Let $X$ be a Hausdorff topological vector space such that $X'$,it topological dual separate points of $X$, and $A\subset X$. I want to know if the following sytatement is true or not?
If $\overline A$ is compact, then $\overline{A}=\overline{A}^w$, where $w$ denotes the weak closure.
My attemp: Since $$A\subset \overline{A}\subset \overline A^w,$$
taking weak closure from sides of the above inclusions, we have
$$\overline{A}^w\subset \overline{\overline{A}}^w\subset \overline{A}^w$$ which gives $\overline{\overline{A}}^w=\overline{A}^w$. From compactness of $\overline{A}$, we know it is weakly compact, and hence weakly closed, so $\overline{\overline{A}}^w=\overline{A}$.