TITLE: Sum of closed subspaces of a Hilbert space is closed
QUESTION [5 upvotes]: Let $M, N ⊂ H$ ($H$ Hilbert), be two closed linear subspaces.
Assume that $\langle u, v\rangle = 0$ $∀u ∈ M$, $∀v ∈ N$. Prove that $M + N$ is closed.
Take a sequence $(g_n)\in M+N$ such that $g_n\to x\in H$.
Then for any $n\geq 1$, $\exists v_n\in N, u_n\in M$ such that $g_n=v_n+u_n$. This implies that the sequences $(v_n)$ and $(u_n)$ converge singularly to the elements $v,u\in H$, and by closeness $v\in N$ and $u\in M$.
By uniqueness of the limit it must hold $x=u+v$ which implies $x\in N+M$.
I did not use any property of orthogonality so I guess this reasoning is wrong. Why? And how should I use orthogonality?

REPLY [2 votes]: Let's do a proof without using limits.
The following fact can be found in any book of Functional Analysis, its proof is easy:

Fact 1: Let $G$ be a subspace of a Hilbert space $H$, then $G^{\bot \bot}=\overline{G}$.

Therefore, we have to show that $(M+N)^{\bot\bot}\subseteq M+N$. In fact, let $x\in (M+N)^{\bot\bot}$, then 
$$\left\langle x,y \right\rangle =0 \qquad \forall y\in (M+N)^{\bot}. \tag{I}$$
Another fact that is easy to show is the following:

Fact 2: $(M+N)^{\bot}=M^{\bot}\cap N^{\bot}.$

Since $M$ is closed then $H=M\oplus M^{\bot}$, then there exists $m\in M$ and $m^{\bot}\in M^{\bot}$ such that $x=m+m^{\bot}$.
But we know that $M\bot N$ then $N\subset M^{\bot}$, then we have to show that $m^{\bot} \in N$. 
In fact, note that if $m^{\bot}=0$ then we finished, there would be nothing to show. So,  we suppose that $m^{\bot} \notin N$ with $m^{\bot}\neq 0$, then since $N$ is closed we have $H=N\oplus N^{\bot}$, so $m^{\bot}\in N^{\bot}$. Therefore, $m^{\bot}\in M^{\bot}\cap N^{\bot}$, then by Fact 2 we have that $m^{\bot}\in (M+N)^{\bot}$. Therefore, by Fact 1 we have 
$$0=\left\langle x,m^{\bot} \right\rangle =\left\langle m+m^{\bot},m^{\bot} \right\rangle=\left\langle m^{\bot},m^{\bot} \right\rangle=\left\|m^{ \bot}\right\|^{2}.$$
Then $m^{\bot}=0$, what contradicts $m^{\bot}\neq 0$.