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Why doesn´t the von Neumann hierarchy to $V_{\omega_1}$ exist in Zermelo set theory if with Scott´s trick you can "count" to $ \omega_1 $

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To see what goes wrong, let's try to prove something even simpler: that $\omega_1$ exists.

We definitely have the set of all well-orderings of $\omega$ (via Powerset and Separation). Now, via Separation and Choice, we can pick a family $\{R_i: i\in I\}$ of well-orderings of $\omega$ such that each well-ordering of $\omega$ is order-isomorphic with exactly one $R_i$. The usual proof that well-orderings are well-ordered by initial segment embedding goes through, so we can build a well-ordering $W$ of order-type $\omega_1$ by "adding all the $R_i$s together." Of course, this is not an ordinal, it's just a binary relation on a set (specifically, on $\omega\times I$) which is a well-ordering and which has "length $\omega_1$" intuitively.

OK, so we can build a thing of length $\omega_1$; why can't we build $\omega_1$ itself?

Well, what we would want to do is show: "For every well-ordering, there is an ordinal of the same length." (Appropriately formalized.) However, this fact uses Replacement!

Specifically, the usual proof goes roughly as: "Suppose there was a well-ordering $R$ of a set $X$, which is not in order-preserving bijection with any ordinal. We can assume that for each $x\in X$, the set $X_x=\{y\in X: y<_R x\}$ is in order-preserving bijection with some ordinal. (Otherwise, if there is some $x\in X$ such that $\{y\in X: y<_R x\}$ is not in order-preserving bijection with any ordinal, then since $R$ is a well-ordering there is a least such $x$; so just replace $X$ and $R$ with $X_x$ and $R\upharpoonright X_x\times X_x$.) So we have a map from elements of $X$ to ordinals. By Replacement, the image of that map exists and is a downwards-closed set of ordinals, and hence an ordinal; and it's easy to show that this ordinal is in order-preserving bijection with $R$."

In $ZC$, however, there is no reason for the class of ordinals corresponding to elements of $X$ to be a set! So the proof breaks down here. In fact, it’s worse: in $ZC$ we can’t even prove that $\omega+\omega$ exists! (And the only way $ZC$ knows $\omega$ exists is by having it "built in" via the Axiom of Infinity.)


Technically, this doesn’t answer your question: all I’ve explained is why the usual proof that $\omega_1$ exists breaks down. I haven’t shown that no proof exists. In order to prove that, really, $ZC$ can’t prove $\omega+\omega$ exists, it suffices to show that $V_{\omega+\omega}$ is a model of $ZC$. (This is a fun exercise!)


EDIT: You may be interested in the paper "Slim models of Zermelo set theory" by Mathias (https://www.dpmms.cam.ac.uk/~ardm/slim.dvi) - it focuses specifically on recursive definitions in Z(C), and where they break. Note that the first sentence of the abstract confirms my statement that Z alone cannot prove that $V_\omega$ exists.

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    $\begingroup$ A good exercise for getting a handle on what sorts of things $ZC$ can and can't do: working in $ZFC$, describe the least ordinal $\alpha$ such that there is no $(X, R)\in V_{\omega+\omega}$ where $R$ is a well-ordering of $X$ with order type $\alpha$. For example, even though $V_{\omega+\omega}$ doesn't contain $\omega+\omega$, it does contain all the binary relations on $\omega$ - in particular, all the countable ordinals are so represented! And the early part of my answer shows that $\omega_1$ is also represented. It's not hard to get $\omega_2$, $\omega_3$, etc. So: how high can you go? $\endgroup$ – Noah Schweber Aug 19 '15 at 3:26
  • $\begingroup$ But in ZC it is not possible neither demonstrate that Vω+1, while the ordinal ω+1 exist. $\endgroup$ – Raul Aug 20 '15 at 6:28
  • $\begingroup$ If you're saying that ZC doesn't prove $V_{\omega+1}$ exists, but does prove that $\omega+1$ exists, this seems right; but what does that have to do with my answer above? $\endgroup$ – Noah Schweber Aug 20 '15 at 14:40
  • $\begingroup$ Because then it is impossible use induction even for ordinals type $ \omega+n$. Is it induction a non propierty of ZC? $\endgroup$ – Raul Aug 21 '15 at 3:06
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    $\begingroup$ It depends what you mean by "induction." ZC is more than strong enough to prove, for any formula $\varphi$: "If $R$ is a well-ordering of $X$, and for all $x\in X$, if $\varphi(y)$ holds $\forall y<_Rx$ then $\varphi(x)$ holds, then $\forall x\in X(\varphi(x))$." But maybe you mean the proof that definition by recursion works - one instance of which is the theorem "$V_\alpha$ exists for every ordinal $\alpha$." This does require replacement, and ZC is not enough to make all recursive constructions work (although some still do). Does that help? $\endgroup$ – Noah Schweber Aug 21 '15 at 3:21

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