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Part of Asaf Karagila's brilliant answer to one of my other questions puzzles me a lot. Namely, I find it hard to understand how there can be a model for ZFC with uncountably many integers. My reasons for being puzzled are entirely based on intuition and possibly cannot be formalised:

I have always thought that it is when we start to approach uncountable structures like the real numbers that mysteries and paradoxes start to appear. I had this idea that as long as we stayed with the countable number structures ($\Bbb N, \Bbb Z, \Bbb Q,\ldots$), there would be a kind of "isomorphism" between these structures across models. For instance, I would expect the prime numbers in one model to be prime numbers of any other model as well. The reason I would think this is that integers are, in a way, more natural than set theory itself. If $S\colon\Bbb N\to\Bbb N$ denotes the successor function from Peano's axioms, then the definition of the number $n$ is $n = S^n(0)$; but this definition is, in a way, circular, as the notation shows. In order to define $n$ in set theory, we need to know what we mean by "applying $S$ $n$ times". This can of course be considered a syntactical issue, like how parantheses are interpreted in logic; but just like the rules of logic, this has to be given in informal language. Rather, the purpose of constructing the natural numbers in set theory is to show that our theory is strong enough to support Peano arithmetic.

But obviously, as Asaf Karagila's answer shows, I'm wrong. But where am I wrong,m and how much am I wrong?

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Gaussler
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    The short answer is that first-order logic isn't "strong" enough to prevent a theory with an infinite model from having models of all infinite cardinalities. See the Löwenheim–Skolem theorem. – hardmath Jul 29 '14 at 11:23
  • But does that mean that, for instance, there is no one-to-one correspondence between prime numbers across models? – Gaussler Jul 29 '14 at 11:38
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    Every model of the natural numbers will have an "initial segment" that consists of what presumably we think of as "the" natural numbers. However we cannot mark out those parts of the model internally (using the first order theory and language). So we can set up a correspondence between (prime) numbers across models, but this is a construction external to the model. – hardmath Jul 29 '14 at 11:51
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    I think the main confusion is the distinction between models of arithmetic and the model of the set theory for which these models of arithmetic are elements of. There are no isomorphism in models of arithmetic. The isomorphism is a part of the bigger universe that contains your models. This bigger universe is some model of set theory (ZFC). Asaf addresses the integer in different models of set theory. – William Jul 29 '14 at 11:55
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    It is a puzzling state of affairs. Going in the other direction, take note of Skolem's paradox, things that ought to be uncountable (e.g. the real numbers) can have countable models, so far as first-order theories are concerned. – hardmath Jul 29 '14 at 11:56
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    @hardmath I like to think of the Skolem's paradox this way: There is a specific set called $\mathbb{R}$. The Cantor theorem states that this specific set $\mathbb{R}$ is uncountable. Nowhere in the proof of Cantor theorem are the algebraic properties of $+$, $\cdot$, $0$, or $1$, etc. fundamentally used. Hence there should be no reason why the first order theory of $\mathbb{R}$ in the language $+$, $\cdot$, $0$, etc should imply uncountability. – William Jul 29 '14 at 12:04
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    @William, I'm not sure I agree that we're not using them, but I cannot say I'm a specialist. But Cantor's argument uses the decimal expansion of the real numbers. The fact that the real numbers have such an expansion can be shown in two ways: Either we define them as decimal expansions, and then it is obvious that $\Bbb R$ is uncountable, independently of its algebraic properties. Or we prove that elements of $\Bbb R$ have decimal expansions, which are essentially power series and thus rely on many algebraic operations. – Gaussler Jul 29 '14 at 12:19
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    @Gaussler Power series is not algebraic. It deals with convergence which is a topological property. The Cantor theorem relates to topology (perfect sets are uncountable). Topological arguments are not first order: they are about collection of subsets of the reals. – William Jul 29 '14 at 12:23
  • @William, I'm not hardmath :-P . But power series are series of sums, so I would say we use $+$ quite a bit. But you may be right that, in the generality you state it (perfect sets are uncountable), there may be a workaround. I just don't quite see it yet. The topology on $\Bbb R$ requires some $+$ and $-$ to be defined. – Gaussler Jul 29 '14 at 12:28
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    And by saying that "Topological arguments are not first order", do you mean that it shouldn't be possible to formalise them in first-order logic? Because I don't quite see why not. Collections of subsets of the reals are elements of $\mathscr P(\mathscr P(\Bbb R))$, which are quite definable within first-order ZFC. – Gaussler Jul 29 '14 at 12:33
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    @Gaussler This is exactly the point was making above. You need to distinguish what language, theory, and model you are in. $\mathbb{R}$ is your model. You can not talk about the topology of $\mathbb{R}$ while still remaining in $\mathbb{R}$ because the objects of a topology are not real numbers but subsets of the real numbers. Hence you need to step outside of $\mathbb{R}$ to some bigger universe containing $\mathbb{R}$, i.e. your model of set theory. You can not talk about the topology using the first order theory of $\mathbb{R}$, you need to use another theory, perhaps $\text{ZFC}$. – William Jul 29 '14 at 12:45

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When we think about models of $\sf ZFC$ we like to think about models which agree with the universe with the very basic things, in particular we expect that $M\models x\in y$ means that the set $x$ is really an element of $y$. And if that happens, then we can easily show that in this case, there is an isomorphism between the integers, and the integers of the model.

But those models are well-founded, they are nice, they are pretty. Not all models are pretty. Many models are intangible, with structure we cannot fathom. The following construction is purely model theoretic.

Suppose that $M$ is a model of $\sf ZFC$, nice or not. We don't care. Pick $X$ to be some set, and extend the language of set theory by adding a constant symbol $c_x$ for each $x\in X$. Now add the following sets of axioms:

  1. If $x\neq y$ add the axiom $c_x\neq c_y$.
  2. $c_x$ is a finite ordinal.

Next note that this theory is consistent by a compactness argument. Any finitely many axioms are satisfied in $M$ by interpreting the constants $c_x$ which appear in these axioms as an approproate finite collection of integers of $M$.

And since the theory is consistent, it has a model $M'$ which is of course a model of $\sf ZFC$. But now each $c_x$ is a different integer. So we have that $|X|\leq|\{m\in M'\mid M'\models m\text{ is an integer}\}|$.

Asaf Karagila
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  • So when such monstrous models for the integers exist, don't we risk that much of the most basic mathematics is actually wrong? For instance, I woundn't be surprised if much of elementary number theory is wrong in that model and hence not derivable from ZFC. – Gaussler Jul 29 '14 at 14:45
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    No, because internally the model can't tell that there are uncountably many integers. Just like a countable model of $\sf ZFC$ can't tell that it only has countably many real numbers. It's one of the delicate points about internal and external properties of the model. – Asaf Karagila Jul 29 '14 at 14:51
  • You're obviously right. My concern is rather that many of the arguments used in mathematics rely on some intuition that we think is easy to formalise if we have to. But much of that intuition could fail in that model. But you proved me wrong several times already, and I am confident that you can do it again. :-) – Gaussler Jul 29 '14 at 15:06
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    But that's the kicker. Our intuition, or at least what we would like to think is our intuition, generally stays true internally to the model. When we start comparing different universes of set theory, sure they would behave the same. Do all groups behave the same? Do all fields? It's just harder for us to think about it that way, because we are not used to working with models of set theory. – Asaf Karagila Jul 29 '14 at 18:44
  • If you start with $(V, \in)$ as your model of ZFC, then the model will only prove true statements. So our intuition will still hold just fine. – Christopher King May 31 '18 at 13:41
  • @PyRulez: Interesting! How do you prove that V is a set? – Asaf Karagila May 31 '18 at 13:43
  • @Gaussler and Asaf Karagila Er, okay its technically a class, not a set. You can do model theory with class sized models though. I'm assuming that we are working in something like Kelly morse set theory, where we can work with classes. – Christopher King May 31 '18 at 13:46
  • @PyRulez: You can also assume that Feferman's theory is in play, and there is a countable transitive set which is elementarily equivalent to the universe (this is given by a schema, of course, but the key point is that there is no increase in consistency). And start with that model as your underlying model. – Asaf Karagila May 31 '18 at 13:48
  • @AsafKaragila That works as well. – Christopher King May 31 '18 at 13:53
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I believe you are referring to the fifth paragraph of Asaf's answer:

I think what Asaf meant is that if $V$ is a model of $\text{ZFC}$. Then there is another model $W$ of $\text{ZFC}$ and a set $x \in V$ such that $W$ thinks $x$ is countable but $V$ thinks $x$ is not countable.

For example if $G \subseteq Coll(\omega, \omega_1^V)$ is generic, then $V[G]$ will think that $\omega_1^V$ is countable but by definition $V$ thinks $\omega_1^V$ is uncountable.

Asaf's sentence "if there is any model of $\mathsf{ZFC}$ then there is such M such that {x∣ $M \models$ is an integer} is an uncountable set, as large as you'd like it to be." is slightly unclear to me. It seems that any reasonable definition of $\mathbb{Z}$ defined from $\omega$ should be absolute. Hence I believe $\{x : M \models x \in \mathbb{Z}\}$ should be the same set as $\mathbb{Z}$ in the ground model which would indeed be countable in the ground model. However, perhaps Asaf's M may not be some transitive extension of the ground model.

Hanul Jeon
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William
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    "It seems that any reasonable definition of ${\bf Z}$ defined from $\omega$ should be absolute." -- only as absolute as $\omega$ is. Which is not absolute at all if you don't assume that the model is transitive. If you do assume that the model is transitive, then of course ${\bf Z}$ is as fixed as $\omega$. In a monster model of ZFC, $\omega$ would have inaccessible cardinality (though only externally). ;-) – tomasz Jul 29 '14 at 13:31
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    @tomasz I suspected that his model was not a transitive extension of the ground model. At the moment, I can not think of any construction where a large set in the original model becomes the naturals in some other model. Do you have an example? – William Jul 29 '14 at 13:43
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    It is a standard construction: if you have an infinite definable set, you can extend the model to have it grow arbitrarily large (by compactness). In a saturated model (of an arbitrary first order theory) every definable set is either finite or has the same cardinality as the entire model. – tomasz Jul 29 '14 at 14:20
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    I think that Asaf meant specifically to talk about non-standard models. I'm not sure, though. I'll have to send him an email and ask. – Asaf Karagila Jul 29 '14 at 14:52