Showing that $M + N$ is a closed subspace of the Hilbert $H$ Exercise :

Let $M, N$ be closed subspaces of the Hilbert space $H$ while it is also assumed that $M \bot N$. Show that the set :
  $$M + N = \{x+y : x \in M, y \in N\}$$
  is also a closed subspace of $H$.

Attempt - Thoughts :
First of all, since $M$ and $N$ are closed, this means that if $(x_n)$ is a sequence of elements belonging in $X$ and $(y_n)$ is a sequence of elements belonging in $N$, then :
$$\lim x_n = x, \quad x \in M$$
$$\lim y_n = y, \quad y \in N$$
Now, since it is $M \bot N$, then it should also be $\langle x, y \rangle = 0$.
Now, defining the Hilbert space norm to be $\|\cdot \| = \sqrt{\langle \cdot, \cdot \rangle}$, it is :
$$\|x+y\|^2 = \|x\|^2 + 2\langle x,y\rangle + \|y||^2$$
But since $x \bot y$, then : $ \|x+y\|^2 = \|x\|^2 + \|y\|^2 \Leftrightarrow \|x+y\| = \sqrt{\|x\|^2 + \|y\|^2}$.
In terms of the definition of a closed set, I would need to prove that for $(x_n) \in M$ and $(y_n) \in N$, then $(x_n + y_n) \in M + N$ and $\lim (x_n + y_n) \in M + N$. 
I can't really see though how I could proceed to any conclusions (or if there's a simpler way etc), so I would really appreciate any hints or elaborations
 A: $M=(M^\perp)^\perp$ and $N=(N^\perp)^\perp$. Then $M+N=(M^\perp\cap N^\perp)^\perp$. As the perpendicular complement of a subspace is always closed, then
$M+N$ is closed.
A: Suppose $\{x_n\} \subset M$,$\{y_n\} \subset N$ and $x_n+y_n\to z$. Then (by orthogonality) $\|(x_n-x_m) -(y_n-y_m)\|^{2}=\|x_n-x_m\|^{2}+\|y_n-y_m\|^{2} \to 0$ so both $\{x_n\}$ and $\{y_n\}$ are Cauchy. If these sequences converge to $x$ and $y$ respectively the $x \in M, y\in N$ and $z=x+y \in M+N$.
A: You can as well run through the definitions. 
Assume $\{x_n+y_n\}_{n=1}^{\infty}\subseteq M+N$ a Cauchy sequence then using $M\perp N$ gives
$$\langle x_n+y_n-(x_m+y_m),x_n+y_n-(x_m+y_m)\rangle=\langle x_n-x_m,x_n-x_m\rangle +\langle y_n-y_m,y_n-y_m\rangle\rightarrow 0$$
as $n,m\rightarrow\infty$ from which You can conclude that $\{x_n\}$ and $\{y_n\}$ are Cauchy sequences and thus have limits $x\in M$ and $y\in N$ respectively since $M$ and $N$ are closed.
Since for the considered sequences one has $\langle x_n-x,x_n-x\rangle\rightarrow 0,n\rightarrow \infty$ and $\langle y_n-y,y_n-y\rangle\rightarrow 0,n\rightarrow\infty$, You can conclude
$$\langle x_n+y_n-(x+y),x_n+y_n-(x+y)\rangle=\langle x_n-x,x_n-x\rangle +\langle y_n-y,y_n-y\rangle\rightarrow 0$$
as $n\rightarrow\infty$ and thus $M+N$ is closed.
A: Worked around the following answer :
Let $(z_n)$ be a convergent sequence in $M + N$. Then we can write $z_n = \underbrace{x_n}_{\in M} +\underbrace{y_n}_{\in N}$.
But $M \perp N $ we have $\| z_n \|^2 = \|x_n \|^2 +\|y_n\|^2$ by the Pythagoreian Theorem, since $\langle x_n,y_n\rangle = 0$ . 
Now, the sequence  $(z_n)$ is a Cauchy sequence, which means that :
$$ \|x_n -x_m\|^2 +\|y_n -y_m\|^2 = \| z_n -z_m\|^2 \rightarrow 0$$
since  $x_n \to x$ and $ y_n\to y$ for some $x\in M $ , $y \in N $  since they are both closed spaces . It then follows that  $$\lim z_n = z = x+y \in N + M $$
thus the space $N+M$ is a closed subspace of $H$.
