I have a question regard unordered sumation.
Definitions: Let $X$ be a set (possible uncountable) and let $f: X\rightarrow \mathbb{R}$ be a function. We say that $\sum _{x\in X}f(x)$ converges absolutely iff ${\; \text{ sup}}\big\{\sum _{x\in F}|f(x)|:F\subset X; \#F<\infty \big\}<\infty$. We define the sumation over $X$ as follows:
$$\sum _{x\in X}f(x):= \sum _{\{x\in X:f(x)\not=0\}}f(x)$$
It is not difficult to show that this is well-defined since ${\{x\in X:f(x)\not=0\}}$ is at most countable
Let $X$ be a countable set and let $f: X\rightarrow \mathbb{R}$ be a function. If $\sum _{x\in X}f(x)$ converges absolutely, we define
$$\sum _{x\in X}f(x):= \sum _{i=0}^{\infty}f(\alpha(i))$$
For some bijective map $\alpha: \mathbb{N} \rightarrow A$.
Lemma: Let $X$ be a set (possible uncountable) and let $f: X\rightarrow \mathbb{R}$ and $g: X\rightarrow \mathbb{R}$ be functions such that $\sum _{x\in X}f(x)$ and $\sum _{x\in X}g(x)$ converges absolutely (2). Then $\sum _{x\in X}f(x)+g(x)$ converges absolutely. Moreover $\sum _{x\in X}f(x)+g(x)=\sum _{x\in X}f(x)+\sum _{x\in X}g(x)$ (2).
Proof:
(1) The absolutely convergent is rather easy and really only matters me the second part.
(2) Let $A=\{x\in X: f(x)\not=0 \text{ or } g(x)\not=0\}$, then we claim that the series $\sum_{x\in A}f(x)$ is convergent. This follows since $A\subset X$ and we know that $\sum_{x\in X}f(x)$ is absolute convergent then former must converges absolutely and so converges conditionally.
Claim 1: $\sum _{x\in A} f(x)= \sum _{x\in X} f(x)$. Since by definition we have that:
$$\sum _{x\in A} f(x)=\sum _{x\in A:f(x)\not=0} f(x) \text{ and } \sum _{x\in X} f(x)=\sum _{x\in X:f(x)\not=0} f(x)$$
It will suffice to show that $\{x\in A: f(x)\not=0\}=\{x\in X:f(x)\not=0\} $. But this follows immediately by construction of $A$.
A similar result holds for $g(x)$.
Claim 2: $\sum _{x\in A} f(x) + g(x)= \sum_{x\in X}f(x) + g(x)$
It will suffice to show that $\{x\in A: f(x) + g(x)\not= 0 \} = \{x\in X: f(x) + g(x)\not= 0\}$. ($\Rightarrow$) is trivial. For ($\Leftarrow$) if $f(x) + g(x)\not= 0$ so either one of them is non zero so is in $A$ and also in the set.
Suppose WLOG that the set $A$ is countable infinite. Then there exists a bijective map $\alpha: \mathbb{N} \rightarrow A$, so
$$\sum _{x\in A} f(x) + g(x)= \sum_{i=0}^\infty f(\alpha(i))+g(\alpha(i))$$
Since by claim 1 we have
$$\sum_{x\in X} f(x)= \sum_{x\in A} f(x)= \sum_{i=0}^\infty f(\alpha(i))\, \text{ and }\, \sum_{x\in X} g(x)= \sum_{x\in A} g(x)= \sum_{i=0}^\infty g(\alpha(i))$$
So we have
$$\sum_{i=0}^\infty f(\alpha(i))+g(\alpha(i))= \sum_{i=0}^\infty f(\alpha(i))+\sum_{i=0}^\infty g(\alpha(i))$$
and we're done. I think this works =)