Locally bounded topological vector spaces

Question

Let ‎$ X $ ‎‎‎be a topological vector space ‎such that every neighborhood of zero contains an infinite dimensional subspace. Then ‎$ X $ is not locally bounded. I do not know, why? We know ‎$ X $ is locally bounded if zero has a bounded neighborhood.

Answer 1

I believe you came across this in Rudin's Functional Analysis at the end of section 3.11. I think his intended argument is a bit unclear there. I was also confused, which is how I came across your post.

He uses the fact that every neighborhood of zero in the weak topology contains an infinite dimensional subspace to prove that the weak topology is not locally bounded. However, the containment of even a finite dimensional subspace will do:

If $U$ is an open set in a topological vector space and if $U$ contains any non-trivial subspace $Y$ (not necessarily infinite dimensional), then $U$ is unbounded.

Pick any non-zero $p \in Y \subset U$. Then $V = U \setminus \{p\}$ is also open. If $U$ is bounded, then $U \subset tV$ for some real $t$.

[I'm using Rudin's assumption that singletons are by definition closed in a TVS; if you're not using this assumption, then the Hausdorfness of the weak topology shows that singletons are closed sets, so my argument here can still be applied to the weak topology on a TVS.]

In particular, $tp \in Y \subset U \subset tV$. But then $p \in V$, a contradiction, so U is unbounded.