Let $A$ be a countably infinite set, and let $f:A\to\mathbb{R}$ be a function that takes elements of $A$ to the reals. Suppose that $\sum_{w\in A} f(w)$ is well-defined (see note below). Also suppose $B_1,B_2,\dots$ are disjoint sets such that $\bigcup_{i=1}^\infty B_i = A$.
We can show that $\sum_{w \in B_i} f(w)$ is well-defined too, since $B_i \subseteq A$.
Can we prove that $$\sum_{i=1}^\infty \sum_{w \in B_i} f(w)$$ converges absolutely? Also, is it always true that $$\sum_{i=1}^\infty \sum_{w \in B_i} f(w) = \sum_{w \in A}f(w)?$$ This clearly works if $A$ is finite, but I'm not sure where to start if $A$ is countably infinite and there can be countably infinite $B_i$'s too (each of countably infinite size).
Note: I am taking the definition of $\sum_{w\in S} f(w)$ to be $\sum_{i=1}^\infty f(w_i)$, where $\{w_i\}$ is an arbitrary sequence composed of elements of $S$, which makes sense if $\sum_{i=1}^\infty f(w_i)$ is absolutely convergent since any rearrangement gives the same value.
EDIT: The proof for this question can be found here. It also questions if the converse is true too.
HINT:
Yes, it's true. Best to use the notion of "summability". Call the family $(f(a))_{a\in A}$ summable if there exists $s$ so that for every $\epsilon > 0$ there exists a finite subset $A_{\epsilon}$ of $A$ so that for every $B$ finite and $B \supset A_{\epsilon}$ we have $|s - \sum_{a \in B} f(a)| < \epsilon$. It's equivalent to your definition but easier to work with. Also, it's equivalent to $\sum_{a\in A} |f(a)| < \infty$. See if that makes it easier for you to prove your result.