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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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    Thank you @Martin Sleziak, I knew this. My interest is how to solve the exercise with the hint.2011-07-03

1 Answers 1

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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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    @leo: Ok, maybe there is a more direct way, but I thi$n$k I did use your hi$n$t, did$n$'t I :-)2011-07-05