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I fail to argue a seemingly trivial thing.

Consider a countable model of ZF/ZFC inside ZF/ZFC and the sets $\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}, ...$ The "outer" and "inner" ZFC agree that those sets are von Neumann ordinals. Now, shame on me, I cannot find an argument why e.g. $\{\emptyset, \{\emptyset\}\}$ cannot be the $\omega$ of that model. The axiom of infinity does not help. It only guarantees that every model is actually infinite (as seen from the outer ZFC), but the axiom does not state that the smallest set with that property is to be called $\omega$. Then dedekind infinity came to my mind. And of course this little $\omega$ is not dedekind infinite. But how can we show that the axioms require $\omega$ to be dedekind infinite (or infinite at all)? I must be blind here.

What is the minimal order type of ordinals in countable models of ZFC? What is the minimal order type such zhat there exists not countable model of ZFC with that order type for its ordinals?

Thank you.

EDIT: It was unclear which membership relation I have considered for the model. However, lets not talk about membership, let us talk about a directed graph $(U, \rightsquigarrow)$ that is a countable model. U is a collection of vertices that express the sets. There is only one vertex that has no ingoing edges. This vertex is the empty set, and that is visible externally. It has outgoing edges to many many other vertices. But there is only one target vertex that has this incoming edge, and no other incoming edges. The natural reflex is to label this vertex the set of the von Neumann ordinal 1, but why cannot we label it $\omega$? It is not in bijection with the empty-set-vertex, and is also the "union" of all smaller vertices. So how to we see externally how to label this vertex?

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2 Answers 2

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The axiom of infinity implies that $$(\forall x)(x\in\omega \rightarrow x\cup\{x\}\in\omega).$$

Since $\{\emptyset\}\in\omega,$ we must have $\{\emptyset\}\cup \{\{\emptyset\}\}\in\omega.$ This set is equal to $\{\emptyset,\{\emptyset\}\}.$

So we know that we have $\{\emptyset,\{\emptyset\}\}\in \omega.$

But if we let $s$ be your set $\{\emptyset, \{\emptyset\}\},$ then $\{\emptyset,\{\emptyset\}\}\not\in s.$ (This is because the only two elements of $s$ are $\emptyset$ and $\{\emptyset\},$ but $\{\emptyset, \{\emptyset\}\}$ is neither of these. If you prefer, you can just use the axiom of foundation to see that $s\not \in s.)$

It follows that $s\ne\omega.$

I'm not sure exactly how related this part of your question is to the first part, but the minimal order type of the ordinals of a well-founded model of ZFC is a quite large countable ordinal. If you look at non-well-founded models of ZFC, then it's not necessarily even clear what "minimal order type" means, because the ordering of the model's ordinals isn't (in the real world) a well-ordering.

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    $\begingroup$ In my opinion this is a misinterpretation of the axiom. It says that there exists a set that has this "inductive" property. We call this set $\omega$, but ZFC has not even a symbol to express that the set with that property is "our" $\omega$. $\endgroup$
    – Damian
    Commented Nov 22, 2016 at 22:57
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    $\begingroup$ @Damian The axiom of infinity says that there is a set with the inductive property. We can prove that there is a minimal such set. Any sentence involving the symbol $\omega$ can be viewed as an abbreviation for a formula of the form $(\forall x)(\text{if }x\text{ is a minimal inductive set, then }\dots ).$ Or, if you prefer, you can use $(\exists x)(x\text{ is a minimal inductive set and }\dots ).$ $\endgroup$
    – Mitchell Spector
    Commented Nov 22, 2016 at 23:10
  • $\begingroup$ I don't deny the existence of a set that actually/externally has infinitely many members, and that you can define that set. But I don't see why a finite model must agree that this is its $\omega$. I also thought that a finite model must have an external well-order for its ordinals. This order type would be larger or smaller, depending on how large the distance between an $\omega_n$-label and the $\omega_{n+1}$-label is (See my EDIT why I'm talking about labels). $\endgroup$
    – Damian
    Commented Nov 22, 2016 at 23:46
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    $\begingroup$ @Damian There is no finite model of ZFC . . . $\endgroup$
    – Noah Schweber
    Commented Nov 22, 2016 at 23:55
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    $\begingroup$ Even a countable model may have non-standard ordinals, even non-standard natural numbers, in which case these orderings that the model thinks are well-ordrerings aren't really well-orderings. (There would then be a non-empty subset A of the ordering without a least element, but no such A would belong to the model.) $\endgroup$
    – Mitchell Spector
    Commented Nov 23, 2016 at 0:16
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One of the things ZFC proves about $\omega$ is: $$\forall x\in\omega(\exists y(x\in y, y\in \omega)).$$ This is not true of the ordinal $1$: there is no element of $1$ which contains $0$ as an element.

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