- #1

- 3

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter creative_
- Start date

- #1

- 3

- 0

- #2

mathwonk

Science Advisor

Homework Helper

2020 Award

- 11,153

- 1,347

- #3

- 3

- 0

Hi mathwonk,

I mean Hausdorff space.

I mean Hausdorff space.

- #4

- 22,089

- 3,297

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 x

[tex]K\subseteq (x_1+\frac{1}{2}K)\cup...\cup (x_n+\frac{1}{2}K)[/tex]

Let M be the finite dimensional subspace spanned by the x

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

- #5

mathwonk

Science Advisor

Homework Helper

2020 Award

- 11,153

- 1,347

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:

- #6

- 3

- 0

Thank you micromass. Your proof makes sense.

Share:

- Replies
- 13

- Views
- 4K