# Definition of nonstandard $\omega$

I would like to know what is THE definition of a "nonstandard $\omega$", or to be exact of "the model $M$ (of set-theory) has a nonstandard $\omega$" (if there is one agreed definition). Is it one of the following statements?

1. $\omega^M$ is ill-founded.
2. There is $x\in M$ such that $M\vDash x\in \omega$ and for every $n\in \omega$, $M\vDash \mathbf{n}<x$ (where $\mathbf{n}$ is the term $1+\dots+1$ $n$ times).
3. $\omega^M\ne\omega$.
4. $\omega^M$ is not order isomorphic to $\omega$. [added following Andres's comment]

I'm quite sure 1 and 2 are equivalent, and I think that also 3. If that is so, can I give either one as the definition?

• That $M$ has a non-standard $\omega$ means that $\omega^M$ is not order-isomorphic to $\omega$. This is really what you should write in 3 rather than simply set-theoretic inequality. This is indeed equivalent to 1 and 2. Commented May 26, 2017 at 13:33
• And this is the "standard" definition? (pun intended) Should I edit the question? Commented May 26, 2017 at 13:38
• Yes, that is the standard definition. In the new version you have, 3 is simply not equivalent to any of the other statements. Set-theoretic equality is too strong a requirement. For instance, start with your favorite model $(M,\in_M)$ of set theory and now consider $(M',\in')$, where $M'=\{(a,0)\mid a\in M\}$ and $(a,0)\in'(b,0)$ iff $a\in_M b$. Clearly, $M$ and $M'$ are the same thing, that they are different as sets is just noise. In particular, if $\omega^M$ is standard, then $\omega^{M'}$ is standard as well, although by design $\omega\ne\omega^{M'}$. Commented May 26, 2017 at 14:15
• You can always collapse whatever part of $M$ is well founded to be transitive. Under this assumption 3 joins the equivalence. (Also, didn't we cover this four years ago? Although that was in the context of PA...) Commented May 26, 2017 at 15:08
• yes, I see. I'm leaving 3 as it is to emphasize these issues for anyone who might also be looking at this question. @Andrés Would you like to turn this into an answer? Commented May 27, 2017 at 5:57

Fix a model $M$ of a theory for which it makes sense to talk about $\omega$ ($M$ does not need to be a model of set theory, it could even be simply an ordered set with a minimum in which every element has an immediate successor and every element other than the minimum has an immediate predecessor; in this case we could identify $\omega^M$ with $M$ itself).
We say that $\omega^M$ is non-standard if and only if (by definition) $\omega^M$ is not order-isomorphic to $\omega$.
By induction (in the ambient theory) there is a unique order embedding from (true) $\omega$ onto a unique initial segment of $\omega^M$. That $\omega^M$ is non-standard thus means precisely that this embedding is not surjective. If $k$ is an element of $\omega^M$ not in its range, then its immediate predecessor is also not in the range, and it follows that $\omega^M$ is ill-founded. (Conversely, if $\omega^M$ is ill-founded, obviously it is not order isomorphic to $\omega$.)
Identifying the elements of the range with their preimages, this gives us your condition 2 as well (without needing to have a $+$ operation in the structure; naturally, if your $M$ is such that we can discuss the $+$ of $\omega^M$ and expect it to satisfy basic first-order properties of the $+$ of true $\omega$, then the image of any $n\in\omega$ under the unique embedding we are discussing is precisely the ${\mathbf n}$ of your formulation of condition 2).
On the other hand, condition 3 (that $\omega^M\ne\omega$) is too strong a requirement, since usually we only care about structures up to isomorphism. It is true, as Asaf pointed out in a comment, that in some (but definitely not all) situations we may want to identify the well-founded part of a model with its transitive collapse (meaning, given $M$ we effectively replace it with an isomorphic $M'$ obtained by replacing the well-founded part of $M$ as indicated, and then proceed to discuss $M'$ exclusively rather than $M$). If this is the case, then yes, of course condition 3 is also equivalent to the other conditions.
But note that even in the context of set theory, even when considering well-founded models, there are natural circumstances when we want to discuss models that are not transitive. For example, it is common in infinitary combinatorics to argue about elementary substructures $M$ of (some initial segment of) the universe. These structures are automatically well-founded, but you do not want to replace them with their transitive counterparts. (It is also true, however, than in this specific example $\omega^M=\omega$, but I hope the point is clear.)