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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 Andres Caicedo, Andreas Thom, Felipe Voloch, S. Carnahan May 10 '11 at 20:15

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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
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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, et.al.). Gerhard "Ask Me About System Design" Paseman, 2011.05.10 –  Gerhard Paseman May 10 '11 at 20:51
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@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
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3 Answers 3

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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