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The sum of an uncountable number of positive numbers

Consider the following question:

For each real number $x$, let $\epsilon_x>0$ be an associated positive number. Is the sum $\sum_{x\in \mathbb{R}} \epsilon_x$ infinite?

I have been puzzling over this for some time. Can someone help? Also I am not sure whether this qualifies as a series, (and if not, what is it called).

Community
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    Yes, because $\mathbb R$ is uncountable. See http://math.stackexchange.com/questions/70194/does-uncountable-summation-with-a-finite-sum-ever-occur-in-mathematics and http://math.stackexchange.com/questions/94050/sum-of-uncountable-many-positive-real-numbers and http://math.stackexchange.com/questions/20661/the-sum-of-an-uncountable-number-of-positive-numbers – Jonas Meyer Feb 02 '12 at 14:45
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    Are you clear on what $\sum_{x \in \mathbf R}$ is supposed to mean? – Dylan Moreland Feb 02 '12 at 16:04
  • This is really close to being a duplicate of the last two questions linked by Jonas Meyer. – Mark Hurd Feb 02 '12 at 16:36

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Remark: I am giving hints because I think this is a fun problem, and worth working out.

Hint 1: Consider $$E_n:= \left\{ x\in \mathbb{R}: \epsilon_x >\frac{1}{n}\right\}.$$

Hint 2: Since each $\epsilon_x>0$ we know that every real number $x$ must lie within some $E_n$. However, there are countably many $E_n$, yet uncountably many real numbers. What can you conclude from this, and what does it tell us about the original series?

Eric Naslund
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    At least one $E_n$ must be uncountable and we can select finitely many elements within it to sum to more then any given positive number $M$. Hence we are done. Is this correct? –  Feb 02 '12 at 16:10
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    @Shahab: That is exactly it, you have shown that an uncountable set of positive numbers always has an infinite sum. – Eric Naslund Feb 02 '12 at 16:19
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I was going to say no: Say $\epsilon_x=e^{-x^2}$ then $\int_{-\infty}^{\infty} \epsilon_x dx$ is finite.

But $\sum_{x\in \mathbb{R}} \epsilon_x$, or even $\sum_{x\in \mathbb{P}} \epsilon_x$ where $\mathbb{P}$ is a non-empty open subset of $\mathbb{R}$, involves an uncountable number of additions, and if every $\epsilon_x$ is positive, no matter how small, you'll always reach infinity eventually.

Of course if $\mathbb{P}$ was a countable infinite subset of $\mathbb{R}$, it would be isomorphic to the integers and so finite series are then possible again.

(Edited as suggested to specify uncountable instead of just infinite.)

Mark Hurd
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Edit: Below is an [excellent] example of something that is definitely false.

It is not a series. It's sort of one of the ideas behind calculus, actually.

If you have certain conditions on how the real numbers $x$ determine the [positive] real numbers $\epsilon_x$, then what you wrote is really $\int_{-\infty}^\infty\epsilon_x dx$.

Whether or not the sum (=integral) is finite is something you need calculus to sort out.

rotten
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    No. The usual interpretation of summing over an infinite set with nonnegative values is taking the supremum over all sums over some finite subset. – Michael Greinecker Feb 02 '12 at 16:16
  • Usual interpretation? Where do you find this? – rotten Feb 02 '12 at 16:21
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    Actually, this is wrong. First of all, noone has called this a series, but that is a minor point. It is perfectly possible to define the sum of an uncountable set of real numbers (one can take it to be the supremum of all finite partial sums), and this will in general be different from the integral you've described here. To give an example, consider a function that is 1 on the interval [0,1] and 0 elsewhere. This has integral equal to 1, but the sum, as I've defined it, is infinite (because the function is strictly positive on an uncountable set). – Martin Wanvik Feb 02 '12 at 16:25
  • For example here (page 11) or here. Just google "uncountable sum". – Michael Greinecker Feb 02 '12 at 16:26
  • @Martin This is true. – rotten Feb 02 '12 at 16:27
  • @Michael I suppose there's a reason I'm an algebraist (and that I failed my measure theory qual!) – rotten Feb 02 '12 at 16:27
  • @rotten: http://en.wikipedia.org/wiki/Series_(mathematics)#Summations_over_arbitrary_index_sets – Martin Wanvik Feb 02 '12 at 16:28
  • @rotten: An integral over the real numbers is very different then a sum. The integral implicitly comes with the Lebesgue measure, which weights each intervals by its length. This is not true for a sum. Note in particular, for the real numbers the measure of any countable set is always zero, that is the integral over a countable number of points is always zero. – Eric Naslund Feb 02 '12 at 16:31
  • I have been sort of the ostrich persuasion. If I really despise something in math - in this case, anything that remotely resembles analysis - I tend to just ignore it and then make stupid claims about it. – rotten Feb 02 '12 at 16:32
  • Dear @Martin: You wrote: "It is perfectly possible to define the sum of an uncountable set of real numbers". You probably meant: "It is perfectly possible to define the sum of an uncountable set of nonnegative numbers". – Pierre-Yves Gaillard Feb 02 '12 at 16:46
  • @rotten: There are many cases of interplay between algebra and analysis. To attack at least some algebraic problems effectively, one may need insights from analysis. – André Nicolas Feb 02 '12 at 17:10
  • @Pierre: I have to admit, I didn't really think carefully about that distinction. You are of course correct; to be sure, the definition still makes sense even if you do allow arbitrary real numbers, but no negative number could ever contribute (provided we define the sum over the empty set to be 0), so the result doesn't exactly deserve to be called a sum anymore. Thanks for setting me straight. – Martin Wanvik Feb 03 '12 at 12:02
  • @Pierre: Although you could make better sense of such a sum by defining it as the integral with respect to the counting measure. I should probably shut up now ;) – Martin Wanvik Feb 03 '12 at 12:23
  • Dear @Martin: Thanks for your reply. Yes, I think (like you, if I understand your comments correctly) that the best is to write $f=f_+-f_-$ with $f_+,f_-\ge0$ (in the usual way), and to define $\sum f(x)$ only if we do not have $$\sum f_+(x)=\infty=\sum f_-(x).$$ – Pierre-Yves Gaillard Feb 03 '12 at 12:45