Claim: If $(x_\alpha)_{\alpha\in A}$ is a collection of real numbers $x_\alpha\in [0,\infty]$ such that $\sum_{\alpha\in A}x_\alpha<\infty$, then $x_\alpha=0$ for all but at most countably many $\alpha\in A$ ($A$ need not be countable).

Proof: Let $\sum_{\alpha\in A}x_\alpha=M<\infty$. Consider $S_n=\{\alpha\in A \mid x_\alpha>1/n\}$.

Then $M\geq\sum_{\alpha\in S_n}x_\alpha>\sum_{\alpha\in S_n}1/n=\frac{N}{n}$, where $N\in\mathbb{N}\cup\{\infty\}$ is the number of elements in $S_n$.

Thus $S_n$ has at most $Mn$ elements.

Hence $\{\alpha\in A \mid x_\alpha>0\}=\bigcup_{n\in\mathbb{N}}S_n$ is countable as the countable union of finite sets. $\square$

First, is my proof correct? Second, are there more concise/elegant proofs?

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    $\begingroup$ This looks fine to me. A slight variant: Suppose there are uncountably many $x_{\alpha} \gt 0$. Then there is $n$ such that $S_{n} = \{\alpha\,:\,x_{\alpha} \geq \frac{1}{n}\}$ is infinite. But this implies that $\sum_{\alpha} x_{\alpha} \geq \sum_{S_{n}} \frac{1}{n} = \infty$. $\endgroup$
    – t.b.
    Feb 6, 2011 at 7:35
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    $\begingroup$ @Theo Your variant hides the crucial step of the proof by asserting "there is $n$ such that $S_n$ is infinite". If you try to explain why $S_n$ should be infinite for at least one $n$, surely the fact that there are infinitely many $a$ such that $x_a>0$ will not be enough to conclude, hence you must use the countable vs. uncountable hypothesis. Plus, your proof is by contradiction where this is not needed... $\endgroup$
    – Did
    Feb 6, 2011 at 9:33
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    $\begingroup$ @Didier: Agreed. Nevertheless this argument always seemed clearer to me, I don't know why. $\endgroup$
    – t.b.
    Feb 6, 2011 at 10:28
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    $\begingroup$ @Theo: Understood. Precisely what I would honestly like to know is why you feel the step Then there is $n$ such that $S_n$ is infinite is natural or clear or intuitive or whatever. $\endgroup$
    – Did
    Feb 6, 2011 at 10:58
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    $\begingroup$ On second thought, I guess a reasonable definition would be the supremum of all finite sums. $\endgroup$
    – Dan Ramras
    Feb 7, 2011 at 0:29

5 Answers 5


Just so the question gets an answer: yes, your proof is correct and is one of several phrasings of the shortest proof that I know.


There is also one question directly relating to this question, that is, how to define the sum of uncountably many numbers (not necessarily positive numbers). The difficulty lies in the fact that there could not be any order of this summation, since there are uncountably many of terms. So, when we talk about the sum of $x_\alpha$, namely, $\sum_{\alpha\in A}x_\alpha$, we are actually saying the following,

For every countable subset of $I\subset A$ with arbitrarily given order, the sequence $(x_\alpha)_{\alpha\in I}$ should be convergent. In other words, the sequence $(x_\alpha)_{\alpha\in I}$ should be absolutely convergent.

A proper definition is given in Paul Halmos' book, Introduction to Hilbert Space and the Theory of Spectral Multiplicity, as follows:

$x=\sum_{\alpha\in A}x_\alpha$ means that for any positive number $\varepsilon$ there is some finite set $I_0$ such that for any finite set (or more generally, countable set) $I\supset I_0$, we have $|x-\sum_{\alpha\in I}x_\alpha|<\varepsilon$.

Note that, the set $\{1,-\frac{1}{2},\frac{1}{3},\cdots,(-1)^{k-1}\frac{1}{k},\cdots, 0,\cdots\}$, where in the end there are uncountably many $0$'s is not convergent any more. But the sequence $\{1,-\frac{1}{2},\cdots\}$ is convergent in the standard sense.

Now we invoke Zorn's lemma, on all countable subsets $I$, with respect to which, the sequence $x_\alpha$ is absolutely convergent, with the inclusion as the order. Note that for any $I_1\subset I_2\subset I_3\subset \cdots$ a chain of countable subsets of $A$, the set $I^*=\bigcup_iI_i$ is also a countable subset of $A$ and by the definition, the sequence with index in $I^*$ is also absolutely convergent. By Zorn's lemma, there should exists a maximal countable subset $I_{max}$. This means that any number $x_\alpha$ with $\alpha\notin I_{max}$ should be $0$, otherwise we can construct another strictly larger countable subset on which the number sequence is absolutely convergent.

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    $\begingroup$ Your proof seems to rely on the assumption that a chain of countable sets must be countable. This is easily seen to be wrong: There are uncountably many sets of the form $(-\infty,r]\cap\mathbb{Q}$. $\endgroup$ May 20, 2012 at 1:51

A more explicit formulation of the statement in the title is this. Let $A$ be some uncountable family of non-negative numbers. If $S$ is to be the "sum of $A$", then by any reasonable definition of "sum", surely $S$ must be greater than or equal to the sum of any finite sub-family of $A$. In that sense we can prove that any for any reasonable definition of "sum", the sum of $A$ must be infinite (unless all but countably many elements of $A$ are zero) by showing:

For any $M$, there exists a finite sub-family $B$ of $A$ such that the sum of $B$ is at least $M$.

Proof: Assume that $A^+$, the positive members of $A$, is uncountable (otherwise the theorem is obviously false). $A^+=\bigcup_n A_n$, where $A_n=\{a\in A | a \geq \frac 1 n\}$. Since the union of countably many finite sets is countable, one of the $A_n$ must be infinite. Grab as many elements as you need from that set to get a sum greater than $M$.


The question is not well-posed because the notion of an infinite sum $\sum_{\alpha\in A}x_\alpha$ over an uncountable collection has not been defined. The "infinite sums" familiar from analysis arise in the context of analyzing series defined by sequences indexed over $\mathbb{N}$, and the series is defined to be the limit of the partial sums. The only objects defined here are (1) finite sums, and (2) limits of sequences indexed by the natural numbers.

If the question is interpreted as asking whether, out of an uncountable collection $X$ of positive reals, one could always form a divergent series, the answer is affirmative, and moreover one can always choose a sequence $(u_n)$ in $X$ with a fixed positive lower bound $u_n>\epsilon>0$. This is because $X$ admits a countable cover consisting of sets $A_r$ of members of $X$ that are greater than $r$ for all rational $r$, and therefore one of the $A_r$ is uncountable and in particular infinite.

Note that over the hyperreals, one does have a notion of infinite sum, when the collection being summed over is hyperfinite (it is necessarily uncountable).

  • 7
    $\begingroup$ The usual notion of (absolute) summation in such a context is that one maps each finite set $F\subseteq A$ to the sum $\sum_{a\in F}x_a$. This defines a map from the family of finite subsets to real numbers. The family of finite subsets is a directed set under reverse inclusion, so this map defines a net. Now the sum is simply the limit of this net. $\endgroup$ Jul 12, 2013 at 8:16
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    $\begingroup$ and moreover in the case OP is interested in, all elements in the sum are non-negative. Then the sum can equivalently be defined as the supremum of all finite sums. $\endgroup$ Jul 12, 2013 at 8:19
  • $\begingroup$ @Michael Greinecker: How "usual" is this notion? Since, as noted already by the OP, this will always be infinite if the collection of real numbers is uncountable, such a theory would be vacuous. $\endgroup$ Jul 12, 2013 at 8:20
  • $\begingroup$ @user72694 No, only if the collection of nonzero numbers is uncountable. You might for example want to look at $L_p$ spaces with counting measure, this amounts to this form of summation. The definition can be found, for example, in Douglas 1972, Banach algebra techniques in operator theory as Definition 1.8. $\endgroup$ Jul 12, 2013 at 10:18
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    $\begingroup$ I think that's what I said, if the OP's collection of numbers is uncountable then the sum will necessarily be infinite as the OP was the first to mention. I doubt it I would find anything to the contrary in Douglas, but thanks anyway. See also my note about the hyperreals (I will add it shortly). $\endgroup$ Jul 14, 2013 at 12:17

As we don't have the notion of 'sum of uncountably many numbers', this question looks strange! But we can transform your question into the following, which makes sense.
Let $(X,\mu)$ be a finite measure space and ${\{A_\alpha\}}_{\alpha \in \Lambda}$ be a collection of pairwise disjoint measurable subsets of $X$, each having positive measure. Can '$\Lambda$' be uncountable?
Answer is no and your proof works here without any issues.

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    $\begingroup$ Isn't the 'sum of uncountably many numbers' defined as the $\sup$ over all finite sums? $\endgroup$
    – them
    Mar 25, 2016 at 21:50

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