Show that if $\sup\big\{\sum\lvert\, f(a)\rvert\big\} < \infty$, then $\{ a \in A : f(a) > 0\}$ is countable. Let $f:A \to \mathbb R$ and suppose that $$
\sup\Big\{\sum_{a\in F}\lvert\, f(a)\rvert : 
F\text{ is finite subset of }A\Big\} < \infty
$$ then the set $\{ a \in A : f(a) > 0\}$ is countable.
My try: Define these element of ${ a ∈ A : f(a) > 0}$, then $∑_{n=1}^\infty\lvert\, f(a_n)\rvert$ is determined, regardless of the method of arrangement.

Assume that the set $P=\{a\in A:f(a)>0\}$ is uncountable. Set $P_n=\{a\in A:f(a)>1/n\}$. Then
clearly
$$
P=\bigcup_{n\in\mathbb N}P_n.
$$
If all $P_n$'s were countable, then so would be $P$, as a countable union of countable sets. Hence, for some $n_0$, then set $P_{n_0}$ is uncountable.
Clearly
$$
\sum_{a\in P_{n_0}}f(a)\ge \sum_{a\in P_{n_0}}\frac{1}{n_0}=\infty.
$$
