# Uncountable models for integers

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?

• 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. Commented Jul 29, 2014 at 11:23
• 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. Commented Jul 29, 2014 at 11:51
• 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. Commented Jul 29, 2014 at 11:55
• 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. Commented Jul 29, 2014 at 11:56
• @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 $\text{ZFC}$. Commented Jul 29, 2014 at 12:45

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 $|X|\leq|\{m\in M'\mid M'\models m\text{ is an integer}\}|$.

• 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. Commented Jul 29, 2014 at 14:45
• 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. Commented Jul 29, 2014 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. :-) Commented Jul 29, 2014 at 15:06
• 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. Commented Jul 29, 2014 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. Commented May 31, 2018 at 13:41

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.

• "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). ;-) Commented Jul 29, 2014 at 13:31
• @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? Commented Jul 29, 2014 at 13:43
• 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. Commented Jul 29, 2014 at 14:20
• 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. Commented Jul 29, 2014 at 14:52