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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).


marked as duplicate by Jonas Meyer, Eric Naslund, Asaf Karagila, Zev Chonoles Feb 7 '12 at 2:21

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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?

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    $\begingroup$ 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? $\endgroup$ – Shahab Feb 2 '12 at 16:10
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    $\begingroup$ @Shahab: That is exactly it, you have shown that an uncountable set of positive numbers always has an infinite sum. $\endgroup$ – Eric Naslund Feb 2 '12 at 16:19

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.)

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    $\begingroup$ You mean uncountable number of additions, not infinite. $\endgroup$ – Eric Naslund Feb 2 '12 at 16:06

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.

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    $\begingroup$ No. The usual interpretation of summing over an infinite set with nonnegative values is taking the supremum over all sums over some finite subset. $\endgroup$ – Michael Greinecker Feb 2 '12 at 16:16
  • $\begingroup$ Usual interpretation? Where do you find this? $\endgroup$ – rotten Feb 2 '12 at 16:21
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    $\begingroup$ 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). $\endgroup$ – Martin Wanvik Feb 2 '12 at 16:25
  • $\begingroup$ For example here (page 11) or here. Just google "uncountable sum". $\endgroup$ – Michael Greinecker Feb 2 '12 at 16:26
  • $\begingroup$ @Martin This is true. $\endgroup$ – rotten Feb 2 '12 at 16:27

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