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Let $(X,\Vert\cdot\Vert)$ be a normed $\mathbb{K}$-vector space and $A \subset X$ be closed and bounded. My problem is how to determine whether $A$ is compact?

I know that a compact subset is always closed and bounded. And that in $\mathbb{R}$ the converse holds due to Bolzano–Weierstraß.

But I think I am missing something. Do there exist subsets in normed vector spaces which are closed and bounded but not compact?

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Let $c_{0}$ denote the infinite-dimensional normed vector space consisting of sequences that converge to $0$. The norm on $c_{0}$ is given by:

$$ \|(a_{1},a_{1},a_{3},\ldots)\|:=\sup_{k\in\mathbb{N}}|a_{k}|. $$

Let $A$ denote the set of all vectors in $c_{0}$ of norm at most one. Clearly $A$ is bounded and it is easy to show that $A$ is closed. To see that $A$ is not compact, let, for each $x\in A$, $B_{1/2}(x):=\{y\in c_{0}:\|x-y\|<1/2\}$. Then, $\{B_{1/2}(x):x\in A\}$ is an open cover of $A$. To see that it has no finite subcover, let, for each $k\in\mathbb{N}$, $e_{k}$ denote the sequence $(0,\ldots,0,1,0,\ldots)$, where the $1$ is in position $k$. Since $\|e_{k}-e_{\ell}\|=1$ if $k\not=\ell$, it follows that no two such vectors can belong to the same $B_{1/2}(x)$ for any $x$.

Note that in general, a set in a metric space is compact if and only if it is complete and totally bounded.

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Yes.

The equivalence of "compact" and "closed and bounded" holds only in finite-dimensional spaces.

In a general, infinite-dimensional normed space (e.g., over the real scalars) complete in its norm (i.e., an infinite-dimensional Banach space), for a set to be compact, it is not enough that it be closed and bounded. It must be closed and totally bounded.

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    $\begingroup$ I think you mean "complete and totally bounded." Here "completeness" is of course with respect to the restricted metric on the subset in question. In a Banach space, closedness suffices, but for general normed spaces it does not. $\endgroup$ – Shalop May 9 at 7:11
  • $\begingroup$ @Shalop Thank you. I edited accordingly. $\endgroup$ – avs May 9 at 7:33
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    $\begingroup$ Sure, let $H$ consist of the set of all sequences of real numbers which are eventually zero. Let us equip $H$ with the $\ell^2$ norm, this means $\|x\| = \sum_n|x_n|^2$. Let $C$ consist of all the sequences such that $\sum_n n|x_n|^2\leq 10$. Then $C$ is closed and totally bounded. Furthermore, if we define $x^i_n:=n^{-2}\cdot 1_{\{i \ge n\}}$, then $(x^i)_i$ is a Cauchy sequence in $C$ but doesn't converge. $\endgroup$ – Shalop May 9 at 7:37
  • $\begingroup$ @Shalop Thank you! $\endgroup$ – avs May 9 at 7:39
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Example. The unit ball $B$ in Hilbert space $l^2$. It is closed and bounded, but not compact. Let the orthonormal unit vectors be $e_n, n=1,2,3,\dots$. Then sequence $e_n$ is a sequence in $B$, but it has no convergent subsequence. This is because all the distances are $\|e_n-e_m\| = \sqrt{2}$ so in fact no subsequence is even a Cauchy sequence.

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