# Closure of models of ZFC under sequences of ordinals

Assume that $M\subseteq N$ are models of ZFC, and, in $N$, $\kappa$ is a regular uncountable cardinal. I'm trying to prove that $N\vDash \ ^{<\kappa} On \subseteq M$ implies $N\vDash ^{<\kappa} M \subseteq M$.

As I understand, the way is to use an encoding by a set of ordinals. This can be done in the usual way as follows: Given $f\colon \kappa \to M$, $f\in N$, take $G$ to be Godel's pairing function. Assume that (In N), $\$ $h\colon \mbox{tc}(\{f\})\to \mu$ is a bijection, for some cardinal $\mu$. Then define --

$A = \{ G( h(a),h(b) ) \colon a\in b \in \mbox{tc}\left( \{ f \} \right) \}$

Now, $A\in N$ is a set of ordinals, and if $A\in M$ then indeed $f\in M$. But $A$ is not necessarily of cardinality $< \kappa$ (If it was, we would be done: $A\in M$ since $N\vDash \ ^{<\kappa} On \subseteq N$). So how can I proceed from here? (The method above proves only that $N\vDash \ ^{<\kappa} H_\kappa \subseteq N$).

Is this the correct direction to dealing with this problem?

I assume $M,N$ are transitive.
Given $f \in {^{<\kappa} M} \cap N$ we can define $g \in {^{<\kappa} On} \cap N$ so that $g(x) = \operatorname{rank}(f(x))$. By assumption $g \in M$. So let $\alpha = \sup \operatorname{im}(g) +1$ and find in $M$ a bijection $G$ from $V_\alpha$ to some cardinal $\lambda$. Then, as $G \circ f \in {^{<\kappa} On} \cap N$, $G \circ f \in M$ but then also $f \in M$ as $G \in M$.