Convex set and weak topology: the Banach hypothesis.

Question

Let X be a Banach space. It is known that if a subset C of X is strongly closed then it is also weakly closed. However, in the proof of the theorem, I can not understand where the hypothesis that X is complete is used. In fact, the strong separation theorem does not require that space X be Banach. Any suggestions?

Answer 1

You are absolutely right. And you can see this more clearly in the more general case as follows:

Suppose $(X,\mathcal{T})$ is a locally convex (Hausdorff) space and $A$ is a convex subset of $X$. Then $A$ is $\mathcal{T}$-closed if and only if it is weakly closed.