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I want to show that there is a finite conjunction $\phi$ of axioms of $ZF$, such that every transitive proper class $M$, which satisfies $\phi$, is already a model of $ZF$.

This is an exercise in Kunen's set theory. There is a hint, it seems to be useful to apply the Reflection principle to the union $M = \cup_{\alpha} M \cap R(\alpha)$. But I don't know with which axioms we can do that (we can only use finitely many!), and why this yields an ordinal which is independent from $M$. Please give me only a hint, because basically I want to solve this on my own, but I don't know how to start with the hint above.

Also, what is the "philosophical" reason that we cannot deduce from this, that $ZF$ is finitely axiomatizable (which is wrong)? I mean I cannot prove that this $\phi$ above proves every axiom, but is there also a deeper reason for this?

EDIT: There was an answer with some hints, but it was deleted... I still don't know how to produce this strange sentence $\phi$.

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Can you formulate the exercise exactly as stated by Kunen? The way I read your formulation, there is no such finite list of axioms. –  François G. Dorais Sep 28 '10 at 18:26
    
It's exactly as I stated it. –  Martin Brandenburg Sep 28 '10 at 20:38
    
François, it is important that it will only be applied to a proper class transtive model---you get all instances of Collection for free this way, by using Collection in $V$, provided that $M$ thinks that all $V_\alpha$'s exist. –  JDH Sep 28 '10 at 21:53
    
Yes, the word "proper" (which was missing earlier) is crucial otherwise you might run out of ordinals... –  François G. Dorais Sep 28 '10 at 22:30
    
Ah, that is indeed a crucial difference. –  JDH Sep 29 '10 at 2:11

1 Answer 1

up vote 6 down vote accepted

Suppose that $M$ satisfies all the easy things like extensionality, pairing, etc., plus $\Delta_0$ separation and suppose also that $M$ thinks for every ordinal $\alpha$ that $V_\alpha$ exists (that is, $V_\alpha^M$ exists in $M$). All this is expressible by a single axiom, but when $M$ is a proper class, it ensures that $M$ satisfies the entire Collection scheme, because for any formula $\varphi(x,y)$ and set $A\in M$ for which we want to collect, we may apply Collection in $V$ to find an ordinal $\alpha$ such that whenever $a\in A$ has $\exists y\varphi(a,y)$ in $M$, then such a $y$ may be found in $V_\alpha$ and hence in $V_\alpha^M$. Thus, $V_\alpha^M$ serves as a collection set inside $M$. Similarly, one can get full Separation in $M$ from $\Delta_0$ Separation (or much less, if you care to optimize it), by applying the Reflection theorem, which allows you to bound all the quantifiers by a suitably large $V_\alpha^M$.

The argument works only when $M$ is a proper class, however, because it appeals to Collection in $V$, and uses that $M$ grows taller than the resulting collection set. This method would break down completely when $M$ is a set, since the collection set in $V$ could be unbounded in $M$.

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I see that I use $V_\alpha$ where Kunen uses $R(\alpha)$. –  JDH Oct 26 '10 at 12:29

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