Perp of a subset and closeness.

Question

Let $V$ be an inner product space. We know that for a subset $S$ of $V$, $S^{\perp}$ is a subspace of $V$. My question is that whether $S^{\perp}$ closed or not in general. I have seen the proof in finite dimensional case but I was unable to understand that does the same goes through in infinite dimention also.

Answer 1

$S^\perp$ is always closed in the norm topology. It is the intersection of the sets $K_a=\{x\in V:(x,a)=0\}$ over all $a\in S$, which is closed as $x\mapsto (x,a)$ is continuous.