What is the meaning of "uncountably infinite" within countable models of set theory?

Question

So, I don't know much about countable models of set theory, other than that they exist. To me, their existence is a very weird thing (and a reason to move away from first-order formulations). Here is one thing that really weirds me out:

We have definitions of infinity, some set is infinite if it can be put into 1-1 correspondence with the natural numbers or a set is infinite if it can be put into 1-1 correspondence with a proper subset of itself (Dedekind infinite, if I recall).

How do these definitions work when we're in a countable model of ZFC? In particular, given that we know that the powerset of a countably infinite set will be uncountably infinite, how are we to understand this given that our model is countable? What does "uncountable" mean in such a context?

Answer 1

Recall that a structure $\mathcal M\models\exists x\varphi(x)$ if and only if there exists some $m\in|\mathcal M|$ such that $\mathcal M\models\varphi(m)$.

Countable is a definable property. $\varphi(a)$ says that $a$ is countable if and only if there exists an injective function whose domain is $a$ and its range includes only "finite ordinals" (transitive sets which are well-ordered by $\in$, and satisfy one of many characterizations of finiteness).

So $\mathcal M\models\varphi(a)$ if and only if there is $f\in|\mathcal M|$ such that $\mathcal M\models f\colon a\to\omega^\mathcal M\text{ is injective}$. If $\mathcal M$ does not know about such $f$, then $\mathcal M$ would think that $a$ is uncountable.

---

The key point is that a model of $\sf ZF$ never knows about sets which lie outside that model. Much like how $\Bbb Q$ does not know about $\pi,e,i$ or $\sqrt 2$ since none of them is a rational number.

Answer 2

Even though $A$ and $B$ are both countable sets in the model, it may be that there is no bijection from $A$ to $B$ which is also in the model.

So "internally" the denizens of that model could say "We have an injection from $A$ to $B$, but no surjection, so $A$ has smaller cardinality", even while external to the model we can see that there is a bijection.

Answer 3

In a countable model, every set of course has an enumeration, but in the case of the powerset of an infinite set no such enumeration is a member of the model! (Sometimes one says there is no "internal" enumeration.)