If I want to prove that $M^{\perp}$is a closed

Question

If I want to prove that $M^{\perp}$is a closed

Can I say because it is the inverse image of $0$ by continuos function ( projection operator )

Answer 1

What you are saying is absolutely correct. There is another way, more simple and basic, to see this. Let $x$ be a limit point of $M^{\perp}$, i.e,. there exists a sequence $(x_n)$ in $M^{\perp}$ such that $x_n \longrightarrow x$. Since inner product is continuous, it follows that for any fixed $y$; $\langle x_n,y\rangle \longrightarrow \langle x,y\rangle$. But $\langle x_n, y\rangle = 0$ for all $y\in M$, therefore, $\langle x,y\rangle = 0$ for all $y\in M$, i.e., $x\in M^{\perp}$. Hence $M^{\perp}$ is closed.