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 $f$.

closed as too localized by Andres Caicedo, Andreas Thom, Felipe Voloch, S. Carnahan♦ May 10 '11 at 20:15
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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 $1/n$. 


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 uncountablymany nonnegative real numbers is finite only if all but countably many of those real numbers are $0$. 


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 nonstandard 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 nonnegative, 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! 

