Can we conclude $rng(f) \subseteq \beta$ for some $\beta \in B \cap \omega_1$?

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Consider the language $\mathcal{L}=\{ \in\}$. Let $\mathcal{A}$ be the $\mathcal{L}$-structure $(V_\theta, \in)$ for some $\theta> \omega$. 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. We know from my previous question that $B \cap \omega_1$ is an ordinal.

Now assume $f \in B$ is a function $f: \omega \to B \cap \omega_1$.

Q1: Does $\mathcal{B}$ correctly "think" $f$ is a function?

Q2: Can we conclude $rng(f) \subseteq \beta$ for some $\beta \in B \cap \omega_1$?

I see $B \cap \omega_1$ must actually be a limit ordinal, so I think the 2nd part will most likely boil down to showing $f$ is bounded.

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Being a function is first-order expressible ("$f$ is a set of ordered pairs such that ..."), and so is absolute between $\mathcal{A}$ and $\mathcal{B}$. Consequently we have:

Whenever $x,y,f\in\mathcal{B}$ with $f:x\rightarrow y$, we have $f[x\cap B]\subseteq B$.

To see this, given $a\in x\cap B$ apply elementarity to the fact - true in $\mathcal{A}$ - that there is an element of $f$ which is an ordered pair with left coordinate $a$.

In particular, this means that if $x\subseteq B$ then $ran(f)\subseteq B$. And this happens, e.g., in the case $x=\omega$ (again by elementarity: each natural number is definable). Now we can again apply elementarity to the statement "$sup(ran(f))<\omega_1$," which is true in $\mathcal{A}$, to get an affirmative answer to your second question.