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Let consider first the space $$V=\{(x_1,x_2,...)\mid \sup_i |x_i|<\infty \}.$$

This is a complete metric space (refer to my course). Now I was wondering if the set $$\mathcal A=\{x\in \ell^\infty \mid \|x\|_{\ell^\infty }\leq 1\},$$ is compact. I know that in a metric space, a set is compact $\iff$ its sequentially compact (i.e. every sequence has a convergent subsequence). I consider the sequence $$x_n=x^i_n,$$ where $x_n^i=(0,0...,0,1,0,...)$ where the $1$ is at the $i^{th}$ position. In other words, $$x_1=(1,0,0,...)$$ $$x_2=(0,1,0,0,...)$$ $$x_3=(0,0,1,0,...)$$

We have that $$\|x_n^i\|_{\ell^\infty }=1$$ for all $n$ and thus $(x_n)_n$ is a bounded sequence. What could be a convergent subsequence ? I guess that there is no convergent subsequence. To me, if a subsequence converge, it must converge to $0$ (I really don't know how to prove this), but in the same time the norm of the limit must be 1 I guess. So it can't have convergent subsequence. Am I right ? So $\mathcal A$ is not compact ?

mathcounterexamples.net
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user380364
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3 Answers3

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According to Riesz's lemma a normed vector space is of finite dimension if an only if its closed unit ball is compact. As your space $V$ is of infinite dimension, $\mathcal A$ can't be compact.

mathcounterexamples.net
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  • Thank you. Is my counter-example a good one ? If yes, how do you show that it has no convergent subsequence ? – user380364 Aug 11 '18 at 09:09
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    Yes. For $m>n$ you have $x_m^n = 0$ hence the sequence can only converge to the always vanishing sequence. In contradiction that $\Vert x_n \Vert_\infty = 1$ for all $n \in \mathbb N$. – mathcounterexamples.net Aug 11 '18 at 09:13
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Hint: if the sequence $(x^i)\subset \ell^\infty$ converges to $x$ in $\ell^\infty$, then it converges component-wise, i.e. $\lim_{i\to +\infty} x^i_n = x_n$ for every $n$.

Rigel
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  • Something doesn't look correct in what you wrote. I have $x_n=x_n^i$, What would be $\lim_{i\to \infty }x_n^i$ if the sequence doesn't depend on $i$ ? The $i$ is just a notation to say $1$ at the $i^{th}$ component. So $x_n^i$ means the $i^{th}$ component. – user380364 Aug 11 '18 at 09:17
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    Your notation is not completely consistent. Anyway, in my notation $x^i \in \ell^\infty$ denotes $x^i = (x^i_1, x^i_2, \ldots, x^i_n, \ldots) = (0, 0, \ldots, 0, 1, 0, \ldots)$ with the $1$ in $i$-th position. – Rigel Aug 11 '18 at 09:20
  • indeed, it looks better :) – user380364 Aug 11 '18 at 09:22
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Note that for $i \neq j$, $||x_i - x_j||_\infty = \max\{|-1|,|1|\} = 1$, so your sequence cannot have a Cauchy subsequence.

There is a fairly common observation here. $[-1,1]^\omega = \mathcal{A}$ and the set of that "cube"'s corners, $\{-1,1\}^\omega$, is a countable set of points with pairwise distance bounded below. With the open cover definition of compact, the collection of open balls of radius $2$ centered on the set of corners is an open cover with no finite subcover (since each center is contained in only one ball in the collection and there are infinitely many corners).

Eric Towers
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