Is R^w a first category topological vector space?

[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

 

[Civilized]

Then the union as n goes from 1 to infinity will be R^w.

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

 

[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.

 

[g_edgar]

What am I missing?

the proof of "nonempty interior"

 

[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?