Projection on uncountable orthonormal set Given an Hilbert space $\mathcal{H}$ and an orthonormal set $\{e_i\}$, using Bessel's inequality I can show that the series:
$$
\sum_{i=0}^n \langle e_i, x\rangle e_i 
$$
and 
$$
\sum_{i=0}^{\infty} \langle e_i, x\rangle e_i 
$$
(summed over a countable set of indices), converge in $\mathcal{H}$. However I find it challenging to extend this result to the sum over an uncountable set of $e_i$'s. Is it possible to prove that given an uncountable set $\{e_i\}_{i\in I}$ the sum
$$
\sum_{i\in I} \langle e_i, x\rangle e_i 
$$
converges in $\mathcal{H}$?
(Following what seems a more or less accepted policy [Best way of asking "check my proof" questions ], I'm posting my own approach as an answer, hoping for feedback on it, as I'm not sure it is correct.)
 A: Yes, that seems correct. There's another way to define that sum that might be considered more elegant: If $X$ is a Banach space and $x_i\in X$ for every $i\in I$ define $$\sum_{i\in I}x_i=x$$to mean that for every $\epsilon>0$ there exists a finite set $E\subset I$ such that $$||x-\sum_{i\in F}x_i||<\epsilon$$for every finite set $F$ with $E\subset F\subset I$.
(The sums over finite subsets of $I$ form a net in an obvious way, and this just defines $\sum_{i\in I}x_i$ as the limit of that net.)
This leaves you with two obvious exercises: Show that this gives the same definition for $\sum_{i\in I}<x,e_i>e_i$, as what you did and also show directly that $\sum_{i\in I}<x,e_i>e_i=x$
A: Given a Hilbert space H, taking a countable set $\{e_i\}$ of orthonormal vectors in H and a vector x in H, the sum:
$$
\sum_{i=1}^{\infty} \big \langle e_i, x \big \rangle e_i = \lim_{n\to\infty}\sum_{i=1}^{n} \big \langle e_i, x \big \rangle e_i
$$
converges to a vector of H. In fact, from Bessel's inequality follows:
$$
\sum_{i=1}^{n} \big |\big \langle e_i, x \big \rangle \big |^2 \le ||x||^2
$$
for every n. The series $\sum_{i=1}^{n} \big |\big \langle e_i, x \big \rangle \big |^2$ converges, and so does $\sum_{i=1}^{n} \big |\big \langle e_i, x \big \rangle \big |$. For absolute convergence then, $
\sum_{i=1}^{n} \big \langle e_i, x \big \rangle e_i$ converges too.
We can define the sum of uncountable positive numbers as:
$$
\sum_{i \in I} p_i = \sup \Big\{ \sum_{i \in J} p_i , J \in \mathcal{P}(I) \wedge J \text{is finite} \Big\}
$$
where $\mathcal{P}(I)$ is the power-set of I. By Bessel's inequality (holding for each of the sums on the finite subset of I) $\sum_{i \in I} \big|\big \langle e_i, x \big \rangle\big|^2 \le ||x||^2$. The series is positive and bounded, thus it will converge. This implies that only a countable number of elements of the sum are different from 0. Thus we can define the uncountable sum $\sum_{i\in I} \big \langle e_i, x \big \rangle e_i$ as the sum over the countable subset of $I$ for which $\langle e_i, x \rangle \ne 0$. Once the sum is restricted over a countable number of vectors, convergence follows from the first part of the proof above.
