How can I show $U^{\bot \bot}\subseteq \overline{U}$?

Question

Let $H$ be a Hilbert space and $U$ a subspace. Let $U^{\bot}$ denote its orthogonal complement.

I had no trouble showing $\overline{U}\subseteq U^{\bot\bot}$. But now I'm stuck for $\supseteq$.

Please could someone help me finish this argument?

This is what I am trying to do:

Let $x \in U^{\bot \bot}$. The goal is to construct a sequence $u_n\in U$ such that $u_n \to x$.

Since $\overline{U} \subset U^{\bot \bot}$ either $x\in \overline{U}$ or $x \notin \overline{U}$. If $x \in \overline{U}$ then we're done. If $x \notin \overline{U}$ I don't know what I can do.

Answer 1

Hint: If $x\not\in \overline{U}$, then by Hilbert projection theorem we have $x=y+z$ with $y\in \overline{U}$ and $0\ne z\perp \overline{U}$.

Since $z\ne 0$ one has $$ \langle x,z\rangle=\langle y+z,z\rangle=\| z\|^2\ne 0. $$ Hence $z\in U^\perp$ does not annihilate $x$ and therefore $x\not\in U^{\perp\perp}$.

Answer 2

For closed subspaces: $$Z=\overline{Z}\implies\mathcal{H}=Z\oplus_\perp Z^\perp$$

For orthogonal decompositions: $$\mathcal{H}=Z\oplus_\perp Z'\implies Z'=Z^\perp$$

Taking both together one gets: $$\overline{U}=\left(\overline{U}^\perp\right)^\perp=U^{\perp\perp}$$ (This works even for the closure on the span of plain sets.)