TITLE: Characterization of normed vector spaces of finite dimension
QUESTION [5 upvotes]: 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 $f:E\to E'$.

REPLY [5 votes]: 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.