TITLE: Uncountable sets in countable models of ZFC
QUESTION [13 upvotes]: If we assume ZFC to be consistent we have, by the Löwenheim-Skolem theorem, the existence of a countable model $\mathcal{U}_0$ of ZFC.
In $\mathcal{U}_0$ there is a infinite ordinal, that is a non-empty limit ordinal. Call the smallest one $\omega$. We can also construct the cardinal $2^\omega := \mathrm{card}(\wp (\omega))$, since the existence of the power set is given by the axioms.
However, the latter is uncountable, but it is a subset of $\mathcal{U}_0$, which is countable; this seems to be a contradiction.
I suspect that this "contradiction" can be resolved by distinguishing between infinity between models of ZFC, but I don't know how to do that.
So my question is: How can I resolve this?
Thanks!

REPLY [7 votes]: The issue is not that $2^\omega$ is uncountable. It is that the set $\{A\mid\mathcal U_0\models A\subseteq\omega\}$ is countable. The fact that $\mathcal U_0$ is a model of $\sf ZFC$ means that in $\mathcal U_0$ there is a object which represents this set; but also that there is no bijection between the object $\mathcal U_0$ "thinks" is $\omega$, and the object representing the set above.
So $\mathcal U_0$ "thinks" there is no bijection between some object and $\omega$, which is exactly the definition for $\mathcal U_0$ "thinks" that some object is uncountable.

The reverse thing is also possible, if there is a model of $\sf ZFC$, then there is one $\mathcal U_1$ such that the set $\{A\mid\mathcal U_1\models A\text{ is a finite ordinal}\}$ is uncountable. So $\omega$, or the set that $\mathcal U_1$ "thinks" is $\omega$—the epitome of countability—is in fact uncountable!