I am stuck with a problem from Conway's book about functional analysis. I have proven the statement for the special case of $I = \mathbb{N}$, but I think my proof isn't transferable to the general case (I'll attach my solution to the special case as an image: solution special case).
So the problem is as follows: Let I be any set and let $l^{2}(I)$ denote the set of all functions $x:I \to\mathbb F$, where $\mathbb{F}$ is either $\mathbb{R}$ or $\mathbb{C}$, such that $x(i)=0$ for all but a countable number of i and $\sum_{i\in I} |x(i)|^{2} < \infty$. For x and y in $l^{2}(I)$ define $\langle x, y \rangle = \sum_{i\in I}x(i)\overline{y(i)}$, then this defines an inner product. Show that $l^2(I)$ is a Hilbert space.
I'd be happy if someone could provide some hints as to how I can modify my proof for the special case such that it is also valid for the general one.
Update:
We will first show that it suffices to prove the special case $I = \mathbb{N}$. If $I$ is countable, then we can enumerate $I$ by $\{i_1, i_2,...\}$ and hence treat $I$ as if it was $\mathbb{N}$. So suppose $I$ is uncountable. Let $\{x_n\}$ be Cauchy in $l^2(I)$. Define $J_n \equiv \{i \in I: x_n(i) \neq 0\}$ for all n in $\mathbb{N}$. Then $J \equiv \{i \in I: \exists n \in \mathbb{N}: x_n(i) \neq 0\} = \bigcup_{n \in \mathbb{N}} J_n$ is countable since it is the countable union of countable sets. For all natural numbers n define $\widetilde{x_n} \equiv \left.x_n\right|_J$, then it immediately follows that $\{\widetilde{x_n}\}$ is Cauchy in $l^2(J)$ and because $J$ is countable we will see later in this proof that $\widetilde{x_n} \to \widetilde{x} \in l^2(J)$. We now define $x\colon I\to \mathbb{F}$, with $\left.x\right|_J \equiv \widetilde{x}$ and $x(i) = 0$ for all $i \in I \setminus J$. Then since $(\sum_{i \in I} |x_n(i)-x(i)|^2)^{1/2} = (\sum_{j \in J} |x_n(j)-x(j)|^2)^{1/2}$ we see that $x_n \to x$ and since $\widetilde{x} \in l^2(J)$ we also see that $x \in l^2(I)$. Hence we have traced back the general case to the special case of $I = \mathbb{N}$.
If $I$ is countable then you can enumerate it as $I = \{i_1, i_2, \ldots\}$, after which you can turn a proof for $\mathbb{N}$ into a proof for $I$ by replacing the appropriate $k$'s by $i_k$'s.
Suppose that $I$ is uncountable. Given your Cauchy sequence $(x_n)$, the set $J = \{i \in I : \exists n, x_n(i) \neq 0\}$ is countable. Restricting each term of your sequence to be a function on $J$ produces a Cauchy sequence in $\ell^2(J)$, so it has a limit, say $\tilde{x} : J \to \mathbb{F}$, by the countable case. Now you just need to extend $\tilde{x}$ to a function $x : I \to \mathbb{F}$ by making the appropriate definition on $I \setminus J$.
(Disclaimer: I didn't look carefully at your proof in the case $I = \mathbb{N}$, so I can't certify whether or not it is correct. I'm just trying to give some idea of how to extend from $\mathbb{N}$ to arbitrary $I$).