Completeness: $\mathcal{l}^2(S)$

Question

Surely, for countable index sets this is just the diagonal trick: $\#S<\infty$
However for arbitrary index sets how do I prove that the limit will actually have only countable non vanishing coordinates?

Answer 1

PhoenueX's comment essentially answers the question, although a better answer to

Why is $x_s=0$ for $x_s\in S\setminus\bigcup_nS_n$?

would be "because I say so". It's our job to come up with a limit, so we manufacture it in any way we can.

Let's start from the beginning. The elements of $\ell^2(S)$ are functions $x:S\to\mathbb R$ (or $\mathbb C$) such that $\sum_S |x(S)|^2<\infty$. This implies, that the support of every $x\in \ell^2(S)$, defined as $\operatorname{supp} x=\{s\in S:x(s)\ne 0\}$, is at most countable.

Suppose $(x_n)$ is a Cauchy sequence. Let $T$ be the union of their supports. Note that $T$ is at most countable. Restricting each $x_n$ to $T$, we get a Cauchy sequence $(\tilde x_n)$ in $\ell^2(T)$. By the countable case, this sequence has a limit $x_T\in \ell^2(T)$. Define $x \in \ell^2(S)$ by $$x (s)=\begin{cases}x_T(s),\quad &s\in T \\ 0,\quad & s\notin T\end{cases}$$ In other words, just extend the function $x_T$ by zero.

Since $\|x_n-x\|_2 = \|\tilde x_n-x_T\|_2$, we have $x_n\to x$ in $\ell_2$, completing the proof.