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Let $I$ be a set $(a_i)_{i\in I}$ be a sequence with positive entries and let $\sup\left\{\sum_{i\in F} a_i \mid F\subseteq I, F \ \text{finite}\right\}<\infty.$ I want to show that in that case the set $J$ of indexes that index numbers that aren't $0$ is at most countably infinite.

Consider this Wikipedia proof: It consists of writing $J$ as $J=\bigcup_{n} A_n$, where $A_n:=\{ i\in J \mid \frac{1}{n} and then showing an inequality for the cardinality of $A_n$.

The problem I have is with the first inequality: $\frac{1}{n} |A_n| \leq \sup\{\sum_{i\in F} a_i \mid F\subseteq I, F \ \text{finite}\}$. At first I thought "it's obvious", since all elements in $A_n$ are greater than $\frac{1}{n}$ so one can bound every $a_j\in A_n$ from below with $\frac{1}{n}$ - but the problem is more subtle I think: Namely at this stage of the proof, we can't exclude, that $A_n$ is not countably infinite; and in that case the sum I can't see any way to prove that inequality!

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    But you start by assuming the RHS is finite. So, $A_n$ cannot be countably infinite.2012-12-22

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You can assume that $|A_n|$ is finite, since otherwise $ \sup\{\sum_{i\in F} a_i \mid F\subseteq I, F \ \text{finite}\} $ would not be finite, which is contrary to your assumptions.

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    @resu Let there be infinitely many elements in $A_n$. Then you can find a sequence $F_k, k\in \Bbb N$ and $|F_k| = k$ of finite index sets with so that the sum over $F_k$ is at least $k/n$, which grows without bounds.${}$Here $F_k$ is chosen so that $i\in F_k$ implies $a_i \in A_n$2012-12-22