Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I am currently reading Chang and Keisler's model theory textbook, and there is something that I don't seem to be able to understand about their proof (through the Omitting Types Theorem) that every countable model $\mathfrak A = \langle A, E\rangle$ of $ZF$ has an end-elementary extension $\mathfrak B = \langle B, F\rangle$ (Theorem 2.2.18 in the third edition of the book). The main argument is fine, but I don't understand the way in which they apply the axiom of replacement in their proof that a certain theory $T$ locally omits certain types. Here's the part I am having trouble with:

First, they note that a formula $\phi(x, c)$ is consistent with $T$ if and only if $$ (\mathfrak A, a)_{a \in A} \models (\forall y)(\exists z)(\exists x)[z \not \in y \wedge \phi(x, z)]. $$

Then they take an element $a \in A$ and a formula $\phi(x, c)$ (a candidate for locally realizing a certain type which we want to omit) and show that $\phi(x, c) \wedge x \in a$ must be consistent with $T$ - so far so good. Then they claim - and here's where I am stuck:

Using the axiom of replacement in ZF, we see in turn that the following sentences hold in $(\mathfrak A, a)_{a \in A}$: $$(\forall y)(\exists z)(\exists x)[z \not \in y \wedge \phi(x, z) \wedge x \in a]$$ $$(\exists x)(\forall y)(\exists z)[z \not \in y \wedge \phi(x, z) \wedge x \in a]$$

and from the second sentence they conclude that for some $b \in A$, $\phi(b, c) \wedge b \in a$ is consistent with $T$, and the theorem follows easily.

My problem is with the passage from the first sentence, which follows at once from $\phi(x, c) \wedge x \in a$ being consistent with $T$, to the second one, in which $x$ is the same for any choice of $y$. I take it that I have to apply the axiom of replacement here - but how? I can see how to use collection to obtain, from the first sentence, something like

$$(\exists u)(\forall y)(\exists z)(\exists x)[x \in u \wedge z \not \in y \wedge \phi(x,z) \wedge x \in a];$$

But that's not what the second sentence says, and it is not enough to conclude the proof.

Chances are that I am forgetting about something bloody obvious, but I've been thinking about this for a couple of days now and I just don't see it. Where is it that I am being stupid?

Thanks!

share|improve this question
    
An alternative proof of the fact uses definable ultrapowers, rather than the omitting types theorem. One constructs a filter on $\text{Ord}^M$ measuring every definable (with parameterrs) class in $M$, such that every class function $f:\text{Ord}\to M$ in $M$ that is bounded on a set in the filter is constant on a set in the filter. The resulting $M$-ultrapower will be an elementary end-extension of $M$. –  Joel David Hamkins Jul 2 '13 at 19:23

1 Answer 1

up vote 4 down vote accepted

The negation of the second formula is logically equivalent to $$\forall x\in a\,\exists y\,\forall z\,(\phi(x,z)\to z\in y).$$ Using collection (and union), this implies $$\exists y\,\forall x\in a\,\forall z\,(\phi(x,z)\to z\in y),$$ which is equivalent to the negation of the first formula.

share|improve this answer
    
Ah, I see - of course you are right. Thanks a lot! –  Cardunculus Jul 2 '13 at 15:59

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.