# Explanation of how models can differ on $\omega$?

Assuming set theory (here, ZF) is consistent, there is a model $V$ of ZF, the universe of all sets.

So, there is a $\omega^V\in V$. A set $A\in V$ is countable iff a bijection $f\in V$ exists between $A$ and $\omega^V$.

By the downward Löwenheim–Skolem Theorem, there is a countable model $M$ of ZFC such that the domain M of $M$ is in $V$ and a bijection $B\in V$ exists between M and $\omega^V$.

Since $M$ is a model of ZFC, there must be some $\omega^M\in M$ and some $\omega_1^M\in M$.

There is no guarantee that $\omega^M$ is $\omega^V$, and no guarantee that $\omega_1^M$ is $\omega_1^V$.

My questions:

1. Can someone explain when/how/why/under what conditions this divergence (between models taking different sets to be $\omega$) happens?
2. When/how/why/under what conditions is there a guarantee that sets will agree on $\omega$?
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Since you seem to keep thinking (and asking) about ZFC, the following critique against using ZFC as foundation for all of mathematics might be interesting to you: math.wustl.edu/~nweaver/indisp.pdf –  Thomas Klimpel Jun 10 '12 at 23:22

To the first question, consider $M$ to be a model of ZFC which is a set in the universe $V$. Suppose that $M$ and $V$ agree on $\omega$, we can take an ultrapower of $M$ by a free ultrafilter on the real $\omega$. This happens in $V$, and externally to $M$.
The result is a model, $N$, which externally speaking has a different version of $\omega$. Namely, $\omega^N$ is the ultraproduct of $\omega^M$ by a free ultrafilter over a countable set, so it is not well-founded and therefore cannot be the real $\omega$.
On the other hand, we can require that the models we work with are $\omega$-models, namely their copy of $\omega$ is isomorphic (from the external, and real point of view) to the real $\omega$. In which case even if the model does not fully agree on how $\omega$ looks like, its agrees on its behavior (again, from an external point of view). If we have a transitive $\omega$-model then as sets $\omega^M$ and the real $\omega$ of $V$ are the same set.
Yet another summary. But first! Your comment here made me laugh out loud for at least 5 minutes. Now to the dry stuff...if $V$ saw $M$ as countable, and $M$ was a standard model of ZFC, would $V$ see every $A\in M$ as at most countable? –  pichael Jun 13 '12 at 8:02
@pichael: No. Suppose we begin with some very very large standard model which knows about $\omega_1$ as a set, etc. Now take a countable elementary submodel which also knows about $\omega_1$ - this is the Skolem-Lowenheim theorem. This model is countable, its membership is the real one because it is a submodel of a standard model; however there is a set in this countable model which the universe knows is uncountable. Note that if we also assume that $M$ is transitive then this is true, as $A\in M$ would imply that $A\subseteq M$. So we have that every set in $M$ is a subset of a countable set. –  Asaf Karagila Jun 13 '12 at 8:13
OK, it's getting clearer. Wiki doesn't help much. So, if $M$ is transitive, it is standard, but not vice versa. Also, if $M$ is a standard model of ZFC and $A$ is uncountable in $M$ then $A$ is uncountable in $V$? –  pichael Jun 13 '12 at 8:22
@pichael: For the first, yes. For the latter no, if $M$ is a countable transitive model of ZFC then every ordinal in $M$ is an ordinal in $V$ and there are only countably many ordinals in $M$. However there is a countable ordinal in $M$ which $M$ thinks is uncountable. –  Asaf Karagila Jun 13 '12 at 8:36
OK, I'll just ask the bigger question: if $V$ sees $M$ as countable and $A\in M$ is uncountable in $M$, what can we say about $A\in V$ -- that is, from $V$'s perspective on $A$? And are there any conditions under which $M$ and $V$ will agree about $A$ in this scenario? –  pichael Jun 13 '12 at 8:45