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Show that the closed ball in $C([0,1])$ of center $0$ and radius $1$ is not compact.

I thought it will be compact since every closed and bounded set in $\mathbb{R}$ is compact?

Why is it not compact and how can I prove it?

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    $\begingroup$ $C([0,1]) \not\subset \mathbb{R}$. $\endgroup$
    – njguliyev
    Nov 8, 2013 at 22:48
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    $\begingroup$ $C([0,1])$ is an infinite-dimensional Banach space. These are different from $\mathbb{R}^n$. You can prove it for example by finding a sequence in the ball that has no convergent subsequence. $\endgroup$ Nov 8, 2013 at 22:48
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    $\begingroup$ find a sequence such that each term differ by a fixed distance. First prove that you can find an $x$ such that $d(x, A) > 1 - \varepsilon$ and $||x|| = 1$. $\endgroup$
    – user40276
    Nov 8, 2013 at 22:49
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    $\begingroup$ Actually, the ball is compact iff the space is finite dimensional, since no infinite dimensional space can be locally compact. $\endgroup$
    – user40276
    Nov 8, 2013 at 22:52
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    $\begingroup$ @StefanSmith: There is a norm on the dual space $X'$, and you can define $B$ to be the unit ball for this norm. Then if you put the weak* topology on $X'$, then it is closed. But this topology isn't the one given by the norm. So you have to be careful, as "the unit ball is closed" is a legitimate statement now. $\endgroup$ Nov 9, 2013 at 16:35

2 Answers 2

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This answer is for posterity, and I hope someone appreciates it. $C([0,1])$ is a metric space, so it suffices to show it has a bounded sequence with no convergent subsequence. Such a sequence is $(f_n)$ where $f_n(x)=x^n$. The boundedness is obvious. The sequence converges pointwise to a noncontinuous function. No subsequence can converge (in the metric of $C([0,1])$, that is, uniformly) because if a sequence of continuous functions converges uniformly to a function, that limiting function must be continuous.

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  • $\begingroup$ I think I see it now. Since it is not continuous then it does not converge uniformly? But then why would it not be compact? Is it because it has no convergent subsequence? $\endgroup$
    – user104235
    Nov 9, 2013 at 0:19
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    $\begingroup$ @user104235 : a metric space is compact if and only if it is sequentially compact, and this argument shows the unit ball in $C([0,1])$ (with the usual norm), which is a metric space, is not sequentially compact, be exhibiting a sequence in it with no convergent subsequence. $(f_n)$ converges pointwise to the function $f$ given by $f(x)=0$ for $0\leq x < 1$ and $f(1)=1$ (sorry, I forgot how to do \cases). $f$ is clearly not continuous. Convergence in the usual metric of $C([0,1])$ is the same as uniform convergence. If a sequence of continuous functions converges uniformly,... $\endgroup$ Nov 9, 2013 at 16:13
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    $\begingroup$ ...it converges uniformly to a continuous function. If any subsequence of $(f_n)$ converged uniformly to a function, it would have to converge pointwise to the same function, namely $f$. Since $f$ is not continuous, this is impossible. How much you like this answer depends on whether you have learned all the facts I have used. I don't want you to unaccept the answer you accepted, which looks good (I have not read it, but no one has complained about it, so it is almost certainly correct), since that would be impolite. $\endgroup$ Nov 9, 2013 at 16:20
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    $\begingroup$ +1: Nice. You can also directly show that some subsequence of your $f_n$ satisfies $\|f_{n_i} - f_{n_j} \| \ge \frac{1}{2}$ for any $i \neq j$, and that the open cover $B(x,\frac{1}{4})$ can have no finite subcover. $\endgroup$
    – copper.hat
    Nov 9, 2013 at 23:18
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    $\begingroup$ @user0 It looks good to me. That is what I had in mind. $\endgroup$
    – copper.hat
    Aug 4, 2023 at 18:42
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Let $f_n$ be zero except on $[\frac{1}{2}(\frac{1}{n+1}+\frac{1}{n}), \frac{1}{2}(\frac{1}{n}+\frac{1}{n-1})]$, where the graph is described by joining the points $(\frac{1}{2}(\frac{1}{n+1}+\frac{1}{n}),0), (\frac{1}{n},1), (\frac{1}{2}(\frac{1}{n}+\frac{1}{n-1}),0)$. Then $\operatorname{supp} f_n = [\frac{1}{2}(\frac{1}{n+1}+\frac{1}{n}), \frac{1}{2}(\frac{1}{n}+\frac{1}{n-1})]$, and $f_n(\frac{1}{n}) = 1$. Hence $\|f_n-f_m\| = \delta_{mn}$.

The collection $\{ B(x, \frac{1}{2}) \}_x$ is an open cover of $C[0,1]$. If we take any finite sub-collection, then at most one of the $f_k$ can be contained in each one, so the finite sub-collection can contain only a finite number of $f_n$. It follows that $C[0,1]$ is not compact.

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  • $\begingroup$ Copper is there another way (easier to see) then the example you provided? Because I would never guessed to try that. $\endgroup$
    – user104235
    Nov 8, 2013 at 23:34
  • $\begingroup$ Well, in some way this example captures the essence of the why it is not compact. In a metric space compactness is equivalent to being complete (analogous to closed in finite dimensions) and totally bounded (analogous to bounded in finite dimensions). The closed unit ball is complete, so compactness fails on total boundedness. The example above shows that $C[0,1])$ is not totally bounded. $\endgroup$
    – copper.hat
    Nov 8, 2013 at 23:41
  • $\begingroup$ Okay then, I will study it more and hopefully I will understand it. Thanks for the help! $\endgroup$
    – user104235
    Nov 8, 2013 at 23:55
  • $\begingroup$ @copper.hat Can you please clarify what $\{B(x,\frac12\}_x$ means in your notation? Is $x$ some function? $\endgroup$
    – sequence
    Nov 1, 2017 at 5:54
  • $\begingroup$ @copper.hat Also, can we use the property of finite intersections of closed sets? Consider $f_n(x) = \frac{1}{n+x}$, which is in $(C[0,1], \|\cdot\|_\infty)$. Let $F_n:=\{f_k\}_{k=n}^\infty$, then $F_n$ is closed. Also, for any $m\in\mathbb{N}$, $\bigcap\limits_{k=1}^m F_n = F_m\ne \emptyset$, but $\bigcap\limits_{n=1}^\infty F_n = \emptyset$. Thus the unit closed ball $B[0,1]$ is not compact in this space. $\endgroup$
    – sequence
    Nov 1, 2017 at 5:57

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