# Finite Dimensional Hausdorff Topological Space

1. Apr 14, 2012

### creative_

How do I prove that a Hausdorff topological space E is finite dimensional iff it admits a precompact neighborhood of zero?

2. Apr 15, 2012

### mathwonk

i think you mean banach space, not hausdorff space, and then this is a well known theorem of riesz: th.5.9.4., p.109, of dieudonne's foundations of modern analysis.

3. Apr 15, 2012

### creative_

Hi mathwonk,

I mean Hausdorff space.

4. Apr 15, 2012

### micromass

Staff Emeritus
It is indeed true that a locally compact Hausdorff topological vector space E is finite dimensional.

Proof: Let K be a compact neighborhood of 0. We can assume that K is balanced. Since (1/2)K is a neighborhood of 0, there are finitely many points x1,...,xn such that

$$K\subseteq (x_1+\frac{1}{2}K)\cup...\cup (x_n+\frac{1}{2}K)$$

Let M be the finite dimensional subspace spanned by the x1,...,xn. Then M is closed. The quotient space E/M is Hausdorff. Since $K\subseteq M+\frac{1}{2}K$, then $\varphi(K)\subseteq \frac{1}{2}\varphi(K)$. So (by induction) $\varphi(2^nK)\subseteq \varphi(K)$.

K is balanced, so $E=\bigcup_n 2^n K$. Thus $\varphi(E)=\varphi(K)$. Thus E/M is compact. Which implies that E/M is one point. Thus E=M.

5. Apr 15, 2012

### mathwonk

what do you mean by zero in a topological space? oh i see micromass, you assumed he meant a vector space. the argument you gave is a variation of the argument for Riesz's theorem that I referred to. That's a nice example of extending an argument to a more general setting. I didn't know that version.

Last edited: Apr 15, 2012
6. Apr 17, 2012

### creative_

Thank you micromass. Your proof makes sense.