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I have a question which appeard during my modifications of Weiestrass theorem.

Let $X$ be a reflexive Banach space. I have a set $A$ which is closed and convex, so from Mazur lemma it is weakly closed. Also $A$ is bounded and $A\subset B(0,r)$. Since $X$ is reflexive, then unit ball is weakly sequentially compact, so also $B(0,r)$ is weakly sequentially compact. My question is as follows: is $A$ also weakly sequentially compact? If so, do you know where I can find the proof?

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Yes, in any topological space, any closed subset of a compact set is compact. Short sketch proof: if you cover a closed subset $A$ of a compact set $B$, you can turn it into an open cover of $B$ by adding one new open set: $B \setminus A$. A finite subcover must be a finite subcover of $A$.

In the case of a Banach space, we have the Eberlein-Smulian Theorem, so $A$ being weakly compact is equivalent to $A$ being weakly sequentially compact.

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