Countable Elementary Submodels $M \preccurlyeq H ( \theta) $.

Question

I am reading through some of Kunen's material on Elementary Submodels and am a little unclear on one proof. Here is a part of the claim:

Let $ \theta$ be an uncountable cardinal, and let $ M$ be a countable with $ M \preccurlyeq H ( \theta) $. Then if $\theta > \omega_{1}$ : $ \omega_1 \in M$ and $ \omega_1 \not\subseteq M$...

I think that it is clear why $ \omega_1 \not\subseteq M$ ( given that $M$ is countable). I am unsure why $ \omega_1 \in M$. Kunen claims that

$\omega_1 \not\subseteq M$ is clear, and $ \omega_1 \in M$ because $\omega_1$ is definable in $ H (\theta) $ ( and in $V$).

I do think that $ \omega_1$ is pretty clearly definable in $H(\theta)$ but I don't see why that implies $ \omega_1 \in M$.

Answer 1

Quite generally, if $A$ is an elementary submodel of $B$, then every definable element of $B$ is in $A$. To prove it, suppose $b$ is definable in $B$, so it's the only element of $b$ that satisfies (in $B$) a certain formula $\delta(x)$ (with just the one free variable $x$). Now the sentence $(\exists x)\,\delta(x)$ is true in $B$, so it's also true in the elementary submodel $A$. Therefore, $A$ has an element $a$ that satisfies $\delta(x)$ in $A$. A second use of the "elementary submodel" assumption shows that this same element $a$ also satisfies $\delta(x)$ in $B$. But $b$ was the only element satisfying $\delta(x)$ in $B$, so we must have $a=b$. Since $a\in A$, it follows that $b\in A$, as claimed.

Answer 2

Because $M \equiv H(\theta)$, $M$ contains a witness $\kappa$ to "there is a least uncountable cardinal", as that sentence is true in $H(\theta)$. Because $ M \preccurlyeq H(\theta)$, the same is true of $\kappa$ in $H(\theta)$. Because $\omega_1 \in H(\theta)$, $H(\theta)$ knows the real "least uncountable cardinal", so $\kappa = \omega_1$.