Direct sum of closed Orthogonal Subspaces Given a sequence $\{\mathscr{H}_n\}_{n=1}^{\infty}$ of closed, orthogonal subspaces of a Hilbert Space $\mathscr{H}$, we define the infinite direct sum to be:
$$
\bigoplus_{n = 1}^{\infty} \mathscr{H}_n = \left \{\sum_{n = 1}^\infty x_n : x_n \in \mathscr{H}_n, \sum_{n = 1}^\infty\|x_n\|^2 < \infty\right \}
$$
The question asks me to prove that this is a closed subspace of $\mathscr{H}$.
The right hand side condition makes sense to me as for orthogonal $x_n$ we have $\|\sum x_n\|^2 = \sum \|x_n\|^2$. For a sum of two elements in the space, we see that:
\begin{align*}
\sum_{n = 1}^\infty\left |x_n+ y_n \right |^2 &\leq \sum_{n = 1}^\infty (|x_n| + |y_n|)^2 \\  &=  \sum_{n = 1}^\infty (|x_n|^2 + 2|x_ny_n| + |y_n|^2) \\ &\leq  \sum_{n = 1}^\infty (|x_n|^2 + |y_n|^2)+ 2\left(\sum_{n = 1}^\infty|x_n|^2\right)^{1/2} \left(\sum_{n = 1}^{\infty} |y_n|^2\right)^{1/2} \\
&<\infty
\end{align*}
Thus, a sum of two elements is also a member of the set. the set is also clearly closed under scalar multiplication. How would I prove it is closed? Exactly why can we take sequences and show they converge in the set?
 A: Let $(y_n)_{ n \in \mathbb{N}}$ be a Cauchy-sequence in $\bigoplus_{n = 1}^{\infty} \mathscr{H}_n$. Write $y_n = \sum_{m=1}^\infty x_{m,n}$ with $x_{m,n} \in \mathscr{H}_m$ and note that by using the orthogonality we obtain
$$\sum_{m=1}^\infty \|x_{n,m} - x_{n',m}\|^2 = \|y_n -y_{n'}\|^2.$$
Thus $(x_{n,m})_n$ is also a Cauchy-sequence and since $\mathcal{H}_n$ is complete (as a closed set of a complete space), we get that $x_{n,m} \rightarrow x_m \in \mathscr{H}_m$. Next, we show that the sum $\sum_{m=1}^\infty x_m$ is convergent. (Here we need to argue that we can interchange the limes and the infinite summation.) For this, note that $\|y_n - y_1\|$ is bounded, say by $M$ and thus
$$\sum_{m=1}^k \|x_{m} - x_{1,m}\|^2 = \lim_{n\rightarrow \infty} \sum_{m=1}^k \|x_{n,m} - x_{1,m}\|^2 \le \limsup_{n \rightarrow \infty} \|y_n-y_1\|  \le M^2.$$
Hence the last series (on the left hand side) is convergent, because it is bounded. By the $\Delta$-inequality we also conclude that $$\sum_{m=1}^\infty \|x_m\|^2 <\infty.$$ Since $\mathscr{H}$ is complete and $(x_m)_m$ are orthogonal, we get that $y = \sum_{m=1}^\infty x_m$  is convergent in $\mathscr{H}$ and by definition we also have $y \in \bigoplus_{n = 1}^{\infty} \mathscr{H}_n$. We can take $N \in \mathbb{N}$ so large that $\|y_n -y_{n'}\| < \varepsilon$ for all $n,n' \ge N$. Thus
$$\sum_{m=1}^k \|x_{m} - x_{n',m}\|^2 = \lim_{n\rightarrow \infty} \sum_{m=1}^k \|x_{n,m} - x_{n',m}\|^2 \le \limsup_{n \rightarrow \infty} \|y_n-y_{n'}\|^2 \le  \varepsilon^2$$
for all $n' \ge N$. Letting $k \rightarrow \infty$ shows that
$$\|y-y_{n'}\|^2 = \sum_{m=1}^\infty \|x_{m} - x_{n',m}\|^2 \varepsilon^2$$
for all $n' \ge N$. Hence $(y_n)_n$ is convegent in $\bigoplus_{n = 1}^{\infty} \mathscr{H}_n$.
A: As in the answer by @p4sch, let $(y_n)_n$ be a Cauchy sequence in $\bigoplus_{n = 1}^{\infty} \mathscr{H}_n$ with $y_n = \sum_{m=1}^\infty x_{m,n}$ and $x_{m,n} \in \mathscr{H}_m$.
For every $m \in \mathbb{N}$ we have
$$\|x_{m,k} - x_{m,j}\|^2 \le \sum_{m=1}^\infty \|x_{m,k} - x_{m,j}\|^2 = \|y_k - y_j\|^2 \xrightarrow{k,j\to\infty} 0$$
so $(x_{m,k})_k$ is Cauchy in $\mathscr{H}_m$. Since $\mathscr{H}_m$ is complete, there exists $x_{m,0} \in \mathscr{H}_m$ such that $x_{m,k} \xrightarrow{k\to\infty} x_m$.
Let $\varepsilon > 0$ and pick $N \in \mathbb{N}$ such that $$k,j \ge N \implies \sum_{m=1}^\infty \|x_{m,k} - x_{m,j}\|^2 = \|y_k - y_j\|^2 < \frac\varepsilon2$$
In particular, assuming $k,j \ge N$ for any $K \in \mathbb{N}$ we have $$\sum_{m=1}^K\|x_{m,k} - x_{m,j}\|^2 < \frac\varepsilon2$$
Letting $k \to \infty$ implies 
$$\sum_{m=1}^K\|x_{m,0} - x_{m,j}\|^2 \le \frac\varepsilon2$$
and since $K$ was arbitrary, it follows
$$\sum_{m=1}^\infty\|x_{m,0} - x_{m,j}\|^2 \le \frac\varepsilon2\tag{$*$}$$
Now we have
$$\sum_{m=1}^\infty \|x_{m,0}\|^2 \le \sum_{m=1}^\infty (\|x_{m,0}-x_{m,j}\| + \|x_{m,j}\|)^2 \le 2\left(\sum_{m=1}^\infty \|x_{m,0}-x_{m,j}\|^2 + \sum_{m=1}^\infty\|x_{m,j}\|^2\right) < +\infty$$
Hence for $r,s \in \mathbb{N}$ we have $$\left\|\sum_{m=r}^s x_{m,0}\right\|^2 = \sum_{m=r}^s\|x_{m,0}\|^2  \le \sum_{m=r}^\infty \|x_{m,0}\|^2 
 \xrightarrow{r,s \to \infty} 0$$
so by completeness $y_0 := \sum_{m=1}^\infty x_{m,0}$ converges in $\mathscr{H}$ and by $(*)$ we have $y_0 \in \bigoplus_{n = 1}^{\infty} \mathscr{H}_n$.
$(*)$ also means $$j \ge N \implies \|y_0 - y_{m,j}\| \le \frac\varepsilon2 < \varepsilon$$ which means $y_{j} \xrightarrow{j\to\infty} y_0$.
