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Sequence A098820 in the OEIS is purely combinatorially defined (all you need to know is integers), and non-decreasing; it is known (Laver, 95) that assuming Large Cardinal axiom I3 the sequence goes to infinity. However, under ZF or ZFC the convergence status is not known.

Now imagine Alice works under ZF+I3 while Bob works under ZF only. Bob starts writing:

  • Pick any $A\in \Bbb N$.
  • Ask Alice to find $n\in \Bbb N$ such that $u_n\geq A$ in her world (we know that there is such an $n$) and communicate it to us.
  • Compute $u_n$ ourselves and see if $u_n\geq A$ here too, which it obviously is, since the definition of the sequence is purely combinatorial and does not rely on any complicated axioms.

How is this not a proof that the sequence goes to infinity under ZF?

My thought is that perhaps I3 is only not known to be inconsistent with ZF, which is quite different from known to be consistent with. So there would be two options: either the sequence provably goes to infinity in ZF, or I3 and ZF are inconsistent. Is this right or am I missing the point?

Arnaud Mortier
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    I think you are missing the point. Even if for each single $n$ you can prove in ZF or even in much weaker theories that $u_n\ge A$, this is not the same as saying that you can prove in ZF the statement $\forall n,(u_n\ge A)$, which is what one is after. – Andrés E. Caicedo May 05 '20 at 20:54
  • @AndrésE.Caicedo I don't understand your comment. We only need one $n$ for which $u_n\geq A$, since we know that the sequence is non-decreasing (this is proved by elementary means that require no special axioms). – Arnaud Mortier May 05 '20 at 20:57
  • Yes, I should have written $u_n\le A$ in both instances. – Andrés E. Caicedo May 05 '20 at 22:28

1 Answers1

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This doesn't work - the issue is with nonstandard natural numbers.

(There's a related comment about complexity - the nonstandardness issue can be gotten around for $\Pi^0_1$ theorems, but your principle is $\Pi^0_2$ - but the nonstandardness issue really lies at the heart of the Alice-Bob idea, so I think it should be mentioned first.)

In the game you're imagining, why can Alice and Bob "share numbers"? Think of them as holding specific models $A\models\mathsf{ZFC}+I_3$ and $B\models\mathsf{ZF}$ respectively; in order for Alice to "see" Bob's questions we need $\omega^B\subseteq\omega^A$, and - far more importantly - in order for Bob to "use" Alice's answers we need $\omega^A\subseteq\omega^B$.

  • To see the issue, suppose Bob is working with the true natural numbers and think about what would happen if we replace Alice with Sam, who holds a model $S\models\mathsf{ZF}+\neg Con(\mathsf{ZF})$. Bob is skeptical of Sam, and asks "What is an example of a (code for a) proof of a contradiction in $\mathsf{ZF}$?" Sam replies, "I'm glad you asked - it's [eldritch horror]." If Sam's response to Bob is a nonstandard natural number, there's not much use Bob can get from it. And of course the true nightmare situation is when two models have incomparable versions of the natural numbers.

So in order for the idea in your post to work as written, we would need the following: for every model of $\mathsf{ZF}$, there is a model of $\mathsf{ZFC}+I_3$ with the same natural numbers. And in fact this is false (unless $\mathsf{ZF}$ is inconsistent): $\mathsf{ZF}$ has a model which thinks (say) that $\mathsf{ZF}$ is inconsistent, but $\mathsf{ZFC}+I_3$ doesn't (regardless of whether it's consistent).

A more careful implementation of the same idea will show that whenever $\varphi$ is $\Pi^0_1$ and $\mathsf{ZFC}+I_3$ proves $\varphi$, then $\mathsf{ZF}$ proves "If $\mathsf{ZFC}+I_3$ is consistent then $\varphi$ is true." However, this still doesn't apply to the above problem, since unboundedness is $\Pi^0_2$.

Noah Schweber
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  • Thanks for your answer! I have only a shallow knowledge about set theory, and it is true that somehow I've always assumed that no matter what model you're in, natural numbers must always be canonically identifiable to those we are used to - because it "has a starting point, and goes one step at a time forwards", it doesn't feel like there should be many ways to do that. But I also had this question at the back of my mind for a long time, and thought that understanding it would be a great way to get a better grasp on it. So thanks. – Arnaud Mortier May 05 '20 at 21:20
  • @ArnaudMortier $\mathsf{ZF}$etc., for all its strength, is still a first-order theory - it's as susceptible to compactness (and hence nonstandard natural numbers) as $\mathsf{PA}$. – Noah Schweber May 05 '20 at 21:27
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    [eldritch horror]. – Andrés E. Caicedo May 05 '20 at 22:36
  • There is still some shady point: the sequence in question here is not defined at all for possibly nonstandard naturals. It's not clear at all how a Laver table with nonstandard size would behave, or even if it exists. Doesn't that mean that Alice's $n$ must be from the initial segment of her naturals and therefore possible to communicate? – Arnaud Mortier May 06 '20 at 15:07
  • @ArnaudMortier The sequence has a definition in the language of set theory. Any model of set theory then parses that definition for all of its natural numbers. Internally to a given model, there's no difference between standard and nonstandard, so nothing like this can help us. – Noah Schweber May 06 '20 at 15:33
  • @ArnaudMortier More generally, consider overspill. If $M\models \mathsf{ZF}$ has nonstandard natural numbers, then if $X\in M$ is a subset of $\mathbb{N}^M$ containing every standard natural number then $X$ contains some nonstandard natural numbers as well. Everything that works for each standard number will also work for at least some nonstandard numbers as well. In particular, if $M$ has nonstandard natural numbers then $M$ has "nonstandard-size Laver tables" - even if it's not clear to us what those should be, $M$ has them almost by accident. – Noah Schweber May 06 '20 at 15:40
  • It may help to consider a less mathematically-intricate example: think about a model $M\models\mathsf{PA}+\neg Con(\mathsf{ZF})$. In $M$ there is some $n$ which $M$ thinks is a code for a proof of $0=1$ from $\mathsf{ZF}$. While to us "it's not clear at all how a [proof] with nonstandard size would behave, or even if it exists," $M$ can't see that there's anything strange about $n$ at all. And continuing the game analogy, if we ask $M$ for a code for a proof of $0=1$ from $\mathsf{ZF}$, $M$ will happily respond with its least such $n$ with no problem at all despite its nonstandardness. – Noah Schweber May 06 '20 at 15:47
  • Thank you very much @NoahSchweber that's great insight. – Arnaud Mortier May 06 '20 at 21:18
  • Someone on Discord told me that the quantifier exchange answer (appearing in Andrés E. Caicedo's commnent on the main post) is clearer and more direct for this Laver table problem. Is the use of nonstandard naturals overkill in this scenario? – C7X Jul 21 '22 at 21:22