Proving uncountable sum as series using nets. I saw this question:
The sum of an uncountable number of positive numbers
Asking about a proof of the following:
Let $A = \{a_i\}_{i\in I}$ be a set of positive numbers. If the uncountable sum converges (as a net) to $a$, then only countably many $a_i$ are nonzero.
And was wondering if this proof any good (I have little experience working with nets):
The closure of $A$ is a closed subset of a metric space and $a$ lies in this closure. Thus there is a sequence in $A$ that converges to $a$.
$$ a = \sum_{i=1}^\infty a_i. $$
If $A$ is not countable, we could find a $a_k \in A$ that is not part of the sequence. If $a_k\neq 0$, then
$$ a_k + \sum_{i=1}^\infty a_i > \sum_{a_i\in A}a_i $$
yielding a contradiction.
I did not immediately see this proof being used anywhere. Sorry if I missed something obvious.
Edit: I missed something obvious: $a$ is not in the closure $A$!
It is, however in the closure of the set of finite sums, so we can find a sequence of finite sums converging to $a$, $(s_i)_{i\in\mathbb{N}}$, where each finite sum features each $a_i$ at most once. Each sum has a finite set of summands, so the union of all summands in the entire sequence is a countable union of finite subsets and is countable. Take the sum of all these summands and we have a countable infinite sum
$$ \sum_{i=1}^\infty a_i. $$
The claim is this converges to $a$. Indeed clearly $\sum_{i=1}^\infty a_i \leq \sum_{a_i\in A}a_i = a$. Conversely, for all $s_j$, there is an $n$ such that $\sum_{i=1}^n a_i \geq s_j$, thus $\sum_{i=1}^\infty a_i \geq a$. The rest of the argument then follows.
 A: You have a nice idea, but it is not properly elaborated. Let us first recall

Let $A = (a_i)_{i \in I}$ be an indexed collection of real numbers. Then $\sum_{i\in I}a_i = a$ means that for each number $\varepsilon > 0$ there exists a finite set $F_\varepsilon \subset I$ such that for each finite set  $F \subset I$ with $F_\varepsilon \subset F$ one has $\lvert a-\sum_{i\in F}a_i \rvert <\varepsilon$.

If all $a_i$ are positive and $I$ is infinite, then clearly we have $0 < \sum_{i\in F}a_i < a$ for all finite $F \subset I$: If there would be a finite $F^*$ with $\sum_{i\in F^*}a_i \ge a$, pick $k \notin F^*$ and let $\varepsilon = a_k > 0$. Then $\sum_{i\in F_\varepsilon \cup F^* \cup \{k\}}a_i \ge \sum_{i\in F^*}a_i + a_k \ge a + a_k = a +\varepsilon$, i.e. $\lvert a - \sum_{i\in F_\varepsilon \cup F^* \cup \{k\}}a_i \rvert \ge \varepsilon$ which is a contradiction.
Now define $B$ as the set of all sums $\sum_{i\in F}a_i$ with a finite $F \subset I$. Then by definition $a  \in \overline B$. Thus there is a sequence of finite $F_n \subset I$ such that $\sum_{i\in F_n}a_i \to a$ as $n \to \infty$. If $A$ is uncountable, there exists $k \in I \setminus \bigcup_{n=1}^\infty F_n$. Convergence implies that $ a - \sum_{i\in F_n}a_i  < a_k$ for some $n$. Therefore
$$a < \sum_{i\in F_n}a_i +a_k = \sum_{i\in F_n \cup \{k \}}a_i < a$$
which is a contradiction. Therefore $A$ cannot be uncountable.
A: It's not true that $a \in \overline{\{a_i\mid i \in I\}}$. So it breaks down right away.
