Consider the language $\mathcal{L}=\{ \in\}$. Let $\mathcal{A}$ be an $\mathcal{L}$-structure whose domain is some "sufficiently large" Von-Neumann universe and which interprets $\in$ in the canonical way. Let $\mathcal{B}$ be an elementary substructure of $\mathcal{A}$. Let $A,B$ denote the domains of $\mathcal{A}, \mathcal{B}$ respectively.
Assume $B$ is countable and let $X= \{ \beta \in B: \beta$ is a countable ordinal$\}$.
Is $X$ itself an ordinal?
I know there's a theorem which says that a set of ordinals will itself be an ordinal if it is transitive, so it seems we just need to show $X$ is transitive.
Here's my first attempt: Let $\beta \in X$ and $\alpha \in \beta$. We want to show $\alpha \in X$. We know that $\alpha$ is a countable ordinal since $\beta$ is. Hence it remains to show $\alpha \in B$.
Best Answer
Perhaps surprisingly, the answer is yes! This is a beautiful result. I'll rephrase it slightly as follows:
Here's a sketch of the proof:
Since $\beta\in M$, $M\preccurlyeq V_\theta$, and $V_\theta\models$ "$\beta$ is countable" (why?), we have $M\models$ "$\beta$ is countable." That is, there is an $f\in M$ such that $M$ thinks $f$ is a bijection between $\beta$ and what $M$ thinks is $\omega$. Both these notions are appropriately absolute, so in reality - and hence in $V_\theta$ - $f$ is actually a bijection between $\beta$ and $\omega$. Now let $n=f(\alpha)$. A priori $n$ merely lives in $V_\theta$, but since $n\in\omega$ we also have in fact that $n\in M$ (each natural number is definable in $V_\theta$ so exists in $M$ by elementarity). Since $M$ sees $n$ and thinks $f$ is a bijection, $M$ must have something it thinks is $f^{-1}(n)$ - but again by basic absoluteness this must be $\alpha$ itself!
In general the set of ordinals in $M$ need not be downwards-closed - this is a feature special to the countable ordinals, via the fact that each individual natural number is definable in $V_\theta$ whenever $\theta$ is infinite.