Show that if $\sum_{i \in I} a_i < \infty$, then $I_0=\{i \in I : a_i > 0 \}$ is countable.

Question

Let $I$ be an indexing set and $a_i \ge 0,$ $i\in I$, real numbers. Show that if $\sum_{i \in I} a_i < \infty$, then $I_0=\{i \in I \mid a_i > 0 \}$ is countable.

How should I approach the problem? I don’t think I can find an injection between $\Bbb N$ and $I_0$. Is there some property of the convergent sum I should consider? I only now that if the sum converges, then the tail tends to zero, but not sure how it is of help here.

I think you mean $I_0={i \in I \mid a_i > 0 }.$ Hint: a union of countably many finite sets is countable. — Calum Gilhooley, Dec 23 '21 at 22:34
What does $\sum_{i\in I} a_i$ mean when $I$ is not countable? — Snaw, Dec 23 '21 at 23:26
@Snaw . When every $a_i\ge 0$ we can define $\sum_{i\in I}a_i=$ $=\sup{\sum_{i\in J}a_i:J\subset I\land |J|<\aleph_0}.$ — DanielWainfleet, Dec 24 '21 at 02:39

score 7 · Accepted Answer · answered Dec 23 '21 at 22:44

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.

Show that if $\sum_{i \in I} a_i < \infty$, then $I_0=\{i \in I : a_i > 0 \}$ is countable.

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