Proving that $M^{\perp} = M^{\perp\perp\perp}$

Question

Let $M$ be a non-empty subset of a Hilbert space $H$.

First, prove that $M \subset M^{\perp\perp}$. I know it must be trivial but I still cannot wrap my head around it. Why can't we just like that claim $M^{\perp\perp} \subset M$? Also, is it true for non-complete inner product spaces?

Second, prove that $M^\perp = M^{\perp\perp\perp}$ . This one I have no idea how to approach.

Answer 1

Observation 1: $A \subset A^{\perp \perp}$. Observation 2: $A\subset B \rightarrow B^{\perp}\subset A^{\perp}$ (these follow straight from the definition of $\perp$). Now apply observation 1 to $A=M^{\perp}$ to get $M^{\perp} \subset M^{\perp \perp \perp}$. Now, again by observation 1, we have $M\subset M^{\perp \perp}$. Applying observation 2 we then get that $M^{\perp \perp \perp}\subset M^{\perp}$.

Answer 2

To get you started, here's a sketch of a proof that $M \subset M^{\perp \perp}$.

By definition, $M^\perp = \{y \in H: \text{for all }x \in M, \langle x, y \rangle = 0\}$.

Thus, we have $$ M^{\perp \perp} = \{z \in H: \text{for all }y \in M^\perp, \langle z,y \rangle = 0\} $$ Now, verify that every element of $M$ satisfies this definition. That is: if $x \in M$, then it is necessarily true that $\langle x,y \rangle$ for every $y \in M^\perp$.

For an example where $M^{\perp \perp} \neq M$, take $M$ to be a set containing only a unit vector in a 2-dimensional Hilbert space.