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More specifically, why are we "allowed" to denote $|\mathbb{N}|<|\mathbb{R}|$ but not $\sum\limits_{n\in\mathbb{N}}1<\sum\limits_{r\in\mathbb{R}}1$?

Can we distinguish between "countable divergence" and "uncountable divergence"?

I apologize in advance for the "rather naive" question.

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    $\begingroup$ A series is the limit of a sequence, but any sequence is countable, so, simply, $\sum_{r\in\mathbb{R}} 1$ means nothing. $\endgroup$ – Jack D'Aurizio Aug 7 '15 at 5:42
  • $\begingroup$ @JackD'Aurizio: So my notation of $\sum\limits_{r\in\mathbb{R}}1$ is by itself wrong? What if the sequence is uncountable (or is that not possible by definition)? $\endgroup$ – barak manos Aug 7 '15 at 5:44
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    $\begingroup$ @barakmanos how do you define $\sum_{r\in \mathbb R} 1$ ? - i do not know any reasonable definition and i would say it is undefined $\endgroup$ – supinf Aug 7 '15 at 6:00
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    $\begingroup$ @Eric: That makes no sense if CH is false. Or in general, since this is a cardinal sum, not an ordinal sum. You can sum orders, but if the index set is not a well-ordering, then the result has no business being well-ordered, and $\Bbb R$ as a linear order is not a well-order. So saying it is any kind of ordinal is wrong. $\endgroup$ – Asaf Karagila Aug 7 '15 at 7:42
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    $\begingroup$ @supinf If $A$ is an index set, and $(x_a)_{a\in A}$ a family of real numbers, then consider the directed (by inclusion) set $\mathscr{F}$ of finite subsets of $A$. Then you have a net $F \mapsto \sum\limits_{a\in F} x_a$ indexed by $\mathscr{F}$ in $\mathbb{R}$. If that net converges, then $\sum\limits_{a\in A} x_a$ is the limit of the net. If we look only at convergence in $\mathbb{R}$, then $\sum\limits_{r\in\mathbb{R}} 1$ doesn't exist (the family $(1)_{r\in\mathbb{R}}$ isn't summable). If we look at convergence in $[-\infty,+\infty]$, it converges to $+\infty$. $\endgroup$ – Daniel Fischer Aug 7 '15 at 7:45
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So, Cardinal Arithmetic, Natural Arithmetic, and Real Arithmetic are all different things. At least for me, addition in each case means something completely different.

Cardinal Arithmetic has to do with addition of sets, which is really closer to (disjoint) union modded out by an equivalence relation (i.e. bijection). In this area, we say that $|A| = |B| \iff$ there exists a bijection between the two sets. So, when we say something like $\aleph_0 + \aleph_0 = \aleph_0$, we are really just saying that there exists a bijection between $\aleph_0 \sqcup \aleph_0$ and $\aleph_0$. We say that $\kappa < \mu$ if there exists an injection from $\kappa$ to $\mu$, but none of those injections are surjections. From these definitions, one can show that we have accurately described an equivalence relation in the class of sets.

Natural Arithmetic, or arithmetic on $\mathbb{N}$ is the arithmetic we learn in school. While we don't learn all the axioms of arithmetic (first of all, because the set of axioms for true arithmetic is not recursive) the basic theory is the pretty much the same as Finite Cardinal Arithmetic. Things change drastically when we consider Non-standard models of arithmetic. In these models of arithmetic, we can add, subtract, multiply and sometimes divide numbers which are larger than all standard natural numbers. We might have a number $\tau$ such that $\tau >n$ for all $n$ in the standard model. Furthermore, $\tau + \tau \neq \tau$. Here, we have that $ \tau <\tau + \tau = 2\tau $ and how you would think Natural Arithmetic might work on very large numbers (with a loose interpretation of large). It should also be stated, however, due to a theorem of Tennenbaum, that none of these arithmetics on non-standard models are computable, i.e. we cannot truly compute anything.

Finally, Real Arithmetic has two separate meanings which I will call First Order Real Arithmetic (FORA), and Measure Arithmetic (MA). Now, FORA is similar to addition, multiplication, division, what have you, on the reals. However, we can only use finite formulas, so infinite sums cannot really be handled by FORA. FORA is the model theory perspective of real arithmetic and in the same way that natural arithmetic can have super big natural numbers, FORA can have super huge large real numbers, as well as super tiny real numbers (known as infinitesimals.

Measure Arithmetic of the real line is where integration takes place. Practically speaking, integration is the way one sums real numbers in the way discrete sums don't really work. The problem is, suppose we want to compute something like $\sum_{i\in [0,1]} 1$. If we simply use integration, or in it's natural notation, $\int_0^1 1dx$, we compute that this sum is equal to 1.

Finally, as Jack D'Aurizio pointed out, infinite sums are defined as the limit of finite sums. There is more to be said about how nice of an infinite cardinal $\aleph_0$ is (practically, since it is the limit of finite cardinals (and we know how to manipulate finite cardinals), it gives us a lot of control over what can happen as things tend to it). But I hope this clears some concepts up!

Edit: Now that I think about it more, even in Logics like $L_{\mu,\kappa}$ you cannot discuss infinite sums over the reals. The major issue is that there is no standard method for dealing with terms of uncountable length. You can, however, discuss formulas with uncountable length.

Addendum: I have just found an old paper by Keisler which deals with a formal system with infinitary functions. Check out section 5 if you are interested. But I should caution that it is pretty notationally dense and difficult to parse.

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  • $\begingroup$ Thanks. Can you please explain formulas with uncountable length? $\endgroup$ – barak manos Aug 7 '15 at 6:09
  • $\begingroup$ Often, the more words in an answer, the more mistakes. This one, however is a rare treat. Great answer. $\endgroup$ – Zach Stone Aug 7 '15 at 6:14
  • $\begingroup$ @barakmanos: So if we have some fixed language $\mathcal{L}$, we have formulas like $x = y$, $x <a$ (where $a$ is a parameter), or for some fixed polynomial, $P(x_1,...,x_n) = 0$. Now, from this finite formulas, we can build an uncountable formula by "putting uncountably finite formulas together". So for instance, we have a formula $(\exists x) \bigwedge_{a \in \mathbb{R}} x \neq a$. This formula (actually a sentence) say that there exists some element who is no "standard real number". Yet, the problem with these types of logics is that the compactness theorem fails (which is really bad). $\endgroup$ – Kyle Aug 7 '15 at 6:19
  • $\begingroup$ Got it, thanks. $\endgroup$ – barak manos Aug 7 '15 at 6:25
  • $\begingroup$ @barakmanos: But it is an interesting question whether or not infinite sequences of terms (i.e. 1+1+...+1+... real number of times) have been studied rigorously. I'll look around and see if I can find anything interest. I have found a paper on 'infinitary predicates', but nothing on finitary terms. It's an old paper (By Keisler), so the notation is a little dated: matwbn.icm.edu.pl/ksiazki/fm/fm52/fm52111.pdf $\endgroup$ – Kyle Aug 7 '15 at 6:30
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Let me just add to Kyle's answer that in cardinal arithmetic you do have a well-defined notion of uncountable sums. Because you're not trying to calculate a real number, rather you're looking for a cardinal.

If you are looking at the sum $\sum_{r\in\Bbb R}1$ as a real-valued sum, then it is indeed a strange thing to write. But looking it at as a cardinal sum, this makes perfect sense, and the cardinal is exactly $2^{\aleph_0}$. Similarly, of course, $\sum_{n\in\Bbb N}1=\aleph_0$ as a cardinal sum, and it's just a divergent series when looking at it as a real-valued sum.

So context is key.

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    $\begingroup$ I thought I had covered everything! There's always a case that slips by... Thanks Asaf! $\endgroup$ – Kyle Aug 7 '15 at 7:41
  • $\begingroup$ You did a good job, and wrote a lengthy answer. Don't beat yourself up. :-) $\endgroup$ – Asaf Karagila Aug 7 '15 at 7:41
  • $\begingroup$ So assuming that the 'larger than' relation can be used over these two cardinals (i.e., it is "legal" to denote $\aleph_0<2^{\aleph_0}$) - why are we not "allowed" to denote $\sum\limits_{n\in\mathbb{N}}1<\sum\limits_{r\in\mathbb{R}}1$? $\endgroup$ – barak manos Aug 7 '15 at 7:48
  • $\begingroup$ @barak: In cardinal arithmetic, we are allowed to, and we should. But cardinal summation is not real-valued, if you mean to ask about real-valued sums, then this is a whole other thing. And the problem arises when there is a notational overload. Which is why I said context is key. $\endgroup$ – Asaf Karagila Aug 7 '15 at 7:51
  • $\begingroup$ Got it, thanks. $\endgroup$ – barak manos Aug 7 '15 at 8:05

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