Let $X$ be a topological vector space. Then how you show $A^\perp$ is closed in $X^*$ under the strong topology?

Question

Let $X$ be a topological vector space and let $A\subseteq X$. Then how you show $A^\perp=\{x^*\in X^*:\langle a,x^*\rangle=0 \textrm{ for all }a\in A\}$ is closed in $X^*$ under the strong topology?

Answer 1

Let $f_a:X^*\rightarrow\mathbb{R}$ by $f_a(x)=\langle x,a\rangle$, $f$ is continuous for the strong topology, so $f_a^{-1}(0)$ is closed.

$A^\perp=\cap_{a\in A}f_a^{-1}(0)$, so it is the intersection of closed subsets, hence it is closed.