Convegence of $\sum_{i\in J}a_i$ implies that index set is countable

Question

Let $J$ be a uncountable set and $\{a_i\}_{i\in J}$ be a set of non-negative real numbers. Prove that $\sum_{i\in J}a_i<\infty$ implies that there is a countable set $H\subset J$ such that $a_i=0$ when $i\in J\backslash H$.

Answer 1

Suppose that $\Omega=\{K\subseteq J: K$ is a finite subsets of $J\}$. Then $$\alpha=\sum_{i\in J}a_i=\sup_{k\in\Omega}\sum_{i\in K}a_i.$$ For any $n\in N$ there is a finite set $k_n\in\Omega$ such that $0\leq\alpha-\sum_{i\in K_n}a_i<\frac{1}{n}$. Set $H=\cup_{n\in N}K_n$, then $H$ is countable. We have $$0\leq\alpha-\sum_{i\in H}a_i\leq\alpha-\sum_{i\in K_n}a_i<\frac{1}{n}$$ which is true for any $n\in N$. Hence $0\leq\alpha-\sum_{i\in H}a_i\leq\lim_{n\to\infty}\frac{1}{n}=0$, and consequently $\alpha=\sum_{i\in H}a_i$. So for all $i\in J\backslash H$ we have $0\leq a_i\leq\sum_{i\in J\backslash H}a_i=0$, i.e., $a_i=0$.

Answer 2

Let $s=\sum_{i\in J}a_i$. Let $H_n=\{\,i\in J\mid a_i>\frac1n\,\}$. Then $|H_n|\le ns<\infty$. Let $H=\bigcup_{n\in\mathbb N} H_n$.