Show that if $\sum_{i \in I} a_i < \infty$, then $I_0=\{i \in I : a_i > 0 \}$ is countable. [duplicate]
Solution 1:
Proof that $I_0=\{i\in I: a_i>0\}$ is countable.
Assume that $\sum_{i\in I}a_i=s<\infty$.
Then, for every $k\in\mathbb N$, the set $$ J_k=\{i\in I: a_n\ge s/k\}, $$ contains at most $k$ elements, since $$ s=\sum_{i\in I}a_i\ge \sum_{i\in J_k}a_i\ge\frac{s}{k}|J_k|. $$ Here $|J_k|$ is the number of elements of the set $J_k$.
But $I_0=\bigcup_{k\in\mathbb N}J_k$.
And hence $I_0$ is countable.