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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?

user104235
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    $C([0,1]) \not\subset \mathbb{R}$. – njguliyev Nov 08 '13 at 22:48
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    $C([0,1])$ 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. – Daniel Fischer Nov 08 '13 at 22:48
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    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 $||x|| = 1$. – user40276 Nov 08 '13 at 22:49
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    Actually, the ball is compact iff the space is finite dimensional, since no infinite dimensional space can be locally compact. – user40276 Nov 08 '13 at 22:52
  • The intuition is that you have infinite directions to walk. – user40276 Nov 08 '13 at 23:08
  • @user40276: What you say is not true. It is true that a locally compact vector space is finite dimensional. It is true that the unit ball of a normed vector space is compact iff the space is finite dimension. But the unit ball of an infinite dimension vector space can be compact. https://en.wikipedia.org/wiki/Banach-Alaoglu_theorem for example. – Najib Idrissi Nov 08 '13 at 23:08
  • @nik Sorry, I mean normed vector space, since I have never studied topological vector spaces. – user40276 Nov 08 '13 at 23:11
  • @nik : how can "unit ball" have a meaning if there is no norm? If this is a stupid question, I would settle for a reference answering this question. I did look at the Wikipedia link, and I found it confusing, and it did not seem to contradict user40276's claim, since it was not apparent to me that the weak* topology on the dual space of a normed vector space can be given by a metric on that dual space. – Stefan Smith Nov 09 '13 at 16:26
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    @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. – Najib Idrissi Nov 09 '13 at 16:35
  • @nik : so, if $X$ is infinite-dimensional, $B$ is not compact if you use the norm on $X'$, but $B$ is compact in the weak* topology on $X'$? – Stefan Smith Nov 09 '13 at 17:46
  • @StefanSmith: Exactly. – Najib Idrissi Nov 09 '13 at 18:15

2 Answers2

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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.

Stefan Smith
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  • 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? – user104235 Nov 09 '13 at 00:19
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    @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,... – Stefan Smith Nov 09 '13 at 16:13
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    ...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. – Stefan Smith Nov 09 '13 at 16:20
  • @user104235 :my answer is essentially an expansion of Daniel Fischer's hint (just below your question) into an answer. – Stefan Smith Nov 09 '13 at 17:40
  • Thank you very much for the clarification! – user104235 Nov 09 '13 at 23:11
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    +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. – copper.hat Nov 09 '13 at 23:18
  • Stefan, one last question since $C([0,1])$ is the space of all continuous, real valued functions on $[0,1]$ then how come you said it converges to a noncontinuous function? – user104235 Nov 11 '13 at 03:39
  • @user104235 : the sequence of functions $(f_n)$ I gave converges pointwise to a noncontinuous function $f$ (which by definition does not belong to $C([0,1])$. Actually any subsequence of $(f_n)$ converges to that same $f$. The convergence cannot be uniform (in other words, a subsequence cannot converge in the metric of $C([0,1])$) because if a sequence of continuous functions converges uniformly to some function $g$, then $g$ has to continuous. – Stefan Smith Nov 11 '13 at 18:43
  • @StefanSmith Thanks for clarification! – user104235 Nov 13 '13 at 03:11
  • @copper.hat Is there a simple way to prove the claim in your comment above? I formulated an inequality in two variables, but it is messy, so before I spend more time to work on it, I thought maybe you could point me in another direction. – user0 Aug 03 '23 at 18:59
  • @user0 It follows from the subsequence I mentioned. Draw a little picture of the $f_n(x) = x^n$ to convince yourself. – copper.hat Aug 03 '23 at 19:48
  • @copper.hat I posted an answer. If you had something simpler yet rigorous in mind, would you please let me know. – user0 Aug 04 '23 at 17:51
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    @user0 It looks good to me. That is what I had in mind. – copper.hat Aug 04 '23 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.

copper.hat
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  • Copper is there another way (easier to see) then the example you provided? Because I would never guessed to try that. – user104235 Nov 08 '13 at 23:34
  • 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. – copper.hat Nov 08 '13 at 23:41
  • Okay then, I will study it more and hopefully I will understand it. Thanks for the help! – user104235 Nov 08 '13 at 23:55
  • @copper.hat Can you please clarify what ${B(x,\frac12}_x$ means in your notation? Is $x$ some function? – sequence Nov 01 '17 at 05:54
  • @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. – sequence Nov 01 '17 at 05:57
  • It means all open balls of radius ${1 \over 2}$ centred on points $x \in C[0,1]$. – copper.hat Nov 01 '17 at 22:33
  • @sequence: Why do you say your $F_n$ is closed? We have $f_n \to 0$ but $0 \notin F_n$. – copper.hat Nov 01 '17 at 22:35