Smallest closed subspace which contains $M$ How can one prove following:
If $M\neq \emptyset$ is any subset of Hilbert space $H$, show that $M^ {\perp \perp }$ is the smallest closed subspace of $H$ which contains $M$.
I know that $M \subset M^ {\perp \perp }$ and that $M^ {\perp }$ and $M^ {\perp \perp }$ are closed spaces. Also,  $\overline{span(M)}$ contains $M$. How can one proceed? Thanks a lot in advance.
 A: There are two canonical ways of doing this. You may assume without loss of generality that $M$ is a subspace.
The first one is to use the usual convexity result:

Lemma. Let $M\subset H$ be closed and convex, and $x\in H\setminus M$. Then there exists $y\in M$, unique, such that $\|x-y\|=\operatorname{dist}(x,M)$.

When $M$ is a subspace, one also gets that $x-y\in M^\perp$. Indeed, for any $z\in M$ we have $\|x-y+z\|≥\|x-y\|$. Squaring, expanding, and canceling, we get
$$
\|z\|^2+2\operatorname{Re}\langle x-y,z\rangle≥0.
$$
As $M$ is a subspace, this can only happen if $\langle x-y,z\rangle=0$ for all $z\in M$. That is, $x-y\in M^\perp$. Knowing this, let $x\in M^{\perp\perp}$. By the Lemma there exists $y\in M$ that realizes the distance and $x-y\in  M^ \perp$. Then $y=x-(x-y)\in M^\perp$. So $y\in M\cap M^\perp=\{0\}$. Then $y=0$ and so $x\in M$, as the distance from $x$ to $M$ is zero. Thus $M^{\perp\perp}\subset M$.

A second way is to know that $H=M+M^\perp$ (but proving this equality usually requires the Lemma above). Then if $x\in M^{\perp\perp}$ we have $x=x_M+x_\perp$. So $$\langle x_\perp,x_\perp\rangle =\langle x_\perp,x\rangle=0,$$ which shows that $x_\perp=0$ and so $x=x_M\in M$.
