Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If $S$ is an uncountable subset of $C[0,1]$, then there is a uniformly convergent sequence $\{f_n\}$ of distinct functions of $S$.

I know how to do this for $C^1[0,1]$ since $S \subset \cup_{m,n \ge 0} \{f: \sup_{[0,1]} |f(x)| \le m, \sup_{[0,1]} |f'(x)| \le n \}$ and then $\{f: \sup_{[0,1]} |f(x)| \le m, \sup_{[0,1]} |f'(x)| \le n \}$ must be uncountable for some $m,n$ (since otherwise $S$ would be countable) and so contains a uniformly convergent sequence by Arzelà–Ascoli.

share|improve this question

1 Answer 1

Suppose $S$ is such that no $x \in C[0,1]$ is the uniform - that is normwise - limit of a sequence $(x_n)$ of distinct elements of $S$.

Let $x \in C[0,1]$, then there is an $\epsilon_x > 0$ such that $U_{\epsilon_x}(x)= \{y \in C\in C[0,1] \mid \|x-y\| < \epsilon\}$ has finite intersection with $S$ (otherwise, for $n \in\mathbb N$ let $x_n \in U_{1/n}(x) \cap S$ pairwise distinct, giving $x_n \to x$). We have $C[0,1] = \bigcup_x U_{\epsilon_x}(x)$. As $C[0,1]$ is seperable metric, it is Lindelöf, therefore there is a countable $A \subseteq C[0,1]$ with $C[0,1] = \bigcup_{x\in A} U_{\epsilon_x}(x)$. Now $S = \bigcup_{x \in A} (U_{\epsilon_x}(x) \cap S)$ is a countable union of finite sets, hence countable.

As your $S$ is uncountable, there is a normwise convergent sequence $(x_n)$ of distinct elements of $S$.

share|improve this answer
    
But the question was looking for a uniformly convergent sequence? –  joriki Sep 20 '12 at 8:02
    
@joriki And (norm-)convergence in $C[0,1]$ is uniform convergence, right? –  martini Sep 20 '12 at 8:02
    
Ah, sorry, I didn't realize it was norm convergence and not pointwise. –  joriki Sep 20 '12 at 8:04

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.