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?
Best 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.