oz. Then y - \{ b\} and y\textsubscript{i} are disjoint. Also, y — i\_b\} contains a point of Ui (namely, the point where y meets y2 n D); therefore y - \{b\} c Ui . Hence b e Ui. Since b e L\{y 1 )*, this is a contradiction. •
THEOREM 1 (J. E. McMillan). \emph{Let} K \emph{be a complete separable metric space, and let} f \emph{be a continuous function mapping} D \emph{into} K. \emph{Let}
X = \{x E CI \emph{there exists an arc} y \emph{at} x \emph{for which} lim f(z) \emph{exists\}} . z\^{}x
\textbf{zEy}
\bigskip
\emph{Then} X \emph{is of type} F\textsubscript{a} \$ .
\emph{Proof} Let \{p\textsubscript{k}\}\textsubscript{k=1} be a countable dense subset of K. Let \{Q(n, be
a counting of all sets of the form
where \emph{0} is a rational number. Let \{U(n, m, k, £)\}\^{}\textsubscript{=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
\textsubscript{Y}= n u u u I\textsubscript{0}(U(n, m, k, £)) Cl A(n, m, k, £).
n=l m=l k=l f = 1
Since I\textsubscript{0}(U(n, m, k, £)) is open, it is of type F\textsubscript{a} . It follows that Y is of type F\textsubscript{a<}5 .
I claim that Y c X. Take any y e Y. For each n, choose m[n], k[n], f [n] with
(2) y € I\textsubscript{0}(U(n, m[n], k[n], f[n])) A A(n, m[n], k[n], £[n]) (n = 1, 2, 3, •••).
For convenience, set U\textsubscript{n} = U(n, m[n], k[n], f[n]). By (2) and Lemma 3, U\textsubscript{n} and U\textsubscript{n+}i have some point z\textsubscript{n} in common. For each n, we can choose an arc y\textsubscript{n} c U\textsubscript{n+}i with one endpoint at z\textsubscript{n} and the other at z\textsubscript{n+1} . Then y\textsubscript{n} c Q(n + 1, m[n + 1]). Also,
and therefore
r r
P 6 R\href{\#\_ftn1}{\textsuperscript{\textsuperscript{\footnote{F. Bagemihl \& G. Piranian, Boundary functions for functions defined in a disk, \emph{Michigan Math, J.,} 8 (1961) 201-207.}}}} : y > 0\} ? 1
H\textsubscript{n} = \{ 0 such that S(r, p) \emph{Q} W. Choose B€ ® so that p €BSs(\textbar{} r, p) . Then JBSS(r, p) Cl w. It follows that
N = U B = U
B€dt(W) B € d(W)
where (J.(W) = £B€® : B" \^{}W\}. Therefore
V\textsubscript{b}\textsuperscript{-1}(W) = E A ‘p'M = E A VJ • 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 \emph{Q} E such thatcpL .. is of class (F (E - M)).
Proof. Let U be any open subset of Y, and let W = (U)’. Set
E\textsubscript{n} = \{x € X : there exists an arc \emph{y} at x, having one endpoint on X\textsubscript{n}, such that \emph{y} - \{x\} -f’\textbackslash{}u)\}
K = \{x € X : there exists an arc \emph{y} at x such that
Y - \{x\} C f"\textsuperscript{1}(w)\}. '
Ob serve that
\^{}(U) cQ E n=l \textsuperscript{n} and : E + Yis a boundary function for f, then \emph{cp} is of honorary Baire class 2 (E, Y).
Next we show that the foregoing corollary is as strong as possible in the sense that if E is any subset of X and (p is a function of honorary Baire class 2 mapping E into a suitable space, then there exists a continuous function in H having f as a boundary function. A proof of this result-- at least for real- or vector-valued functions was outlined by Bagemihl and Piranian [2, Theorem 8], in the case
where E = X. Although the construction given here is carried out much \emph{l}
more explicitly than the construction given by Bagemihl and Piranian, my treatment differs from theirs in only two aspects that are of any significance. First of all, the proof of the theorem for arbitrary subsets E of X depends on Lemma 6 of the Introduction. Secondly, Bagemihl and Piranian say in the last line of their proof that there is "no difficulty now in extending f continuously to the whole of D in such a manner that 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\textsuperscript{-1}(\{v € R\textsuperscript{3} : \textbar{}v\textbar{} = \textbar{} J)
L = g\textsuperscript{-1}(\{v € R\textsuperscript{3} : \textbar{}v\textbar{} > \textbar{} \})
F = g\textsuperscript{-1}(\{vGR\textsuperscript{3} : \textbar{}v\textbar{} £y \}).
\bigskip
Let g\textsubscript{0} = gl\^{}. H is homeomorphic to R , so by [5,.Lemma 2.9,p. 299],- g can be extended to.a continuous function \textsuperscript{s}o
g\textsubscript{1} : H -> \{v G. R\textsuperscript{3} : \textbar{} v \textbar{} = . •
3
Define f\^{} : H -> R - \{0\} by setting <
fjCz) = g(z) if z € L
f\textsubscript{x}(z) = . g\textsubscript{x}(z) if z € F.
Then, since F and L are closed, f\^{} is continuous on H. It is easy to 3 2
verify that h(t) \emph{— 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\textbar{}\textsubscript{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\textbar{}\textsubscript{R} is continuous, so that f ® (h\textbar{}\textsubscript{R N}) = (f <» h) \textbar{}\textsubscript{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 \_2, T uedcw)}}
\href{U\%E2\%82\%ACd\%28W\%29}{\texttt{U€d(W)}}
Thus Y a boundary function for
C + 1(E, Y).
Proof.
Observe
that
connected metric space,
Y) where
f. Then
Let U be any open subset of
C = AUB.
For each x
and let V =
g >\_ 1, E a subset of
ip is of
Y - U.
= V\textsuperscript{-1}(V)
choose an arc y\textsubscript{v}
Baire class
Set
at x such
lim z->x f(z) \textsuperscript{z €Y}x
\textsuperscript{Y}x
\textsuperscript{Y}x
\textbar{}z - x\textbar{} £ 1\}
- ’ \{x\} c f\textsuperscript{-1}(V)
Notice that if x G A and
We will say that
have subarcs y ' and y ' 'x 'y
if
\emph{if}
\emph{x € B.}
y G B, then
y„ meets y.
\textbf{' X}
in
respectively such
n’ \textsuperscript{311(1 Y}x' \textsuperscript{n Y}y* *
L = \{x€ A : (Vn)(3y)(y € C, a
M = \{x € a
= \{x'€
Let
and
: (Vn).(3y)(y€ C
: (3n)(v meets no x
: (9 n)(y meets no
L \^{}ULh
M = M\textsubscript{a} U .
and
y " *•
provided that y and y x y
that x £ y\textsubscript{x}’ C H\textsubscript{n},
meets y in H )\} y ' x n'
meets y
in\textsubscript{Hn})\}
Y\textsubscript{y} (with y \textbar{} x) in H\textsubscript{n})\}
Y\textsubscript{y} (with y f x) in HJ
\bigskip
Observe that L\textsubscript{\&}, 1\^{}, M\textsubscript{\&}, 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 \emph{y}
(with y
x) in . Then n >\_n(x) implies that y\textsubscript{x}
Let
• meets X\textsubscript{n}, and, if x € M, n > n(x)\}.
Then K K . for each n, and C = ( 7 X . n n+1 ’ n
n=l
We next show that for each positive integer n and each x there exists a nondegenerate closed interval I® such that
x € C L\textsubscript{a} (X - K\^{}) . By the definition
y € C (y \texttt{1} x) such that y meets y\textsubscript{x} in
interval having its endpoints at
x and
y-
of L , there exists a’
Let I\textsuperscript{n} be the closed x
Let t be any point of
We must prove that t £ L\textsubscript{\&} u (X - assume t £ K . Then y. meets X n \textsuperscript{1}1 n
that y. must meet either y or y ’t 'x ’•
K ) . n'
If t i K .we are done. \textasciitilde{} n’
and hence it is clear from
rigorized by means of
Theorem 11.8
in (This argument can ; on p. 119 in [11].) But,
then (because t € K\textsubscript{n})
n >\_ n(t), so
Therefore
t £ M. Now
Hence y
C - B = A.
' I\textsuperscript{n}. x
So
Figure 5
be
if t € M,
that this situation\^{}is impossible.
x £ L £ A, so, since y intersects y , a • ’ 'x 'y*
Similarly, since y intersects y or y
■x 'y
\bigskip
tec
= A.
Thus t £ A - M = L , and we have shown that a
\bigskip
(X -
K ).
Let W n
x€ L
For each n,
a
L c W n C \emph{Q} [L u (X - K )] A c, Cl 11 d 11
and therefore
\bigskip
\bigskip
Vac n=l co
■ \{p\textbar{} [L u (X - K\textsubscript{n})]\} n c '
n=l co
- I\textsuperscript{L}\textsubscript{a}\textsuperscript{u (X} -C[ \textsuperscript{K}n\textsuperscript{)] nC}
CO.
= (L a C) u (C - U K ) = L u )- (-») = 0 and (+<») - (+00) = 0. In other respects we adhere to the usual conventions regarding arithmetic operations that involve -co or +00. Let
\textsuperscript{T(}-\textsuperscript{a}0’ \textsuperscript{b}o’ \textsuperscript{a}l» \textsuperscript{b}P = < : 0 £ y £ 1 and
(\textsuperscript{a}l - \textsuperscript{a}\textsubscript{o})y + a\textsubscript{o} £ x £ (b\textsubscript{x} - b\textsubscript{Q})y + b\textsubscript{Q}\}.
A set of this form will be called a \emph{closed trapezoid}. We also consider <\textbar{}>to be a closed trapezoid. A set S will be called a \emph{trapezoid} if there exists a closed trapezoid T such that T\textsuperscript{1} \emph{Q} S ST, where T\textsuperscript{1} denotes the interior of T relative to H\^{}. Every trapezoid is Lebesgue-measurable, though not necessarily Borel-measurable. *
If s, s’ are disjoint line segments having endpoints \^{}a\textsubscript{Q}, 0\^{} , and , respectively, where a\^{} £a\^{}' (i = 0, 1), then let
T(s, s’) = T(s’, s) = T(a\textsubscript{o},,a\textsubscript{o}’; a\textsubscript{p} a\textsubscript{J}’).
If s = s’, then we let\_T(s, s’) = T(s’, s) = s. In what follows we will use the symbol X\textsubscript{Q} as an alternative designation for the x-axis X. This will enable us to make statements about X\^{} (1=0, 1) (where Xj, denotes, as before, \{ Y such that f(z) = 0 almost everywhere and ipis a boundary
function for f.
Proof. If \{y \} \textsubscript{Y}th® family of arcs described in Theorem 10, let X X £ a f(z) =0 if z is in no y
\textbf{X}
f(z) = 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 X\textsubscript{q} and the other on X\^{}, and let SUz . Assume that S is a Borel set.
£ £
If m (S n X\textsubscript{Q}) and m (S n X\^{}) are known, what lower bound can be given for m(S)? . A solution to this problem might be helpful in attacking a problem of Bagemihl, Piranian, and Young [3, Problem!].
\bigskip
\section{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) \textbackslash{}1959) 29-34.
\href{\%3Csup\%3E-\%3C/sup\%3E}{\texttt{-}}.. ,.
4. s. Banach; Uber analytisch darstellbare Operationen in. abstrakten .Raumen, Fund. Math,, \emph{.!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\Slash{} \textsuperscript{. 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., \emph{14} (1965) 589-612.
9. T. J. Kaczynski, On. a boundary property of continuous functions, ' Michigan Math. J., 13 (1966) 313-320. \textsuperscript{.}
10. J.E. McMillan, Boundary properties of functions continuous in a disc, Michigan Math. J.,'3 (1966) 299-312. \textsuperscript{.}
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,
\bigskip
\part{Ted's Work as an Assistant Professor of Mathematics at the Uni. of California}
\chapter{7. 1968 - Note on a Problem of Alan Sutcliffe}
\textbf{Original PDF}: \href{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
\textbf{Reprinted from the} Mathematics Magazine
\textbf{Vol. 41, No. 2,} \textbf{March,} \textbf{1968}
\textbf{NOTE ON A PROBLEM OF ALAN SUTCLIFFE}
T. J. KACZYNSKI, The University of Michigan
If n is an integer greater than 1 and \emph{ah,} • • , a\textbf{\textsubscript{b}} a\textbf{o} are nonnegative integers, let
(ah, • • • , ai, ao)\textsubscript{n} denote ahn\textsuperscript{h} + • • • + ain + ao.
Thus if O;;;ai;;;n-1 (i = O, • • •, h), then \emph{a}\emph{\textsubscript{h},\^{}\^{},} a\textbf{\textsubscript{1}}\textbf{,} a\textbf{o} are the digits of the number (ah, • • •, a\textsubscript{b} ao)n relative to the radix n. Alan Sutcliffe studied the prob- ,
lem of finding numbers that are multiplied by an integer when their digits are reversed \emph{(Integers that are multiplied when their digits are reversed,} this Magazine, \textsuperscript{?}
\bigskip
1964] \textbf{MATHEMATICAL NOTES} 653
Similarly,
(3) 6* = - 1.
Taking \emph{q} = characteristic of \emph{F} (g-l=O), choose \emph{t} and \emph{r} as specified in the lemma. Using relations (1), (2), (3), we have
(Z + \emph{ra} + i)(r» + 1 + \emph{ria + lb)} = r(Z\textsuperscript{2} + r* + l)a + (Z\textsuperscript{2} + r\textsuperscript{2} + 1)6 = 0.
One of the factors on the left must be 0, so for some numbers \emph{u, v, w, u} 0 (mod \emph{q)}\emph{,} we have \emph{w+va-\textbackslash{}-ub} \emph{= Q,} or \emph{b =} \emph{— u\textasciitilde{}\textsuperscript{l}va — u\textasciitilde{}\textsuperscript{l}w.} So \emph{b} commutes with \emph{a,} a contradiction. We conclude that S is not a generalized quaternion group, so 5 is cyclic.
Thus every Sylow subgroup of \emph{F*} is cyclic, and \emph{F*} is solvable (\footnote{P. T. Church, Ambiguous points of a function homeomorphic inside a sphere, \emph{Michigan Math. J.,} 4 (1957) 155-156.}, pp. 181— 182). Let \emph{Z} be the center of F* and accnme \textsuperscript{7}\emph{F*.} Then \emph{F*\Slash{}Z} is solvable, and its Sylow subgroups are cyclic. Let \emph{A\Slash{}Z} ("fit.. ZC\^{}) be a minimal normal subgroup of \emph{F*\Slash{}Z. A\Slash{}Z} is an elementary abelian group of order \emph{p\textsuperscript{k}} \emph{(p} prime), so since the Sylow subgroups of \emph{F*\Slash{}Z} are cyclic, \emph{A\Slash{}Z} is cyclic. Any group which is cyclic modulo its center is abelian, so \emph{A} is abelian. Let \emph{x} be qny element of \emph{F*, y} any element of \emph{A.} Since \emph{A} is normal, \emph{xyx\textasciitilde{}'(E.A,} and (1 \emph{+x)y} = z(l +x) for some z£Z. An easy manipulation shows that \emph{y} \emph{— z = zx — xy = (z — xyx\textasciitilde{}')x.}
If \emph{y — z} \emph{= z — xyx\textasciitilde{}'} \texttt{0, then y = z = xyx\textasciitilde{}l, so x and y commute. Otherwise, x} (z — xyx\textasciitilde{}\textsuperscript{l})\textasciitilde{}'(y<\Slash{}em> — z). But \emph{A} is abelian, and z, \emph{y, xyxr'\^{}A,} so \emph{x} commutes with \emph{y.} Thus we have proven that \emph{A} is contained in the center of F*. a contradiction.
\textbf{References}
\textbf{1.} E. Artin, Uber einen Satz von Herrn J. H. M. Wedderburn, Abh. Math. Sem. Hamburg, 5 (1927) 245.
\textbf{2.} L. E. Dickson, On finite algebras, Gottingen Nachr., 1905, p. 379.
\textbf{3.} M. Hall, The theory of groups, Macmillan, New York, 1961.
\textbf{4.} Miller, Blichfeldt and Dickson, Theory and applications of finite groups, Wiley, New York, 1916.
\textbf{5.} B. L. van der Waerden, Moderne Algebra, Ungar, New York, 1943.
\textbf{6.} J. H. M. Wedderburn, A theorem on finite algebras, Trans. Amer. Math. Soc., 6 (1905) 349.
\textbf{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
\bigskip
y e A(n + 1, m[n+ 1], k[n + 1], f [n 4- 1]) C U\textsubscript{n+1} c Q(n 4-1, m[n+ 1]),
\emph{2tt} \emph{i} 1
and therefore each point of y\textsubscript{n} has distance less than from y. Now
\^{}+\textasciitilde{}F 0 \textsuperscript{a}s n -* °°; hence, if we set y = \{y\} U U\textsubscript{n=}i y\textsubscript{n}, then \emph{y} is an arc with
one endpoint at y.
Since U\textsubscript{n} and U\textsubscript{n+1} have a point in common,
\textsuperscript{f-1}(\textsuperscript{* S}(\^{}> \textsuperscript{p}k[n])) \textsuperscript{and rl}(\textsuperscript{S}(\^{}Tl’ Pk[n+1]))
have a common point, and hence
\textsuperscript{S}(\^{}>Pk[n]') \textsuperscript{and S}(\^{}+l’ Pk[n+1])
have a common point. Therefore, if p is the metric on K, then
< \textbackslash{} \textsuperscript{1} . \emph{\textsuperscript{1}} \Slash{} \emph{1} , \^{}Pk[np Pk[n+1] — 2\textsuperscript{n} 2\textsuperscript{n}”\^{}
\chapter{8. March 1969 - Boundary Functions for Bounded Harmonic Functions}
\textbf{Original PDF}: \href{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. \href{https://web.archive.org/web/20190815201940/http:/homepages.rpi.edu/\%7Ebulloj/tjk/tjk6.html}{Boundary functions for bounded harmonic functions.} \emph{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)
\section{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.
\section{Article by Ted}
BOUNDARY FUNCTIONS AND SETS OF CURVILINEAR CONVERGENCE FOR CONTINUOUS FUNCTIONS
BY
T. J. KACZYNSKI
Let \emph{D} be the open unit disk in the complex plane, and let \emph{C} be its boundary, the unit circle. If \emph{x} e C, then by an \emph{arc at x} we mean a simple arc \emph{y} with one end point at \emph{x} such that \emph{y-\{x\}\^{}D.} If \Slash{}is a function mapping \emph{D} into some metric space \emph{M,} then the \emph{set of curvilinear convergence} of \emph{f} is defined to be
\{x e C: there exists an arc \emph{y} at \emph{x} and there exists a point \emph{pEM} such that \Slash{}(z) -> \emph{p} as \emph{z x} along \emph{y\}.}
If is a function whose domain is a subset \emph{E} of the set of curvilinear convergence of \Slash{}, then<\Slash{}> is called a \emph{boundary function} for \Slash{} if, and only if, for each x \emph{e E} there exists an arc \emph{y} at x such that \Slash{}(z) -><\Slash{}> (x) as z -> x along y. Let S be another metric space. We shall say that a function<\Slash{}> is of \emph{Baire class} 1(S, \emph{M)} if
(i) domain \^{} = S,
(ii)range and
(iii) there exists a sequence of continuous functions, each mapping S' into \emph{M,} such that \emph{ , and, since \textbar{}\Slash{}(p) — p\textbar{} <2\textsuperscript{1\Slash{}2}— 1 =ctn fw, the part of this line which lies in \emph{H\textsubscript{o}} is contained in \emph{s(p,} l.fw). We show that \emph{h\textsubscript{Y}} approaches along this line. By substituting \emph{(f(p)—p)y+p} for \emph{x} in the expression for A\textsubscript{n}(,v, \emph{y),} one obtains
A\textsubscript{n}(x,y) = [(1 - \emph{ny)} V 0]
(8) 17
\textsuperscript{r}n + 4 + 2(-
?n-
If \emph{P\^{}s\textsubscript{n},} then \emph{f(p)\^{}l\textsubscript{n},} and one can verify directly that (8) vanishes. Ifp>s\textsubscript{n}, then \emph{f(p) r\textsubscript{n},} and again one can verify directly that (8) vanishes. Thus A\textsubscript{n}(x, \emph{y)} vanishes along that part of the line (7) which lies in \emph{H.}
Solving (7) for \emph{f(p),} we find that, along the given line,
\Slash{}(p) = \emph{(x-(l-y)p)ly,} and hence
\emph{p} = \Slash{}\emph{(\Slash{}■(?)) = \Slash{}}((x-(i \emph{-y)p)ly)- .}
Therefore (if 0 +\textasciitilde{} and -» as u -> -<». Consequently there exists precisely one number u(x, y) that satisfies the equation
(29) u(x, y) - h*( \textsuperscript{y)} ) = 0.
I claim that u(x, y) is a continuous function on
= \{ -
set
A\textsubscript{n}(x, y) =
[(1 - ny) V 0] [(1 - ——\textsuperscript{1} . X A\Slash{}
n n
I r + A - 2s
\textsuperscript{1} n n n
. .s -x
2 -2— y
Then
\textsuperscript{A}n
is continuous in H. Observe that A\textsubscript{n}(x, y) = 0 when y >\_
Using this fact, it is:easy to show that, if we set
f = f + E A , \textsuperscript{0} n=l \textsuperscript{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 ■ \textsuperscript{n}
obtain
(31) A\textsubscript{n}(x, y) =
[(1 - ny) V 0] [(l- —i-j- \textbar{}r\_ <• 1 4 2(1 - 1) (s - p) n n \textsuperscript{3}
- 2h(p) \textbar{} ) V O] v\textsubscript{n<}
If p £ s , then h(p) £ \&\textsubscript{n}, and one can verify directly that (31) vanishes. If p > s\textsubscript{n}, then h(p) £ r , and again one can verify directly that (31) vanishes. Thus A\textsubscript{n}(x, y) vanishes along that part of the line (30) lying in H.
Solving (30) for h(p), we find that, along the given line, h(p) = ,
and hence p = h (h(p)) = h ( -—-Xl-XlE) . Therefore, if 0 < y < 1, p = u(x, y) . So, if \^{}x, y\^{} satisfies (30) , n £ 2, and £ y £ \textasciitilde{}, then
f\textsubscript{0}(x, y) = (yn(n+l) - n)g\textsubscript{n}(p) + ((n+1) - yn(n+l))g\textsubscript{n+1}(p).
Since the coefficients of g\textsubscript{n}(p) and g\textsubscript{n+}j(p) in the above expression add up to 1 and since both coefficients lie in [0, 1], f (x, y) lies on the line segment joining g\textsubscript{n}Cp) to g\textsubscript{n+}\^{}Cp)» \textsuperscript{an}\^{} it follows that f\textsubscript{Q}(x, y) approaches ip(p) \textsuperscript{as} y\^{} approaches p along the line (30). this line lying in H, f(x, y) show that f approaches (s\^{}) that lies in H. Again, we first consider the value of along the
Since each A\textsubscript{n} vanishes on the part of also approaches ip(p) along the line.
Let s\^{} be any point of N. We along the part of the line r. A
(32) x = (-------- x---- - sjy +
given line. Substituting the value of x given by (32) into the expres
sion for A , we obtain n’
(33) A\textsubscript{n}(x, y) =
\textsubscript{[C}l-ny) V 0j[(l - -1- \textbar{}r\textsubscript{n} - r\textsubscript{m + +} 2(1 - 1) (s\textsubscript{n}- \textbar{}) V 0]v\textsubscript{n}.
n n
If s < s , then £ < r < £ < r , and one can verify directly that m n m m r-. n n’
(33) vanishes. If s < s , then £ < r < £ < r , and again one can n m n n — mm
verify that (33) vanishes. Thus, for n \textbar{} m, \^{}(x, y) = 0 when \textsubscript{6j}R\textsuperscript{2}:Olies on the line (9) and in \emph{H.}
If we take \emph{n=m} in (10), we obtain
A\textsubscript{m}(x,y) = [(1-wy) v 0]z\textsubscript{m}.
Therefore A\textsubscript{m}(x,y) approaches z\textsubscript{m}=\^{}(s\textsubscript{m})-\^{}(s\textsubscript{m}) along the given line.
Take any