# Functional analysis question

1. Jul 11, 2009

### redrzewski

This is from Rudin, Functional Analysis 2.1. Not homework.

If X is an infinite-dimensional topological vector space which is the union of countably many finite-dimensional subspaces, prove X is first category in itself.

What about this example? Take R^n (standard n-dimensional space of reals) as each of the finite-dimensional subspaces. Then the union as n goes from 1 to infinity will be R^w.

R^w is infinite-dimensional, and it will contain closed sets that have non-empty interior. R^w seems like it will satify the axioms of the topological vector space. Hence this would contradict the problem.

What am I missing?

thanks

2. Jul 11, 2009

### Civilized

This is not true, just as $R \cup R \neq R^2$.

3. Jul 11, 2009

### redrzewski

I was assuming R U R^2 U R^3 = R^3, etc.

I'm getting hung up on if this construction actually is a union of finite dimensional sets. On the one hand, given any set in the union, it has finite dimension.

On the other hand, since every set is a superset of the ones of lower dimensions, and that the total union has infinite dimension, it seems like one of the sets in the union must have infinite dimension.

Last edited: Jul 11, 2009
4. Jul 12, 2009

### g_edgar

the proof of "nonempty interior"

5. Jul 13, 2009

### redrzewski

I don't understand the comment. Are you saying that R^w is 1st category, and that I need to prove it?

But the closure of any open ball B(0,r) in a metric for R^w has non-empty interior, right?