Let $\left(\mathcal{X}, \tau\right)$ be a locally convex topological space and $K$ a compact subset. Is there a reference for the proof that on $K$ the relative topology coincides with the relative weak topology?
In other words, if $U\in \tau$ is an open set, then we know that $K\cap U$ is (by definition) open with respect to the relative topology in $K$. The claim is that there exists a weakly open set $V$ (i.e. open in terms of the weak topology on $\mathcal{X}$) so that $K\cap U=K\cap V$. Why is this true?
Proposition $\mathrm{A.3.12}$ of Bogachev's "Gaussian measures" uses this fact and, since it has been some time since I last dealt with topology, I am not really sure if it can be proved with elementary tools or if it follows from some more advanced theorem.
I also found some references to this on StackExchange, e.g. see the comments here, but no proof.
Thanks in advance!
This is elementary, so I will give a proof instead of a reference.
$K$ is also weakly compact (because any weak open cover is also a strong open cover). The identity map from $(K, strong)$ to $(K, weak)$ is a continuous bijection between compact Hausdorff sapces. Hence, it is a homeomorphism, which proves your result.