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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 $f$.

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closed as too localized by Andrés E. Caicedo, Andreas Thom, Felipe Voloch, S. Carnahan May 10 '11 at 20:15

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

There has to be something eluding me, why did you tag your question with the large cardinals and set theory tags? Does the following argument not work? $S=\Cup_{n\in\mathbb{N}\setminus \lbrace 0\rbrace} S_n$ where $S_n=\lbrace s\in S\mathrm{~s.t.~} f(s)>\frac{1}{n}\rbrace$, thus one of them is non denumerable and taking a denumerable subset of said $S_{n_0}$ will yield an infinite sum. – Olivier Bégassat May 10 '11 at 18:45
I changed the tags. The question is certainly not about large cardinals, nor even about set theory as that is understood on this site. – Pete L. Clark May 10 '11 at 19:04
Even though you already have your answer, I'm going to close since this particular question isn't quite what we want here. Also, it is phrased in a style suspiciously resembling a homework problem. – S. Carnahan May 10 '11 at 20:20
It looks like something from Concrete Mathematics (Knuth, Gerhard "Ask Me About System Design" Paseman, 2011.05.10 – Gerhard Paseman May 10 '11 at 20:51
@David Roberts: the identically $0$ function does not fit the conditions imposed by the OP. (Maybe that's part of your joke; if so, okay, but I didn't get it...) – Pete L. Clark May 11 '11 at 2:33

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 $1/n$.

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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 $0$.

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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!

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