Convergence of projections onto a nested sequence of subspaces of a Hilbert space My professor said something in class which I am not sure how to prove.
Given a sequence of subspaces $\mathcal{H}_n$ of a Hilbert space $\mathcal{H}$ such that $\mathcal{H}_n\subset\mathcal{H}_{n+1}$. And defining $\mathcal{H}_\infty$ to be the smallest closed subspace that contains $\bigcup_n \mathcal{H}_n$. 
$\forall X \in \mathcal{H}$ $\mathcal{P}_n X \to \mathcal{P}_\infty X$ 
Where $\mathcal{P}_n$ is defined to be the projection in $\mathcal{H}_n$.
He told us that the proof uses the orthogonality of the set $\{\mathcal{P}_1X, (\mathcal{P}_2-\mathcal{P}_1) X,(\mathcal{P}_3-\mathcal{P}_2) X,...\}$ and the fact that $||\sum_{i=1}^\infty(\mathcal{P}_{i+1}-\mathcal{P}_i) X||<\infty$. I have already been able to prove the orthogonality of the set nevertheless I don't know how to show that $||\sum_{i=1}^\infty(\mathcal{P}_{i+1}-\mathcal{P}_i) X||<\infty$.
Furthermore, I don't see how the two conditions above are useful in the proof.
 A: Stating the goal as $\|\sum_{i=1}^\infty(\mathcal{P}_{i+1}-\mathcal{P}_i) x\|<\infty$ is misleading. If the expression in parentheses makes sense (the sum converges to some vector $y\in \mathcal H$), then of course $\|y\|<\infty$. There are no  vectors of infinite norms in a Hilbert space. On the other hand, if the series does not converge, then the norm of what are we talking about?
I guess the hint suggests that one should show  $$\sum_{i=1}^\infty \|(\mathcal{P}_{i+1}-\mathcal{P}_i) x\|^2 < \infty \tag1$$ which together with orthogonality indeed implies the convergence of the series $\sum_{i=1}^\infty(\mathcal{P}_{i+1}-\mathcal{P}_i) x$. One can prove (1) by using Pythagorean theorem: 
$$
x - \mathcal P_1x = (x-\mathcal P_n x) + (\mathcal P_n x - \mathcal P_{n-1}x) + \dots + (\mathcal P_2x-\mathcal P_1x)
$$
where all terms of the sum on the right are orthogonal to each other, hence 
$$
\|x - \mathcal P_1x\|^2 = \|x-\mathcal P_n x\|^2 + \|\mathcal P_n x - \mathcal P_{n-1}x\|^2 + \dots + \|\mathcal P_2x-\mathcal P_1x\|^2
$$
showing that the partial sums of (1) are bounded.

But I don't think the hint is useful. This would be my proof: write $x=y+z$ where $y =  \mathcal P_\infty x$ and $z = x - y$. Then $z\perp \mathcal H_\infty$, hence $z\perp \mathcal H_n$ for all $n$, hence  $\mathcal P_n z=0$. It remains to prove that $\mathcal P_n y\to y$. Since $\bigcup_n \mathcal H_n$ is dense in $\mathcal H_\infty$, for every $\epsilon>0$ there exist $N$ such that $\operatorname{dist}(y, \mathcal H_N)<\epsilon$. Then for all $n\ge N$ 
$$
\|y - \mathcal P_ny \| = \operatorname{dist}(y, \mathcal H_n) 
\le \operatorname{dist}(y, \mathcal H_N) <\epsilon 
$$
as required by the definition of the limit. 
