Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Suppose $f_n$ is a sequence of real valued functions on $[0,1]$ which converges pointwise to zero.

  1. Is there an uncountable subset $A$ of $[0,1]$ so that $f_n$ converges uniformly on $A$?

  2. Is there a subset $A$ of $[0,1]$ of cardinality the continuum so that $f_n$ converges uniformly on $A$?

Background: Egoroff's theorem implies that the answer to (2) is yes if all $f_n$ are Lebesgue measurable. It is not hard to show that the answer to (1) is yes if you change "uncountable" to "infinite".

Motivation: I thought about this question while teaching real analysis this term but could not solve it even after looking at some books, googling, and asking some colleagues who are much smarter than I, so I assigned it as a problem (well, an extra credit problem) to my class. Unfortunately, no one gave me a solution.

ADDED 11-12-10: Thanks for all the great answers. I accepted Jonas' answer since it was the first one.

share|improve this question
5  
This is the best question I have ever seen. –  Will Jagy Nov 12 '10 at 4:02
6  
Of course, for your questions it is only the cardinality of [0,1] that is important, not any topological or measure-space structure we might usually put on it. I assume you phrased it this way to highlight the contrast with Egoroff's theorem? –  Yemon Choi Nov 12 '10 at 4:04
3  
convergeness ?? –  Will Jagy Nov 12 '10 at 4:25
1  
My one remaining desire, an answer from Jonas Meyer, has been fulfilled. Now I rest. –  Will Jagy Nov 12 '10 at 4:40
4  
The paper "Small Combinatorial Cardinal Characteristics and Theorems of Ergorov and Blumberg" by Krzysztof Ciesielski and Janusz Pawlikowski, Real Analysis Exchange, p905-912, might be useful here. It seems to imply that the answer to your question is "yes" for some SUBSEQUENCE of the $f_n$, subject to the consistency with ZFC of certain hypotheses about Ramsey ultrafilters and dominating numbers. The paper is on Project Euclid. Here's a link: projecteuclid.org/… –  SJR Nov 12 '10 at 5:15

5 Answers 5

up vote 16 down vote accepted

I did some Googling and came up with something that looks relevant, Theorem 10 quoted below from Morgan's Point set theory. It cites works of Sierpiński from the late 1930s, but I can't tell what works are cited because the preview won't let me see that page in the references.

The existence of a linear set having the power of the continuum that is concentrated on a denumerable set is equivalent to the existence of a pointwise convergent sequence of functions of a real variable that does not converge uniformly on any uncountable set.

share|improve this answer
9  
The existence of a set of reals of the cardinality of continuum which is concentrated on a countable set is a known consequence of the continuum hypothesis. –  Péter Komjáth Nov 12 '10 at 4:44
8  
Thanks, Jonas, and everyone else for the information. If any of you offers a class in Googling for Math I will enroll. –  Bill Johnson Nov 12 '10 at 13:23

A simple diagonalization argument gives a counterexample if you assume the continuum hypothesis. Let a modulus of convergence be a sequence $\delta:{\mathbb N}\to(0,\infty)$ which converges to 0, and say that $\delta>\delta'$ if $\delta(n)>\delta'(n)$ for sufficiently large $n$. By diagonalizing, for any countable set $\{\delta_n\}$ of sequences there is some $\Delta$ (which still converges to 0) such that $\Delta>\delta_n$ for all $n$.

We say that a sequence $x_n$ converges to $x$ with modulus $\delta$ if $|x_n-x|<\delta(n)$ for all $n$. We then have that $f_n$ converges to $f$ uniformly iff there is a single $\delta$ such that $f_n(x)$ converges to $f(x)$ with modulus $\delta$ for all $x$.

Now assuming the continuum hypothesis, the set of all moduli and the set of all points in $[0,1]$ are both in bijection with $\omega_1$; let $(\delta_\alpha)_{\alpha<\omega_1}$ and $(x_\alpha)_{\alpha<\omega_1}$, respectively, be enumerations. For each $\alpha$, let $\Delta_\alpha$ be a modulus such that $\Delta_\alpha>\delta_\beta$ for all $\beta<\alpha$ (since there are only countably many such $\beta$). Now let $f=0$ and define $f_n(x_\alpha)=\Delta_\alpha(n)$. Then $f_n(x_\alpha)$ converges to $f(x_\alpha)=0$ for all $x_\alpha$. However, for any fixed modulus $\delta=\delta_\beta$, $f_n(x_\alpha)$ cannot converge with modulus $\delta$ for $\alpha>\beta$ by construction, so there are only countably many points on which $f_n$ converges uniformly with modulus $\delta$.

share|improve this answer
    
Nice, simple argument, Eric. –  Bill Johnson Nov 12 '10 at 15:30

To add a bit to Jonas Meyer's answer.

Sierpiński's result was first published in C.R. Soc. Sc. Varsovie 1928, p. 84-87. It was reproduced in his monograph "Hypothèse du continu" (Lwów, 1934, p. 52):

Proposition $C_9$. Il existe une suite infinie convergente de fonctions d'une variable réelle
$f_1(x)$, $f_2(x)$, $f_3(x),...$ qui convergent non uniformément sur tout ensemble indénombrable.

Sierpiński effectively derived it from the statement which is implied by the continuum hypothesis (Ibid. p. 36):

Proposition $C_1$. Il existe un ensemble linéaire $N$ de puissance du continu qui admet un ensemble au plus dénombrable de points communs avec tout ensemble (linéaire) parfait non-dense.

share|improve this answer
    
Proposition $C_9$. There exists a converging infinite sequence of functions of a real variable $f_1(x)$, $f_2(x)$, $f_3(x),...$ which converges non-uniformly on every uncountable set. –  Andrey Rekalo Nov 12 '10 at 5:26
    
Proposition $C_1$. There exists a set of reals $N$ of the cardinality of continuum such that its intersection with every perfect nowhere dense set of reals is at most countable. –  Andrey Rekalo Nov 12 '10 at 6:32
    
Thanks for the reference. I didn't know Sierpiński's monograph was available online. –  Andres Caicedo Nov 12 '10 at 22:56

This paper by R. Pinciroli claims that the stronger statement:

$f_n$ converges uniformly on sets of outer measure arbitrarily close to 1

is undecidable in ZFC.

share|improve this answer
2  
So the answer is both yes and no to question 1. Is it provable in ZFC that sets of positive outer measure have cardinality the continuum? –  Bill Johnson Nov 12 '10 at 13:35
    
Bill, it is independent of ZFC. See this answer of Joel David Hamkins: mathoverflow.net/questions/8972/… or this expository note by Briggs and Schaffter: jstor.org/stable/2320153. –  Jonas Meyer Nov 12 '10 at 15:31
    
Thanks, Jonas. That is quite interesting. –  Bill Johnson Nov 12 '10 at 16:42

Also, Shinoda (1973) proved that if Martin's Axiom and $\neg CH$ hold, then the answer to 1) si affirmative.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.