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Is $C([0,1])$ (I guesss with the max-norm) a compact space?

I have to know that because I want to apply Arzela Ascoli.

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It is a real vector space... –  Qiaochu Yuan Feb 14 '13 at 22:05
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If it were, then so would be the subspace of constant functions, which is isometric (hence homeomorphic) to $\mathbb{R}$ (usual norm, non compact). –  1015 Feb 14 '13 at 22:07
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It’s not even locally compact; see this question. –  Brian M. Scott Feb 14 '13 at 22:16
    
The subspace of functions $[0,1] \to [0,1]$ where $|f(x)-f(y)| \leq |x-y|$ is compact, that follows from Arzela-Ascoli. –  sdcvvc Feb 14 '13 at 22:27
    
@scevvc: your set is unbounded, you need to assume a tiny bit more. –  GEdgar Feb 14 '13 at 22:31
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3 Answers

up vote 2 down vote accepted

Consider $f_n(x) = \begin{cases} 0 & x \in [0,\frac{1}{2}-\frac{1}{n}) \\ 1+n(x-\frac{1}{2}) & x \in [\frac{1}{2}-\frac{1}{n}, \frac{1}{2}) \\ 1 & x\in [\frac{1}{2},1] \end{cases}$.

(Since you were considering Arzela Ascoli.)

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That's a lot of work when you could consider $f_n(x)=n$. –  Chris Eagle Feb 14 '13 at 22:20
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I realize that, but the $f_n$ are bounded, which shows that it goes beyond considering constant functions, and it is easy to see that the $f_n$ are not equicontinuous, so it is really not a lot of work at all. –  copper.hat Feb 14 '13 at 22:21
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No, of course not. No (nontrivial) normed space is compact. The sets $\{v\mid\|v\|<n\}$ form an open cover with no finite subcover.

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No it is not, but you don't need $\rm C([0,1])$ to be compact to apply Ascoli-Arzela, but that $[0,1]$ is compact !

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