# Models having the same standard system

In the paper "A natural model of the multiverse axioms", Gitman and Hamkins give the following construction:

Given a countable computably saturated model of ZFC, $M$, they show that there is an $N\in M$ which satisfies $Th(M)$ and $M$ thinks that $N$ is $\omega$-nonstandard.

Then they claim that "since $\omega^M$ is an initial segment of $\omega^N$ and $M$ is $\omega$-nonstandard, it follows that $M$ and $N$ have the same standard system."

First, I want to make sure I understand why $\omega^M$ is an initial segment of $\omega^N$: since $M$ thinks of itself as standard, and of $N$ as non-standard, $M\vDash\forall n<\omega (n\in\omega^N)$, so by looking from the outside we get $\forall n<\omega^M (n\in\omega^N)$. Is that right?

Second, and more importantly, I don't understand why it follows that they have the same standard system. Is it something immediate that I miss, or a well-known result?

First, I want to make sure I understand why $\omega^M$ is an initial segment of $\omega^N$: since $M$ thinks of itself as standard, and of $N$ as non-standard, $M\vDash\forall n<\omega (n\in\omega^N)$, so by looking from the outside we get $\forall n<\omega^M (n\in\omega^N)$. Is that right?

That's exactly right.

Second, and more importantly, I don't understand why it follows that they have the same standard system. Is it something immediate that I miss, or a well-known result?

Let $n\in \omega^M$ be nonstandard. Then for any $a\in M$, $a\cap \omega=(a\cap n)\cap\omega$, so to determine the standard system of $M$, we can restrict to just subsets of $n$. But $M$ thinks $n$ is finite, so $M$ thinks $N$ must contain all subsets of $n$. So the entire standard system of $M$ must be in the standard system of $N$. Conversely, if $a$ is an element of $N$, then $a\cap\omega^M\in M$ and this has same intersection with the true $\omega$ as $a$, so the standard system of $N$ is contained in the standard system of $M$.

(Here I am abusing notation slightly and identifying $\omega$ with the canonical initial segment of $\omega^M$ which is isomorphic to it and $\omega^M$ with the canonical initial segment of $\omega^N$ which is isomorphic to it. Since the element relation of $M$ is not actually $\in$ and the element relation of $N$ is not actually the element relation of $M$, things like "$a\cap \omega$" or "$a\cap\omega^M$" which I wrote don't make sense literally.)

Actually, more strongly, you can prove that if $M$ and $N$ are arbitrary $\omega$-nonstandard models of ZFC (so we're not assuming $N\in M$) and $\omega^M$ is isomorphic to an initial segment of $\omega^N$ over the language of arithmetic, then $M$ and $N$ have the same standard systems. To prove this, let $m\in\omega^M$ be any nonstandard element. Note that for any $a\in M$, $M$ thinks $a\cap m$ is a finite set. It follows that $M$ can encode the set $a\cap m$ with an element of $\omega^M$, by the usual method of encoding finite sets of natural numbers in arithmetic. Identifying $\omega^M$ with its isomorphic initial segment of $\omega^N$, this element will also encode the set $a\cap m$ in $\omega^N$, so $a\cap m$ must also be a set in $N$. It follows that the standard system of $M$ is contained in the standard system of $N$. The converse can be proved the same way, since the natural number encoding a subset of $m$ in $N$ will still be an element of the initial segment $\omega^M$.

• It seems that you are using only the fact that $M$ is nonstandard and thinks that $N$ contains at least all of it's $\omega$, so it would work also if $N$ is not thought by $M$ to be nonstandard? May 25, 2017 at 20:02
• Yes, this would work even if $M$ thinks $N$ is standard. May 25, 2017 at 20:08
• I think you're confused about what everything means here: $a$ is an element of $M$. When we refer to "elements" of $N$, we mean elements $a\in M$ such that $a\in^M N$. When we think of $N$ as a structure in the language of set theory, the underlying set of this structure is really $\{a\in M:a\in^M N\}$, not the literal set $N$ itself. The literal set $N$ could be any set at all, since $\in^M$ may be totally unrelated to the true $\in$ relation. May 25, 2017 at 21:19
• A-ha, I see. Indeed I did not understand it this way, thanks. I see now that this is hidden in the way they identify $N$ - it is constructed using the completeness theorem in $M$, so indeed everything there is an element of $M$. May 25, 2017 at 21:27
• Hmmm, that's true. I've added another argument that uses only that fact. May 27, 2017 at 17:21