Is $\{ w \in H : (h,w) = 0 \}$ a compact subset of $H$ for a fixed $h$?

Question

Let $H$ be a infinite-dimensional Hilbert space. We are given an element $h \neq 0 \in H$ (fixed). Is the following set $$A:=\{w \in H : (h,w) = 0\}$$ compact in $H$?

I think it is not enough information to say it is compact in general...

Answer 1

No. If $w_1,w_2\in A$ and $\alpha\in\mathbb{C}$ are given, then $$(h,w_1+\alpha w_2)=(h,w_1)+\overline\alpha(h,w_2)=0.$$ That is, $A$ is a subspace, hence not bounded and therefore not compact.

Answer 2

$A$ is a vector subspace of $H$ which is never compact is its dimension is greater or equal to $1$ which is the case here.

Answer 3

No.

It is covered by open sets $\{x\in H\mid (x,x)<n\}$ for $n=1,2,\dots$.

There is no finite subcover.