# The sum of an uncountable number of positive numbers

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?