I wanted to know the approximate sum of real numbers from 0 to 1.

Please tell me how we can find it.


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    $\begingroup$ there are infinitely many(uncountable) numbers greater than $1/2.$ $\endgroup$ – abel Jun 28 '15 at 17:05
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    $\begingroup$ It is difficult to take a sum of an uncountable number of numbers. Even if you restricted the sum to positive rational numbers in top half of the interval, you would get an infinite result. $\endgroup$ – Henry Jun 28 '15 at 17:05
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    $\begingroup$ Something like $\int_{x=0}^1 x$ is more appropriate, I think. $\endgroup$ – davcha Jun 28 '15 at 17:13

The sum diverges: $$\large\sum_{a\in[0,1]}a=\sup_{\substack{\text{finite}\\S\subset[0,1]}}\sum_{a\in S}a\geq \sup_{N\in\mathbb{N}}\sum_{n=2}^N\left(\frac{1}{2}+\frac{1}{n}\right)\geq\sup_{N\in\mathbb{N}}\left(\frac{N}{2}\right)=\infty$$

Observe this definition from Folland's real analysis book:

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    $\begingroup$ Under this definition with $f(x) \geq 0$ for all $x \in X$, we get $\sum_{x \in X} f(x) = \infty$ whenever there are uncountably many $x \in X$ such that $f(x)>0$. $\endgroup$ – Michael Jun 28 '15 at 17:17
  • $\begingroup$ Indeed, that's Folland's next statement :) (image link) $\endgroup$ – Zev Chonoles Jun 28 '15 at 17:20
  • $\begingroup$ Interesting. I have never seen such a definition. Perhaps it is only useful for that fact itself...which can be interpeted as saying that the sum is always infinity except at times when standard definitions of sums apply. =) Yet your quote says "We shall occasionally encounter uncountable sums," I wonder when such things would actually be encountered. $\endgroup$ – Michael Jun 28 '15 at 20:27
  • $\begingroup$ @Michael Re: I wonder when such things would actually be encountered. I think that this might be a good question on the main site. (However, there were already some posts about this type of sum, so if you decide to post a separate question about this, you should search first.) I do not know where they are used in Folland's book. $\endgroup$ – Martin Sleziak Jul 2 '15 at 8:00
  • $\begingroup$ One application I have seen is the definition of the Hilbert space $\ell_2(A)$. If you already learned about Hilbert spaces, you have certainly seen $\ell_2=\ell_2(\mathbb N)$, where the inner product is given by $\langle x,y \rangle = \sum\limits_{i\in\mathbb N} x_iy_i$. If we allow summation over arbitrary sets, then we can define $\ell_2(A)$ using almost the same construction; in this case, the inner product will be $\langle x,y \rangle = \sum\limits_{i\in A} x_iy_i$. It can be shown that this is indeed a Hilbert space and that every space is isomorphic to $\ell_2(A)$ for some set $A$. $\endgroup$ – Martin Sleziak Jul 2 '15 at 8:01

The sum is infinite. We don't really have a good definition for $$\sum_{x\in [0,1]}x,$$ But it seems reasonable that he following relation should hold for any proposed definition. $$\sum_{x\in\mathbb [0,1]} x\geq \sum_{n\in \mathbb N} \frac{1}{n}=\infty.$$

  • $\begingroup$ Of course, thanks. $\endgroup$ – Alex S Jun 28 '15 at 17:08
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    $\begingroup$ I will not downvote, but of course $\sum_{x \in [0,1]} x$ has no meaning since we cannot sum over an uncountably infinite set. $\endgroup$ – Michael Jun 28 '15 at 17:09
  • $\begingroup$ Good point. I have edited my answer to reflect this. $\endgroup$ – Alex S Jun 28 '15 at 17:13
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    $\begingroup$ @Michael: I have seen a commonly accepted definition of such a sum as long as every term of the sum is non-negative. Please see my answer for example. $\endgroup$ – Zev Chonoles Jun 28 '15 at 17:15
  • $\begingroup$ @Michael: See for example this post and other posts linked to it for some further references. $\endgroup$ – Martin Sleziak Jun 30 '15 at 11:43

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