# Compactness in normed vector spaces.

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?

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.

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.

• 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. – Shalop May 9 at 7:11
• @Shalop Thank you. I edited accordingly. – avs May 9 at 7:33
• 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. – Shalop May 9 at 7:37
• @Shalop Thank you! – avs May 9 at 7:39

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.