Folland 5.49(b): Bounded subsets in $X$ (resp. $X^*$) are nowhere dense in the weak (resp. weak*) topologies

Question

I'm reviewing for quals and I came across a problem on weak topologies I can't seem to put together (it could be because it's 11pm!):

Let $X$ be an infinite dimensional Banach space. Then every (norm) bounded subset of $X$
is nowhere dense in the weak topology, and every (norm) bounded subset
of $X^*$ is nowhere dense in the weak* topology.

Could this be a consequence of the Baire category theorem, or just a simple direct argument? It's a bit tricky because the notion of weakly open sets is not very easy for me to conceptualize. For reference the problem suggests using a preceding exercise:

If $E\subset X$ (resp. $X^\ast$) is norm bounded, then so is its weak
(resp. weak*) closure.

I was able to prove this comfortably. This problem is interesting because, as Folland notes, it implies that, e.g., $X^\ast$ is not locally compact in the weak* topology because bounded sets, although compact by Alaoglu's theorem, have empty interior. (Am I interpreting this correctly?)

Answer 1

To be nowhere dense means that the closure has empty interior. Since the weak closure is norm bounded (as you say you can prove), it suffices to prove that every norm bounded set has empty interior in the weak topology.

And indeed, just note that all the basic open sets of the weak topology are unbounded, as is a classical fact about the weak topology in infinite dimensional spaces. (For this point: A basic open set is the collection of all $x$ making $|f_i(x-x_i)| < \epsilon_i$ for finitely many functionals $f_i,$ points $x_i,$ and reals $\epsilon_i.$ If this set is non-empty, fix a point $x$ in it. In infinite-dimensions, you can always find a point in the kernel of finitely many functionals all at once; say this point is $k.$ Then $x+tk$ is in this basic open set for each scalar $t,$ and so you're unbounded by varying $t.$)