Another question about Cohen's article

Question

This is another question related to Cohen's Discovery of Forcing:

Consider the following passage from p. 1091:

Suppose $M$ were a countable model. Up until now we have not discussed the role countability might play. This means that all the sets of $M$ are countable, although the enumeration of some sets of $M$ does not exist in $M$. The simplest example would be the uncountable ordinals in $M$. These of course are actually countable ordinals, and hence there is an ordinal $I$, not in $M$, which is countable, and which is larger than all the ordinals of $M$.

What I understand: If $M$ is countable it cannot contain all countable ordinals since the set of all countable ordinals, $\omega_1$, is uncountable. Hence there exists at least one countable ordinal $\alpha$ not in $M$.

What I don't understand: Why does it hold that $\alpha$ is bigger than all the countable ordinals in $M$? It seems to me that one could pick countably many countable ordinals $\alpha_n$ so that above any given countable ordinal $\beta$ there is an $\alpha_n$. (I am thinking of the real numbers where the natural numbers would be a countable set such that above any natural number we can find a natural number). But for some reason a construction like this does not work in this case. Why not?

Answer 1

The reason is cofinality. $\omega_1$ is a regular ordinal (at least if we assume choice), so for any countable collection of countable ordinals there is another countable ordinal above all of them.

Answer 2

There is something rather specific to certain countable well-founded models that may help you here, and it appears frequently enough that you may want to be aware of it anyway. To avoid a sea of notation and technicalities, let's in fact look only at models $M$ obtained by, for example, taking a countable elementary substructure of some $V_\alpha$. This is more or less the situation being discussed, since if $M$ is just an arbitrary model, then its "ordinals" do not need to be ordinals in $V$, and do not need to correspond to true ordinals ($M$ could be a non-standard, ill-founded model, for example).

OK, so: $M$ is elementary in $V_\alpha$ for some appropriately large $\alpha$. We want $M$ to be a model of $\mathsf{ZF}$, or at least to "look a lot" like a model of $\mathsf{ZF}$, and the simplest way of arranging this is to let $\alpha$ be gigantic (much larger than $\omega_1$, than $\omega_2$, etc).

Now: Suppose that $X\in M$ and $M\models$"$X$ is countable". Then, by elementarity, $V_\alpha\models$"$X$ is countable", so $X$ truly is countable, and in $V_\alpha$ we can find a surjection $f:\omega\to X$. By elementarity, in $X$ there is such a function as well, call it $g$. So $M\models$"$g$ is a function, $g:\omega\to X$, $g$ is surjective".

By elementarity again, $g$ is truly a surjective function $g:\omega\to X$. Now, the point is that $\omega$ and all finite ordinals are definable, so they all belong to $M$ (in fact, elementarity grants us that anything in $V_\alpha$ that is explicitly definable belongs to $M$). So, for each $n<\omega$, $n\in M$, so $M\models$"$n\in\mathrm{dom}(g)$", and whatever $M$ thinks $g(n)$ is, it is truly $g(n)$ in $V_\alpha$ (and $V$). But this means that $X=\{g(n)\mid n<\omega\}$ is not only an element of $M$, it is in fact a subset of $M$.

Notice this is specific to countable objects. For example, the reals of $M$ are a subset of the reals of $V$, but many reals are not in $M$, simply on cardinality grounds. Anyway, we have shown that, in our setting, if $X$ is countable, and $X$ belongs to $M$, then $X$ is in fact a subset of $M$. In particular, this applies to any countable ordinal that belongs to $M$. This means that the first ordinal not in $M$ is countable, as only countably many ordinals belong to $M$, and the countable ordinals in $M$ are closed under "initial segments".

On the other hand, $\omega_1\in M$, but there are many countable ordinals not in $M$, and $\omega_2\in M$, but there are many ordinals between $\omega_1$ and $\omega_2$ missing from $M$ as well. Similarly, for any uncountable set $X$ that belongs to $M$, there will be infinitely many elements of $X$ in $M$, but there will also be uncountably many elements of $X$ not in $M$. This suggests that it may be easier to analyze the properties of the model $M$ if we insist that $M$ is transitive, so we do not have to distinguish between "internal" and "external" elements of the elements of $M$ (if $X\in M$, and $y\in X$, then $y$ is internal if $y\in M$, and $y$ is external otherwise).

Naturally, once we decide to take $M$ countable, then $M$ is no longer an elementary substructure of a $V_\alpha$, and there will be countable ordinals in $M$ that $M$ thinks are uncountable. Still, in this case, the first ordinal not in $M$ would be countable, but for a different reason than the one described above.

The second volume of Discovering modern set theory by Just and Weese discusses these elementary substructures in detail. I turns out that (even though we do not typically use them in forcing arguments) elementary substructure arguments are very useful in set theory, precisely because of the interplay between internal and external objects.