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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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    Could you make the question more self-contained by quoting the inequality that you are asking about?2012-12-22
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    Is $J=\{i\in I|a_i\not=0\}$ ?2012-12-22
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    @Amr yes $ \ \ \ \ \ \ $2012-12-22
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    In your last line you say that the sum of the RHS is undefined. What are you refering to ?2012-12-22
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    @TrevorWilson sigh, ok, done.2012-12-22
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    If any of the sets $A_n$ is infinite (countable or not) then the sum on the right hand side, which is defined as the supremum of the finite sums, is already infinite. The unordered sum of a set of nonnegative numbers (such as each $A_n$) is always defined, although it may be infinite.2012-12-22
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    An alternate proof if you want: $J = \bigcup n A_n$ if $J$ was uncountable, by the pigeon hole principle some $A_k$ is uncountable. So $\sum_I a_i \geq \sum_{a_i \in A_k} a_i \geq \sum_{a_i \in A_k} \frac{1}{k} = \infty$.2012-12-22
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    @ Amt, indeed I wasn't careful there, I fixed it know.2012-12-22
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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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    Could you give a more formal proof, why the $\sup$ would not be finite, for infinite $A_n$ ?2012-12-22
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    @resu: if there are $kn$ elements in $A_n$ then the supremum is at least as big as $kn\cdot 1/n = k$. We know the supremum is finite; call it $S$. Then there cannot be more than $Sn$ elements in $A_n$, because of the first sentence.2012-12-22
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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