Infinite direct sum and closed linear subspace 
Suppose that $\left\{ \mathcal{H}_n \, \big| \, n\in\mathbb{N} \right\}$ is a set of orthogonal closed subspaces of a Hilbert space $\mathcal{H}$. We define the infinite direct sum
  $$\bigoplus_{n=1}^{\infty} \mathcal{H}_n := \left\{ \sum_{n=1}^{\infty} x_n \, \big| \, x_n \in \mathcal{H}_n \, \text{and} \, \sum_{n=1}^{\infty} \|\ x_n \|^2 < \infty \right\} . $$ 
  Prove that $\oplus_{n=1}^{\infty} \mathcal{H}_n$ is a closed linear subspace of $\mathcal{H}$. 
Since for any S ⊂ H, $S^⊥$ is a closed linear subspace of H, I want to prove this theorem by showing that $$\bigoplus_{n=1}^{\infty} \mathcal{H}_n = S^⊥ $$
  Does anyone have an idea about how to prove this? Thanks a lot.

 A: Let $M=\displaystyle\bigoplus_{n=1}^{\infty}\mathcal{H}_{n}$, assume that $x_{m}\in M$ and $x_{m}\rightarrow x$, we are to show that $x\in M$.
Now each $x_{m}\in M$ can be written as $x_{m}=\displaystyle\sum_{n}y_{mn}$, $y_{nm}\in\mathcal{H}_{n}$, then  $\|x_{m}-x_{m'}\|^{2}=\left\|\displaystyle\sum_{n}(y_{mn}-y_{m'n})\right\|^{2}=\displaystyle\sum_{n}\|y_{mn}-y_{m'n}\|^{2}\rightarrow 0$ as $m,m'\rightarrow\infty$. But $\|y_{mn}-y_{m'n}\|^{2}\leq\displaystyle\sum_{n}\|y_{mn}-y_{m'n}\|^{2}$, so $\|y_{mn}-y_{m'n}\|\rightarrow 0$ as $m,m'\rightarrow\infty$. Since $\mathcal{H}_{n}$ is closed, $y_{mn}\rightarrow y_{n}$ for some $y_{n}\in\mathcal{H}_{n}$.
Now we have
\begin{align*}
\left\|x_{m}-\sum_{n}y_{n}\right\|^{2}&=\left\|\sum_{n}(y_{mn}-y_{n})\right\|^{2}\\
&=\sum_{n}\|y_{mn}-y_{n}\|^{2}\\
&=\sum_{n}\lim_{m'\rightarrow\infty}\|y_{mn}-y_{m'n}\|^{2}\\
&\leq\liminf_{m'\rightarrow\infty}\sum_{n}\|y_{mn}-y_{m'n}\|^{2},
\end{align*}
which can be controlled by arbitrarily small, so $x=\displaystyle\sum_{n}y_{n}$. We also have 
\begin{align*}
\displaystyle\sum_{n}\|y_{n}\|^{2}&=\sum_{n}\lim_{m\rightarrow\infty}\|y_{mn}\|^{2}\\
&\leq\liminf_{m\rightarrow\infty}\sum_{n}\|y_{mn}\|^{2}\\
&=\liminf_{m\rightarrow\infty}\|x_{m}\|^{2}\\
&=\|x\|^{2}\\
&<\infty.
\end{align*}
