Uncountable models for integers

Question

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?

Answer 1

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.

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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.

Answer 2

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