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I have to verify if the $C([0,1])$, space of all continuous functions defined on interval $[0,1]$ with supremum metric is compact.

As I know, we have to check if every sequence of functions $f_{n}(x)$ has subsequence that $f_{n_{k}}(x)$ is convergent. In this metric of course conervgence implies uniform convergence, so there won't be a problem of showing continuity of the limit function. But I really don't know if we can make subsequence convergent.

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4 Answers 4

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The sequence $f_n(x)=x^n$ does not have a convergent subsequence since $f_n$ converges pointwise towards $f(x)=0$ if $x\neq 1$ and $f(1)=1$ which is not continuous. Hence, the space is not compact.

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    $\begingroup$ Of course! Thank you very much, I was blind or something :) $\endgroup$
    – janusz
    May 24, 2016 at 11:15
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    $\begingroup$ ...is not *continuous. $\endgroup$
    – ahorn
    May 24, 2016 at 11:15
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    $\begingroup$ While this is a good example, notably they were asking about all of $C[0,1]$, which is not even bounded. Any compact set in a metric space must be bounded. $\endgroup$
    – Ian
    May 24, 2016 at 11:16
  • $\begingroup$ @Tsemk Arostide the solution you have done is it for the supremum metric asked in the question ? $\endgroup$ Feb 26, 2022 at 15:02
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There is a general result:

Theorem The unit ball in a normed space is compact if and only if the space is finite dimensional.

An immediate consequence of this is that the unit ball in $C([0,1]$ is not compact and hence (as a closed subset of a space being non-compact implies the space itself is not compact) the whole of $C([0,1])$ is non-compact.

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  • $\begingroup$ "as a subset of a space being non-compact implies the space itself is not compact" This is not true. The set $(0,1)$ is not compact with Euclidean metric, but $[0,1]$ is. $\endgroup$
    – Kooranifar
    Dec 21, 2021 at 15:26
  • $\begingroup$ oh yeah - probably I meant to say "as a closed subset of a space being non-compact implies the space itself is not compact" $\endgroup$
    – Josh R
    Dec 31, 2021 at 0:02
  • $\begingroup$ I see, it makes sense then. $\endgroup$
    – Kooranifar
    Dec 31, 2021 at 2:25
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Let be $f_n:[0,1]\to\mathbb{R}$ with $f_n(x):=n$ for all $n\in\mathbb{N}$. Clearly, $f_n\in C([0,1])$ for all $n\in\mathbb{N}$ but $(f_n)_{n\in\mathbb{N}}$ is unbounded and can't contain any convergent subsequence. Hence, $C([0,1])$ is not compact.

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A Banach space is finite-dimensional if (and only if) it is locally compact. It's easy to see that $C[0,1]$ is a Banach space with respect to the supreme norm. And, trivially, the property of compactness implies local compactness. To sum it up, there is not infinite-dimensional Banach space (such as $C[0,1]$ ) that is also compact.

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  • $\begingroup$ I do not understand your third sentence at all (perhaps you meant to switch compactness and local compactness in that sentence.) This is also a higher level view than I think is appropriate for the OP. $\endgroup$
    – Ian
    May 24, 2016 at 11:17
  • $\begingroup$ @Ian Absolutely! $\endgroup$
    – user335721
    May 24, 2016 at 11:27

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