$H$ Hilbert space, $S\subset H$ implies $S^\perp$ closed

Question

Let $H$ be a Hilbert space, $S\subset H$. Then $S^\perp$ closed.

I'm not sure how to approach this problem.

Let $M=\mbox{span}(S)$, $P=\mbox{span}(S^\perp)$ then $S\subset M$, $S\perp P$. Let $x\in H$, then $x=m+p$, with $m\in M$, $p\in P$. Then $H=M\oplus P$. Then there exists a sequence $0=\langle m_n,p_n \rangle\to \langle m,p \rangle=0$. Thus $M$ and $P$ are closed.

How can we conclude that $S^\perp \subset P$ is closed?

Answer 1

Let $s_{n}$ be a sequence in $S^{\perp}$, converging to $x \in H$. we need to show that $x \in S^{\perp}$, ie $<x,s>=0$ for all $s \in S$. $<x,s>=<\lim s_{n},s>=\lim <s_{n},s>=0$, by continuity of inner product!

Answer 2

You should observe that $S^\perp$ is a nonvoid intersection of closed sets.