$M$ and $N$ are subspaces of a Hilbert space. If $M\subset N$, show that $N^{\perp}\subset M^{\perp}$. Show also that $(M^{\perp})^{\perp}=M$.
I know that the orthogonal complement of $X$ is the set $X^{\perp}=\{x\in H : x{\perp}X\}$ where $X$ is any subset of a Hilbert space $H$. I'm not sure how to proceed. Any hints or solutions are greatly appreciated.
Best Answer
If $x$ $\in$ $N^{\perp}$ then $x$ $\perp$ $N$ and since $M \subset N$ then $x \perp M$ therefore $x \in M^{\perp}$
Now let $x \in M$, then $x \perp M^{\perp}$ and therefore $x \in M^{\perp \perp}$. That means $M \subset M^{\perp \perp}$. Since $M^{\perp \perp}$ is a closed and linear subspace containing M that means $\overline{\lt M \gt} \subset M^{\perp \perp}$
Suppose now that there is a $x \in M^{\perp \perp} \setminus \overline{\lt M \gt}$. There exists a $y \in \overline{\lt M \gt}$ so that $x - y \perp \overline{\lt M \gt}$. Then $x - y \perp M$ and therefore $x - y \in M^{\perp}$. Also $x - y \in M^{\perp \perp}$ since it is a linear combination of elements of $M^{\perp \perp}$. This means that $x - y = 0$ and therefore $x = y$ and $x \in \overline{\lt M \gt}$ which is false.