Clarifying "$\omega$-models" of ZF(C)

Question

Preliminaries: To be extra clear, I'm assuming (hopefully correctly) the following:

• The "finite von Neumann ordinals" always means the sets $\emptyset , \;\; \{\emptyset\} , \;\; \{\emptyset , \{\emptyset\}\} , \;\; \{ \emptyset , \{\emptyset\} , \{\emptyset , \{\emptyset\}\}\} \ldots \;$ In other words: $\emptyset , \; S(\emptyset), \; S(S(\emptyset)), \; S(S(S(\emptyset))) \ldots$, where $S$ is the successor function $S(x) := x \cup \{x\}$
• The Axiom of Infinity always guarantees the existence of an inductive set, i.e. a set $I$ such that $\emptyset \in I$ and $x \in I \implies (x \cup \{x\}) \in I$

My question is what exactly is meant by an "$\omega$-model" or "$\omega$-standard model"? What I think I can infer (with less than 100% confidence) from what I've read is:

• An "$\omega$-model", or "$\omega$-standard model", is a model in which there exists a "smallest" inductive set (the "$\omega$ of the model") whose elements are exactly the finite von Neumann ordinals (all the finite von Neumann ordinals but nothing else). That is, the $\omega$ of the model is the "standard" $\omega$.
• Thus also a non-$\omega$-model is a model in which every inductive set contains some other elements in addition to the finite von Neumann ordinals. So for such a model, the "smallest" inductive set is some "non-standard $\omega$".

Is this right?

Answer 1

The finite ordinals aren't defined as in your first bullet because that's not a definition, and even if you try to make it into one, it is circular. There are lots of different ways one can define it... one nice way that doesn't use the axiom of infinity is that a finite ordinal is an ordinal (i.e. a transitive set that is well-ordered by the $\in$ relation) whose reverse ordering is a also well-order. But more often, we take the axiom of infinity for granted and just define a finite ordinal as a member of $\omega$, where $\omega$ is defined as the minimal inductive set, or perhaps as the least limit ordinal. Since here we're concerned mostly with what the set of natural numbers is, as opposed to what a natural number is, let's just take a typical route and define $\omega$ as the minimal inductive set and define a natural number to be any element of $\omega.$

This is a definition in ZF, so any model $(M,E)$ of ZF, there is a set $\omega^M\in M$ such that $(M,E)\models \mbox{"$\omega^M$ is the minimal inductive set."}$ Moreover, $(M,E)\models \mbox{"$E$ is a well-ordering on $\omega^M$"},$ since it's provable in ZF that $\in$ is a well-ordering on $\omega.$ $(M,E)$ is an omega-model if and only if any of the following equivalent statements hold

1. $E$ actually is a well-ordering on the set of $E$-elements of $\omega^M.$
2. The relation $E$ on the set of $E$-elements of $\omega^M$ is isomorphic to $(\omega,\in).$
3. The $E$-elements of $\omega^M$ are exactly $\{(S^M)^n0^M : n\in \omega\},$ where $0^M$ is $(M,E)$'s empty set and $S^M$ $(M,E)$'s ordinal successor function. The sets of the form $(S^M)^n0^M$ are called the model's standard natural numbers.

What happens in a non-omega model is that the model's standard natural numbers are followed by a bunch of other things the model thinks are natural numbers, and they are ordered like a bunch of copies of $\mathbb Z.$ As you indicate in your last bullet, the model still satisfies that $\omega^M$ is the minimal inductive set... it's just simply the case that all of the model's inductive sets contain more than just the model's standard naturals (if you "read the fine print", there's no guarantee otherwise).

On a side note, I say "the model's standard naturals", but one should realize that unlike everything else I attribute to the model, this is not something we can define within the model... it is an external definition. And this has to be the case, since otherwise, we could just use this as the definition of the natural numbers and every model would be an $\omega$-model.

Answer 2

I believe that you mean the correct thing, but you need to be a bit careful with this.

Given a model $\mathcal M=(M,E)$ of ZF(C), we always have $\mathcal M\models"\omega\text{ exists and it contains exactly the finite ordinals}"$ (where $\omega$ is defined as the intersection of all inductive sets), the problem is that we may have $x\in M$ such that $"x\text{ is a finite ordinal}"$ but $\mathcal M\not\models"x\text{ is a finite ordinal}"$ or maybe $"x\text{ is not a finite ordinal}"$ but $\mathcal M\models"x\text{ is a finite ordinal}"$.

Additionally, we may even have $x\notin M$ such that $"x \text{ is finite ordinal}"$, so $\cal M$ doesn't even know $x$ exists.

Lastly you can have a weird situation where $"x,y \text{ is finite ordinal and }x<y"$ but $\mathcal M\models "x,y \text{ is finite ordinal and }x>y"$

A model if not an $\omega$-model if one of the above is true for your model.

$\mathcal N=(N,E)$ is an $\omega$-model exactly when the above doesn't happen, that is, the $\omega$ of $\mathcal N$ is the real $\omega$, and the finite ordinals of $\mathcal N$ are exactly the real finite ordinals and they are ordered like the real finite ordinals.