Sums of uncountably many real numbers - MathOverflow [closed] most recent 30 from http://mathoverflow.net 2013-05-18T18:25:15Z http://mathoverflow.net/feeds/question/64526 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/64526/sums-of-uncountably-many-real-numbers Sums of uncountably many real numbers David Corwin 2011-05-10T18:31:04Z 2011-05-10T19:04:00Z <p>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 $f$.</p> http://mathoverflow.net/questions/64526/sums-of-uncountably-many-real-numbers/64527#64527 Answer by Andreas Blass for Sums of uncountably many real numbers Andreas Blass 2011-05-10T18:38:14Z 2011-05-10T18:38:14Z <p>No. $S$ is the union of the countably many sets <code>$A_n=\{s\in S:f(s)&gt;1/n\}$</code>, so some <code>$A_n$</code> must be infinite (in fact uncountable). Thus, your sum contains infinitely many terms all of which are at least $1/n$.</p> http://mathoverflow.net/questions/64526/sums-of-uncountably-many-real-numbers/64528#64528 Answer by David Corwin for Sums of uncountably many real numbers David Corwin 2011-05-10T18:38:45Z 2011-05-10T18:38:45Z <p>Actually, I just realized how to solve the problem. The answer is that it is not possible.</p> <p>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.</p> <p>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 $0$.</p> http://mathoverflow.net/questions/64526/sums-of-uncountably-many-real-numbers/64533#64533 Answer by Pete L. Clark for Sums of uncountably many real numbers Pete L. Clark 2011-05-10T19:03:10Z 2011-05-10T19:03:10Z <p>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 <a href="http://www.math.uga.edu/~pete/3100supp.pdf" rel="nofollow">these notes on sequences and series</a> (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.</p> <p>Note that this fact comes up sometimes in practice. In <a href="http://math.stackexchange.com/questions/3777/is-there-a-function-with-a-removable-discontinuity-at-every-point" rel="nofollow">this math.SE question</a> 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 <em>does not use</em> the kind of <strong>uncountable pigeonhole principle argument</strong> that you need to answer the current question. And I got a very nice answer!</p>