Uncountable sets in countable models of ZFC

Question

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!

Answer 1

The contradiction is inside the definition of "countable". A set is countable if there exists a surjection from $\mathbb{N}$ to our set. The function that would make our inside-the-model set countable doesn't exist inside of the model, so inside of the model, the set is uncountable.

Answer 2

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.

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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!