Orthogonal complement of a Hilbert Space I have this problem:
Let $S$ be a subset of a Hilbert $H$ and let $M$ be the closed subspace generated by $S$. Show that


*

*$M^{\perp} = S^{\perp}$

*$M = (S^{\perp})^{\perp}$

*if $V$ is a subspace of $H$, then $H = \overline{V}\oplus V^{\perp}$.


I have some doubts, because $H$ don't have finite dimension. For example, for 1. its clear that $S \subseteq M$ and then $M^{\perp} \subseteq S^{\perp}$. Later, if $x\in S^{\perp}$ then $\langle x, a\rangle = 0$, for all $a\in S$. Now in finite dimension I know how justify that $\langle x, b\rangle = 0$, for all $b\in M$, but in a Hilbert I really don't know. Thanks in advance for your help

It should be well known that for a  closed subspace $M$ of a Hilbert space $H$, $H = M \oplus  M^{\perp}$.
From $ S \subset M$ follows  $M^\perp \subseteq S^\perp $ so 
$H = M^\perp \oplus M \subseteq S^\perp \oplus M $.
Then $S^\perp $ cannot be a proper subset of $M^\perp $  so $S^\perp = M^\perp $. 
$M^\perp $ is closed (intersection of kernels of continuous $x \mapsto (m, x),m \in M $), so  $ H = (M^\perp)^\perp \oplus M^\perp = M \oplus M^\perp $. $M$ cannot be proper subspace of $(M^\perp)^\perp$   so $M=(M^\perp)^\perp$. Therefore $M= (M^\perp)^\perp = (S^\perp)^\perp$.
If $V$ is a closed subspace of $H$ then from $V \subset \overline V, V^\perp \supseteq {\overline V}^\perp $, so $ H = \overline V \oplus (\overline V)^\perp \subseteq \overline V \oplus V^\perp$ wherefrom $H= \overline V \oplus V^\perp$. 
