Claim: If $(x_\alpha)_{\alpha\in A}$ is a collection of real numbers $x_\alpha\in [0,\infty]$

such that $\sum_{\alpha\in A}x_\alpha<\infty$, then $x_\alpha=0$ for all but at most countably many $\alpha\in A$ ($A$ need not be countable).Proof: Let $\sum_{\alpha\in A}x_\alpha=M<\infty$. Consider $S_n=\{\alpha\in A \mid x_\alpha>1/n\}$.

Then $M\geq\sum_{\alpha\in S_n}x_\alpha>\sum_{\alpha\in S_n}1/n=\frac{N}{n}$, where $N\in\mathbb{N}\cup\{\infty\}$ is the number of elements in $S_n$.

Thus $S_n$ has at most $Mn$ elements.

Hence $\{\alpha\in A \mid x_\alpha>0\}=\bigcup_{n\in\mathbb{N}}S_n$ is countable as the countable union of finite sets. $\square$

First, is my proof correct? Second, are there more concise/elegant proofs?

**Answer**

Just so the question gets an answer: **yes**, your proof is correct and is one of several phrasings of the shortest proof that I know.

**Attribution***Source : Link , Question Author : Benji , Answer Author :
Pete L. Clark
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