# A question about a detail of Eberlein-Šmulian Theorem: the relation between weak and strong closures of a set

Theorem:

Let $$B$$ be the closed unit ball of a Banach space $$X$$. Then $$B$$ is weakly compact if and only if it is weakly sequentially compact.

Forward direction:

We first assume $$B$$ is compact. Kakutani's Theorem tells us that $$X$$ is reflexive. Every bounded sequence in $$X$$ has a weakly convergent subsequence. Since $$B$$ is weakly closed, $$B$$ is weakly sequentially compact.

Why can we assume $$B$$ is weakly closed? I'm guessing this follows from $$B$$ being strongly closed but I'm unclear on the relationship of the two.

I think weak closure implies strong closure: $$B$$ is weakly closed so its compliment is weakly open. Thus $$B^C$$ is certainly open in a stronger topology - such as the norm topology. Thus $$B$$ is closed in the norm topology. I don't see how the other direction can be true though because we are "coming from a finer topology".

• The assumption in the forward direction is that $B$ is weakly compact. The author probably should have put "weakly" in that first sentence. – Nate Eldredge Jun 29 at 21:24
• @NateEldredge Indeed, or $X$ is finite-dimensional and everything trivialises. – Henno Brandsma Jun 29 at 21:30
• I did not see through to the consequences as Henno did, but I was able to pick this up! The author does this at least twice in this proof. – yoshi Jun 29 at 21:33
• In discussions where several topologies are involved, I prefer to be very specific in what topology some notion is meant. It matters greatly. – Henno Brandsma Jun 29 at 21:41
• the relation is quite clear: weakly closed implies strongly closed, but the reverse only holds for some sets, like convex ones, see here. That's another way to see that $B$ (which is strongly closed and convex) is weakly closed too. – Henno Brandsma Jun 30 at 9:11

$$B$$ weakly compact implies $$B$$ weakly closed, merely by $$X$$ being a Hausdorff space in the weak topology (which actually holds because we can separate points by functionals, Hahn-Banach etc.). This is just a special case of "compact sets are closed in Hausdorff spaces"! (general topology is actually useful, see?)
From this question (and general Banach space theory) we see that a strongly closed and convex subset is in fact weakly closed too. This applies to the closed unit ball $$B$$; if you already know the result you could use that as well. Of course, I like the general topology approach better, as it uses less.