Sums of uncountably many real numbers Suppose $S$ is an uncountable set, and $f$ is a function from $S$ to the positive real numbers. Define the sum of $f$ over $S$ to be the supremum of $\sum_{x \in N} f(x)$ as $N$ ranges over all countable subsets of $S$. Is it possible to choose $S$ and $f$ so that the sum is finite? If so, please exhibit such $S$ and $f$.
 A: Actually, I just realized how to solve the problem. The answer is that it is not possible.
Suppose the sum is finite. Let $S_n$, for positive integer $n$, be the set of $x \in S$ such that $f(x) \ge \frac{1}{n}$. Then for each $n$, $S_n$ must be finite, if the sum is finite. But $S = \bigcup_n S_n$, meaning that $S$ is at most countable.
In other words, the sum of uncountably-many non-negative real numbers is finite only if all but countably many of those real numbers are $0$.
A: This is a standard result in undergraduate analysis, although it is admittedly somewhat hard to find in the standard references.  The following is a very non-standard reference: see the last exercise in II.9.4 of these notes on sequences and series (see p. 69...for now; page numbers are subject to change).  They occur in the context of a larger discussion on unordered summation, which is what you are looking into above.  The general definition of unordered summability is a bit more complicated (it is a nice special case of convergence with respect to a net, although one needn't use the term), but in the case where the values of the "$S$-indexed sequence" are non-negative, it coincides with what you have given: see Proposition 82.
Note that this fact comes up sometimes in practice.  In this math.SE question I set as a challenge to give a proof of the following fact -- there is no function $f: \mathbb{R} \rightarrow \mathbb{R}$ with a removable discontinuity at every point -- which does not use the kind of uncountable pigeonhole principle argument that you need to answer the current question.  And I got a very nice answer!
A: No.  $S$ is the union of the countably many sets $A_n=\{s\in S:f(s)>1/n\}$, so some $A_n$ must be infinite (in fact uncountable).  Thus, your sum contains infinitely many terms all of which are at least $1/n$.
