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.

  • $\begingroup$ It's not clear and even false (!) that $a$ is in the closure of $A$. So your proof attempt breaks down right at the start. $\endgroup$ – Henno Brandsma Feb 21 at 16:28

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.


It's not true that $a \in \overline{\{a_i\mid i \in I\}}$. So it breaks down right away.

  • $\begingroup$ Oh sorry! That should be that $a$ is in the closure of $$ \left\{ \sum_{j=1}^n a_{i_j} \;|\; n\in\mathbb{N}\ \right\} $$ $\endgroup$ – Mentastin Feb 21 at 16:39
  • $\begingroup$ @Mentastin That notation is unprecise. And a sequence of those is not the nice series you present as well. $\endgroup$ – Henno Brandsma Feb 21 at 16:42
  • $\begingroup$ That is completely true! I need to think whether the argument makes any sense anymore. I was at first thinking about infinite sums of vectors, hence the confusion. In that context taking the closure of the span makes more sense. Thank you for taking the time to answer though :) $\endgroup$ – Mentastin Feb 21 at 16:45
  • $\begingroup$ I have updated the proof. I hope it it now correct. I'm afraid it is much less clear now though. $\endgroup$ – Mentastin Feb 21 at 17:35

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