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Theorem 2.2.18 in Chang and Kiesler uses omitting types to show that any countable model of ZF has an elementary end extension. Can we control the countable order type of such a model? for example, if $ X \prec H_ {\omega_2}$ can we have an elementary extension $Y \prec H_{\omega_2}$ such the $order type(Y)$ is bounded by some ordinal?

Any help or reference would be appreciated.

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I must be missing something: $H_{\omega_2}$ (or any its elementary submodel, for that matter) is not a model of ZF. –  Emil Jeřábek Mar 19 '12 at 11:45
    
You are correct. Please refer to Joel's answer for a discussion. –  Eran Oct 17 '12 at 8:42
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1 Answer

up vote 8 down vote accepted

There are several interesting issues arising in your question.

First, you ask about models of ZF, but then mention elementary substructures of $H_{\omega_2}$, which of course is not a model of ZF, because it does not satisfy the power set axiom. In the context of $H_{\omega_2}$, you probably intend to discuss the theory $\text{ZFC}^-$, meaning the theory of ZFC without the power set axiom, and this makes for a very interesting question. (Meanwhile, please see my recent paper with Gitman and Johnstone What is the theory of ZFC without power set? for reasons to be careful about how this theory is described---in particular, one should be sure to include collection rather than merely replacement; they are no longer equivalent without power set.)

Second, in the context of set theory, there are distinct concepts of end-extension.

  • One model $(M,E)$ end-extends to another $(N,F)$ if $(M,E)$ is a substructure of $(N,F)$, meaning that $M\subset N$ and $E=F\cap M\times M$, and also $a\mathrel{F}b\in M$ implies $a\in M$. That is, the sets of $M$ do not gain new elements in $N$. In other words, $M$ is a transitive subclass of $N$.

  • One model $M$ top-extends to another model $N$, if $M$ is a rank initial segment of $N$. In other words, $M$ end-extends to $N$ and also all new objects of $N$ have ordinal rank larger than any ordinal of $M$. For example, this is the situation when $V_\alpha\prec V_\beta$, but in the general case, top-extensions needn't have that the height of $M$ is an ordinal of $N$.

Set theorists often use the term "end-extension" to refer to top-extensions, so one must take care, as the concepts are distinct for models of set theory. For example, a nontrivial forcing extension is an end-extension but not a top-extension. When it comes to elementary extensions of models of ZF, however, even in the case of nonstandard models, then the two notions coincide. This is because if $M\prec N$ is an elementary end-extension, then by elementarity, the $V_\alpha^M$ remains the $V_\alpha$ of $N$ for any ordinal $\alpha$ in $M$, and furthermore $N$ cannot add any new ordinals below an ordinal of $M$. Thus, all new sets of $N$ must have rank larger than any ordinal of $M$ and so $N$ is a top-extension. So for models of ZF, the order type of an elementary end-extension must be strictly larger.

The corresponding fact is not true for models of $\text{ZFC}^-$, where the rank initial segments $V_\alpha$ of the von Neumann hierarchy do not exist as sets. To see this, start with GCH in $V$ and then force to add $\aleph_2$ many Cohen reals to form the extension $V[G]$, and consider the model $M=H_{\omega_2}^{V[G]}$. Now add one more Cohen real $c$ over $V[G]$, and consider $N=M[c]=H_{\omega_2}^{V[G][c]}$. I claim that $N$ is an elementary end-extension of $M$. First, it is clearly an end-extension, since both are transitive sets in $V[G][c]$. Second, for elementarity, observe that the two-step forcing $\text{Add}(\omega,\omega_2)*\text{Add}(\omega,1)$ is isomorphic to $\text{Add}(\omega,\omega_2)$ by absorbing the final factor into an earlier stage. Furthermore, this isomorphism can be arranged to fix any $\omega_1$ sized initial segment of the forcing. Now, any fact true in $N$ about some parameters in $V$ and an $\aleph_1$ sized piece of $G$, but not involving $c$, is forced by a condition, and thus will be true whether as computed in $H_{\omega_2}^{V[G]}$ or in $H_{\omega_2}^{V[G][c]}$, by using that isomorphism. And so the inclusion $M\subset N$ is elementary.

The following theorem shows that the question of whether one must go to strictly larger order types is actually independent of ZFC.

Theorem.

  1. It is relatively consistent with ZFC that every countable $X\prec H_{\omega_2}$ has a nontrivial end-extension $X\prec Y\prec H_{\omega_2}$ with exactly the same ordinals.

  2. It is also relatively consistent with ZFC that whenever $X\prec Y\prec H_{\omega_2}$ and $X\neq Y$, then $Y$ has strictly higher order type than $X$.

Proof. For statement 1, use the model $V[G]$ as described above. The point is that for any countable $X\prec H_{\omega_2}^{V[G]}$, we can find an $X$-generic Cohen real $c$ in $V$, and it will follow that $X\prec X[c]$ by the argument given above.

For statement 2, use $L$ and the fact that $H_{\omega_2}^L=L_{\omega_2}$. If $X\prec Y\prec L_{\omega_2}$ and these are end-extensions, then because there is a definable well-ordering, each new object of $Y$ not in $X$ has an ordinal position $\alpha$ in the canonical $L$ order, and this ordinal is in $Y$ but not in $X$. So by the end-extension property, it cannot be smaller than any ordinal of $X$, and so the order type of $Y$ is strictly higher than the order type of $X$. QED

I've written too much already, but one can say much more about bounding the order-type of $Y$ over $X$, by fixing suitable Skolem functions and then arguing that there is a club of closure points. And there are also interesting things to say for the case of extensions of models of full ZF rather than $\text{ZF}^-$.

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Thanks so much for your detailed answer. I'll read it thoroughly and try to learn from it. regards. –  Eran Mar 19 '12 at 22:20
    
Joel, you may be interested in this question. Perhaps there is an easy way of playing with end-extensions to address it as well? –  Andres Caicedo Jan 14 at 17:40
    
@Andres, I posted an answer there using the standard system. –  Joel David Hamkins Jan 14 at 19:21
    
Thank you! ${}$ –  Andres Caicedo Jan 14 at 19:52
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