It is well known that:
A closed subspace $S\subseteq H$ and $H$ is Hilbert space, then $H = S\oplus S^{\perp}$ and $ S^{\perp}$ is also closed.
I'm thinking that since $S^{\perp} = H\setminus S$ and $S$ is closed, so $S^{\perp}$ is also open?
hmm, sure.$S^{\perp}\cap S = {0}\neq \emptyset$, the previous equality is invalid.