# Saturated extensions of ZFC models

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?

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If $a$ is a natural number of $M$, it is also a natural number of $N$, but not vice versa, so being a natural number is not absolute, only upwards absolute. What you are adding is a new natural number. Of course, $M$ doesn't see it. Similarly, "finite" is not absolute. There are finite objects in $N$ that $M$ has no way of seeing. If $a$ is in $M$ and $N$ thinks $a$ is finite, then of course it is finite in $M$, but that is weaker than saying that "finite" is absolute. –  Andres Caicedo Mar 9 '11 at 20:53
@Sumac: Your belief that "natural numbers and $\omega$ are absolute" would be correct if you were dealing with transitive models or, more generally, if $M$ were transitive from the point of view of $N$. But in the situation you consider, there is no transitivity available. –  Andreas Blass Mar 9 '11 at 21:14
@Andreas Blass, Andres Caicedo : Thank you very much for your fast response. Indeed I forgot to notice that since $N$ won't be transitive (even if $M$ is) then most of the concepts like "finite" etc won't be absolute. Just to be sure I understand though... If we had extended a countable transitive model of ZFC $M$ to some $M[G]$ through forcing, since $M[G]$ is also c.t.m. of ZFC, we wouldn't be able to add any natural numbers in $M[G]$. –  Sumac Mar 9 '11 at 22:12
@Sumac: Yes, that's correct; in this case we have absoluteness of "finite" and "natural number". But the extension $M[G]$ is not countably saturated, and (unless $M=M[G]$) it is not an elementary extension of $M$. –  Andres Caicedo Mar 9 '11 at 23:00
@Andreas and Sumac: isn't the problem really well foundedness, not transitivity? –  Justin Moore Mar 10 '11 at 1:49

I'm addressing both the question and the comments, but possibly this question should be closed. First let's be clear what we mean by a model of set theory. A model is a set $M$ (or perhaps a class) with a binary relation $E$. $(M,E)$ satisfies Foundation if for every $x$ in $M$, there is an $E$-minimal $y$ in $M$ such that $y E x$. $M$ is well founded if for every $x \subseteq M$, there is an $E$-minimal $y$ in $M$ such that $y$ is in $x$. If $E$ is $\in$, then $M$ is transitive if every element of $M$ is a subset of $M$. Well foundedness is implicit in transitivity.
Furthermore, any well founded model $(M,E)$ of ZFC is isomorphic to a transitive set (or class) with the membership relation. This is the Mostowski collapse at work. So transitivity is not the issue here. The failure of finiteness to be absolute is tied up in the ill foundedness of the model in question.