# Characterization of normed vector spaces of finite dimension

I have this problem:

Let $E$ be a normed vector space. $S=\{x\in E : ||x||=1\}$. Show that if $S$ is compact then $\dim E$ is finite.

This follows directly from the Riesz's lemma, but in the notes of the course, the hint for this exercise is:

"The set $\{x\in E: a\leq ||x||\leq b\}$ is homeomorphic to the set$[a,b]\times S$, for $b>a>0$."

How can I use this to solve the problem?

Here a homeomorphism is a function $f$ between metric spaces $E$ and $E'$, bijective, and such that $f$ and $f^{-1}$ are continuous. And then, $E$ is homeomorphic to $E'$ if there exist a homeomorphism $f:E\to E'$.

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@John M: If $\mathrm{dim}E$ is infinite, then using Riesz's lemma one can get a sequence of points in $S$ with pairwise distances bounded below by $\frac{1}{2}$, which shows that $S$ is not (sequentially, and hence plain old) compact because it contains a sequence with no convergent subsequence. – Jonas Meyer Jul 3 '11 at 6:40
@Jonas Meyer: You're right. It does follow directly, so I repudiate my own comment. – John M Jul 3 '11 at 6:50
See e.g p.19 of Fabian, Habala et al: Banach Space Theory; books.google.com/… – Martin Sleziak Jul 3 '11 at 12:09
Thank you @Martin Sleziak, I knew this. My interest is how to solve the exercise with the hint. – leo Jul 3 '11 at 15:24

This is a standard result for (Hausdorff) topological vector spaces: They are finite dimensional iff they are locally compact. The hint in your exercise can be used to show that E is locally compact iff S is compact. Because if S is compact, then so is $$\overline B_{1} := \{ \|x\| \le 1 \} \cong S \times [0, 1]$$ and therefore every closed epsilon ball $$\overline B_{\epsilon} := \{ \|x\| \le \epsilon \}$$ is compact.

So, we can assume that E is locally compact and have to show that this implies that it is finite dimensional. This is a standard excercise.

Define for every $\epsilon \gt 0$ $$B_{\epsilon} (x_0) := \{ \|x - x_0\| \lt \epsilon \}$$

Since $\overline B_{1}$ is compact, you have for every open cover consisting of $B_{\epsilon} (x_0)$ for all $x_0 \in \overline B_{1}$ a finite subcover for some points $x_1, ..., x_n$. If you choose your $\epsilon$ small enough, you'll be able to show that $\overline B_{1}$ is contained in the linear span of these points.

But, again, this is a standard argument you'll find in any book about topological vector spaces, like, for example in Helmut Schäfer's book "Topological Vector Spaces" in paragraph 3 ("Topological Vector Spaces of Finite Dimensions") of the first chapter.

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van Beek, Only a little thing. You don't use exactly the hint that I post, but this is useful to me, thanks. – leo Jul 3 '11 at 15:37
$\overline{B}_1$ is merely a continuous image of $S \times [0, 1]$, no? Of course, this is enough for what you go on to say. – Dylan Moreland Jul 3 '11 at 15:47
@leo: Ok, maybe there is a more direct way, but I think I did use your hint, didn't I :-) – Tim van Beek Jul 5 '11 at 6:43