# On compactness of weak closure of a subset in a locally convex space

Let $$X$$ be a complete Hausdorff locally convex topological vector space, let $$X''=(X',\beta(X',X))'$$ be its standard bidual. Let $$\sigma(X,X')$$ be the weak topology on $$X$$ and let $$\sigma(X'',X')$$ be the weak* topology on $$X''$$.

Let $$A$$ be an absolutely convex (=balanced+convex) closed subset of $$X$$ such that its $$\sigma(X'',X')$$-closure in $$X''$$, that is, $$\overline{A}^{\sigma(X'',X')}$$ is compact in $$(X'',\sigma(X'',X'))$$. Is it true that the weak closure $$\overline{A}^{σ(X,X')}=A$$ is compact in $$(X,\sigma(X,X'))$$?

I think that this should be true since $$(X,\sigma(X,X'))$$ is a topological subspace of $$(X'',\sigma(X'',X'))$$. But I am unable to verify this. Can any body give a clue, or a hint, or an answer?

No. The unit ball of $$c_0$$ is weakly closed but not weakly compact. Its closure in the bidual is the unit ball of $$\ell^\infty$$ which by Alaoglu is weak$$^\ast$$ compact.

• Thanks for the answer. It seems that the space $c_0$ is a good source of counter-examples. Sep 22, 2019 at 19:59
• Any non-reflexive Banach space would work here. Sep 22, 2019 at 20:01