Hi, I would like some help on something that in some ways has already been touched in the past here. I saw these relevant questions being answered, but I am still unable to understand some things. I would be grateful if someone could help me.

Let $M \models$ ZFC that is countable and let $p(x) = {0 \in x, \ldots, n \in x, \ldots} \cup {x \in \omega }$ be a type where $0, \ldots, n, \dots$ are the finite ordinals of $M$, and $\omega$ the first limit ordinal of $M$. Now let $N$ be an $\omega_1$-saturated model s.t. $M \prec N$. Now $p(x)$ obviously is fin. satisfiable in $M$, therefore it is satisfiable in $N$, by let's say some set $a$. Both $M$ and $N$ are ZFC models, therefore natural numbers and $\omega$ are absolute for $M, N$. So how is it possible to have an ordinal $a$ (since $a \in \omega$) which is finite (in $N$ and therefore in $M$) but different from all $0, \dots, n, \ldots$?

In the same way, given any countable set $b$ of finite sets of $M$, I can always find a saturated model $N$ s.t. $M \prec N$, with some $c \in_N b$ and $c$ infinite. Is this correct?

1more comment