$v \subseteq H \implies V^\bot$ is a closed subspace of the Hilbert space $H$

Question

Exercise :

Show that if $H$ is a Hilbert space and $V \subseteq H$, then $V^\bot$ is a closed subspace of $H$.

Attempt :

I thought of two possible approaches. One would be a classic one, getting an arbitrary sequence $v_n \in V^\bot$ and showing that $\lim v_n = v \in V^\bot$. The other one would be showing that $V^\bot = (V^\bot)^{\bot\bot}$, which implies that $V^\bot$ is a closed subspace of $H$.

For my first approach, let $v_n \in V^\bot$ and $v_n \to h \in H$ . Then, for all $v \in V$, it is :

$$\langle x,v \rangle = \langle \lim v_n, v\rangle = \lim\langle v_n, v\rangle =0 \implies x \in V^\bot $$

That's really straightforward and simple.

I am wondering how would one complete my thought for my second possible approach, meaningly showing that $V^\bot = (V^\bot)^{\bot\bot}$. I seem a little rusty on that.

Answer 1

Presumably you already know that (1) $X\subseteq (X^\bot)^\bot$ for all sets $X$ and (2) if $X\subseteq Y$ then $X^\bot\supseteq Y^\bot$. (If not, just check that both facts follow immediately from the definition of ${\ }^\bot$.)

Apply (1) with $V$ as your $X$ to get $V\subseteq(V^\bot)^\bot$ and then use (2) to get $V^\bot\supseteq((V^\bot)^\bot)^\bot$.

On the other hand, if you apply (1) with $V^\bot$ as your $X$, then you get $V^\bot\subseteq((V^\bot)^\bot)^\bot$.

So you have inclusions in both directions, and therefore $V^\bot=((V^\bot)^\bot)^\bot$.

Answer 2

Hint: An arbitrary intersection of closed subspaces is a closed subspace.