Is there a sum of an uncountable set of Real numbers? [duplicate]

This question already has an answer here:

The addition of Real numbers is commutative, so instead of saying we can find the sum of a sequence $$\{a_1,...,a_n\}$$ of real numbers that are pairwise not equal, we can say that there is a sum of a finite set $$A=\{a_i:1\leq i \leq n\}$$ of real numbers, since the order of summation does not matter.

The same thing is true for infinite sequence. Instead of saying we can find the sum of a infinite sequence(series) $$\{a_1,...\}$$(i.e. $$\sum_{i=1}^\infty a_i$$) of real numbers that are pairwise not equal, we can say that there is a sum of an infinite set $$A=\{a_i: i \in \mathbb{N}\}$$ of real numbers, since the result of summation is NOT changed if we put $$a_i$$ in different order.

Question: Can we define a sum of a uncountable subset of real numbers? Does the order of summation matter in this case?

marked as duplicate by mrtaurho, José Carlos Santos real-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Feb 13 at 11:00

• I think one way to define such sums over say $\mathbb{R}$ by taking supremum over finite subsets. – user428700 Feb 12 at 9:07
• Note that for an infinite sequence the order of summation does not matter IF the sum is absolutely convergent. Otherwise you can get any sum you want through reordering. An uncountable set of real numbers will not have a finite sum unless at most countably many of the numbers are not zero. – quarague Feb 12 at 9:07
• @quarague I know the first part of your comment. But could you write an answer about the second part? – Jethro Feb 12 at 9:12
• Isn't that more or less what an integral is in concept? I understand that we use countable sums to define them if we're talking about Riemann integration, but still . . . . – Robert Shore Feb 12 at 9:53
• @RobertShore Not really - intuitively, an integral is an uncountable sum of infinitesimals (= the areas of the "infinitely thin" rectangles). Here the OP is asking about a genuine uncountable sum of reals. – Noah Schweber Feb 12 at 14:10

In a general sense, the answer is no: we certainly get nothing new at the level of absolute convergence, and we get little concrete new even when we look at ordinal-length sums.

First, let's look at the situation where all the reals we're summing are nonnegative (so convergence and absolute convergence coincide, and we don't have to worry about order in the usual setting). If $$(a_i)_{i\in I}$$ is a family of nonnegative reals and $$I$$ is uncountable, one of the following must occur: either all but countably many $$a_i$$s are zero, or there is some $$\epsilon>0$$ such that $$\{i: a_i>\epsilon\}$$ is uncountable.

• Why? Well, for each positive rational $$q$$, let $$X_q=\{i: a_i>q\}$$. Then we have that $$I=[\bigcup_{q\in\mathbb{Q}_{>0}} X_q]\cup\{i: a_i=0\}.$$ This expresses the uncountable set $$I$$ as a union of countably many sets (since there are only countably many positive rationals), and so one of those sets must be uncountable; if it's $$\{i:a_i=0\}$$ then we're in the first case, and if it's one of the $$X_q$$ then just take $$\epsilon$$ to be such a $$q$$.

Now it's easy to argue that - if each of the $$a_i$$s is nonnegative - we need to set $$\sum_{i\in I}a_i=+\infty$$ (or "diverges" if you prefer) unless all but countably many of the $$a_i$$s are zero: if uncountably many of the $$a_i$$s are nonzero, then we get infinitely many $$a_i$$s which are above some fixed positive real, and that goes to infinity.

So when everything has the same sign, uncountable sums reduce to classical countable sums. What about when different signs are allowed?

The usual definition of sum over an arbitrary index set looks at the set of all finite partial sums. However, this doesn't take into account order, and so gives the answer "undefined" for (the unordered versions of) sums which conditionally converge. We might reasonably hope that we could bring order into the situation somehow to get a more interesting picture.

One natural candidate for this is to use arbitrary ordinals (this bit quickly gets technical; I hope I've made it somewhat understandable). We can define (in the sense of "give values to or set to 'undefined'") "sums of length $$\gamma$$" for an arbitrary ordinal $$\gamma$$ by recursion on $$\gamma$$:

• Base clause: The empty sum is zero.

• Successor clause: If $$(a_\eta)_{\eta<\theta+1}$$ is an $$(\theta+1)$$-length sequence of reals and we've already analyzed all $$\theta$$-length sums, then $$\sum_{\eta<\theta+1}a_\eta=(\sum_{\eta<\theta}a_\eta)+a_{\theta}$$ if $$\sum_{\eta<\theta}a_\eta$$ exists, and is undefined otherwise.

• Limit clause: If $$\lambda$$ is a limit ordinal, $$(a_\eta)_{\eta<\lambda}$$ is a $$\lambda$$-length sequence of reals, and we've already analyzed every $$\theta$$-length sum for every $$\theta<\lambda$$, we set $$\sum_{\eta<\lambda}a_\eta$$ to be $$L$$ if $$(i)$$ for all sufficiently large $$\theta<\lambda$$ the sum $$\sum_{\eta<\theta}a_\eta$$ is defined and $$(ii)$$ the limit of the values of the partial sums $$\sum_{\eta<\theta}a_\eta$$ for $$\theta<\lambda$$ exists and equals $$L$$ (this can be defined precisely via topology; I'm ignoring that issue for now), and undefined otherwise.

This looks attractive for a bit, and is not fully trivial. However, it still isn't really that different from the usual $$\mathbb{N}$$-length sum. For example, general results from descriptive set theory indicate that - looking at the least uncountable ordinal $$\omega_1$$ - if $$(a_\eta)_{\eta<\omega_1}$$ is a "reasonably definable" $$\omega_1$$-sequence of reals and $$\sum_{\eta<\omega_1}a_\eta$$ as defined above exists, then it in fact equals the values of "most" of the partial sums (specifically: there is a club of $$\theta$$ on which $$\sum_{\eta<\theta}a_\eta$$ is defined and equals $$\sum_{\eta<\omega_1}a_\eta$$). This is all going a bit advanced, but the point is that we're not getting a lot of new behavior in concrete cases. So this explains why even the above notion, which avoids the usual trivialities, doesn't really show up.

• Incidentally, you might be a bit surprised that I referred to the construction of the ordinal-length summation notation as "recursion" as opposed to "induction." This is reflecting a bit of a pet peeve of mine: induction is a proof method, recursion is a construction method. This distinction is silly a lot of the time, but actually can become important in set theory, and this is a set-theory-flavored question. So that should explain my choice of language; at the same time, it really is a super minor issue, so don't read too much into it. – Noah Schweber Feb 12 at 14:40
• Some $\eta$s should be $\theta$s in your successor case. – Andrés E. Caicedo Feb 12 at 15:07
• @AndrésE.Caicedo Arghlbarghl. Fixed! – Noah Schweber Feb 12 at 15:08
• Where is this approach treated in detail, Noah? In particular I am wondering what sort of definability you require for the club argument. – Andrés E. Caicedo Feb 12 at 15:10
• @AndrésE.Caicedo I've actually never seen it explicitly before - it's just a natural construction. Of course more large cardinals give you "tameness" for broader notions of definability, with (unless I'm missing something basic) ZFC alone handling pretty much everything that would actually occur outside of logic - the point being to consider the countably many sets of the form $\{\theta:\sum_{\eta<\theta}a_\eta<q\}$ for $q\in\mathbb{Q}$, hit each individually with whatever club dichotomy you have, and then take the intersection of the countably many clubs. Beyond that, I'd have to think a bit. – Noah Schweber Feb 12 at 15:17