I got stuck on this problem and got no clue to solve it. Can anyone one here help me? I really appreciate.
Let $X$ be a set and $\mathcal{A}$ the collection of all finite subsets of $X$, directed by inclusion. Let $f: X \rightarrow \Bbb{R}$ be an arbitrary function and for $A \in \mathcal{A}$, let $z_{A} = \sum_{x\in A}f(x)$. Then the set $< z_{A} >$ converges in $\Bbb{R}$ iff $\{x: f(x) \neq 0 \}$ is a countable set $\{x_{n}\}_{n \in \Bbb{N}}$ and $\sum_1^\infty|f(x_{n})| \lt \infty$, in which case $z_{A} \rightarrow \sum_1^\infty f(x_{n})$
Thanks so much
The $\Leftarrow$ direction is pretty straightforward, I think.
For the $\Rightarrow$ direction, try the contrapositive. There are two cases to consider; the case that the function has uncountable support, and the case where the function has countable support, but does not converge absolutely on that support. You have to show that $\langle z_A\rangle$ does not converge in either case.
The second condition is again pretty straightforward: if the series does not converge absolutely, then there is a sequence $x_{n}$ for which the sequence of partial sums $\sum_{n=1}^{k}f(x_{n})$ does not converge, and you're done.
The more interesting case is when the function has uncountable support. Use the pigeonhole principle to conclude that there is some rational $q$ such that the set $$\big\{x\in X \,\big\vert\, \vert f(x)\vert >q\big\}$$ is infinite. It should be easy from there.