# Proving $Con(ZF^-+\neg AC)$ with atoms(urelements).

Greets

This exercise, II.9.10 from Kunen's Set Theory (2011 ed.), has bothered me for a time:

Work in $\mathsf{ZFC^-}$ and assume that there is an infinite set $T$ such that $x=\{x\}$ for all $x\in T$. Let $M$ be the class of all $y\in WF(T)$ such that for some finite $A\subseteq T$, $\hat\pi(y)=y$ for all permutations $\pi$ of $T$ such that for some finite $A\subseteq T$, $\pi\upharpoonright A=id_A.$ Prove that $T\in M$, and that $M$ is a transitive model for $\mathsf{ZF^-}$ plus $(\mathsf{AC})^{WF}$ plus "$T$ cannot be well-ordered".

Where $WF(T)$ is the class constructed as follows:

1. $R(0,T)=T$
2. $R(\alpha+1)=\mathcal P (R(\alpha,T))$
3. $R(\gamma,T)=\bigcup_{\alpha<\gamma}R(\alpha,T).$

Let $WF(T)=\bigcup_{\alpha\in ON}R(\alpha,T).$ $\hat\pi$ is the natural extension of $\pi$ to all of $WF(T)$; so that $\forall x,y\in WF(T)[x\in y\leftrightarrow \hat\pi(x)\in\hat\pi(x)].$

I couldn't prove that $M$ is a model of $\mathsf{ZF^-}$, and not even that $M$ is transitive.

However, I made a different contruction yielding the same results:

I considered a sort of $L$ inside $WF(T)$, call it $L(T)$; the same construction of $L$ but beginning with $T$ instead of $\emptyset$. Then you get a rank for each element of $L(T)$.

Then you prove by induction, on the rank, that for each $x\in L(T)$, there is some finite $A\subseteq T$ which is fixed by all $\hat\pi$ such that $\pi\upharpoonright A=A(\clubsuit).$ And also one proves that $\hat\pi$ is surjective restricted on each level, onto the same level, whenever $\pi$ is a permutation of $T$ $(\star)$.

Now one can easily see that $L(T)$ is transitive; for any $\alpha$, $L(\alpha,T)\subseteq L(\alpha+1,T)$, and that comprehension holds in $L(T)$; using $(\star)$. As the construction is with levels, one can show that for any set $y$ with $y\subseteq L(T)$ there is $z\in L(T)$ such that $y\subseteq z$. Putting toghether these facts one shows that $L(T)$ is a model of $\mathsf{ZF^-}$.

Using $(\clubsuit)$ one proves that $T$ cannot be well-ordered in $L(T)$; and as $(WF)^{L(T)}=WF$, we obtain $(AC)^{WF}$.

Now, is the model $M$ given in the exercise the same as the one I constructed?; recalling some model theory this sounds plausible, but I can't see why.

Can the exercise be done in a more direct way?

Thanks

• I believe that I have written some exposition on permutation models in several answers, by the way. Sep 30, 2013 at 11:16
• I also much prefer Jech's introduction to permutation models (you can find it somewhere in his Set Theory book on chapter 15; or chapter 4 on the small AC book). Sep 30, 2013 at 11:19

It seems to me that there is a mistake in the definition of $M$. Consider for example $A$ an infinite co-infinite subset of $T$, it is clear that no such subset should be in $M$ (because given a finite subset of $T$ we can always find a permutation not preserving $A$ while fixing the finite subset).

However the set $\mathscr A=\{\pi A\mid\pi\in\operatorname{Sym}(T)\}$ is definitely closed under any permutations. The current definition of $M$ allows $\mathscr A$ to be included, but none of its members can be in $M$.

If however you define $M$ by induction, requiring that $A\in M_{\alpha+1}$ if and only if there is such finite set etc. and $A\subseteq M_\alpha$, then you can prove the transitivity of $M$.

Then you can verify that the axioms of $\sf ZF^-$ hold; moreover you can easily verify that $WF^M=WF$; which is a model of $\sf ZFC$.

• What I did was what you wrote on the third line, as a set $X$ is definable over $M_{\alpha}$ with some parameters $a_1,\ldots,a_n$, if and only if all permutations fixing $\{a_1,\ldots,a_n\}$ pointwise, also fix $X$. Sep 30, 2013 at 14:53
• @Camilo: That may end up a strictly smaller universe, though. Sep 30, 2013 at 15:11
• But shouldn't $M_{\alpha+1}\subseteq \mathcal{P}(M_{\alpha})$, if you change your definition with this condition, we obtain the same model I thought of. Sep 30, 2013 at 15:28
• This it the same difference between taking $L$ and taking $V$. By taking only definable subsets you lose a lot of undefinable subsets. Note for example, that the pure part of $M$ as described by me is exactly all the pure sets in the universe; whereas by taking only definable sets you end up with $L$. I'm not saying it's a mistake, I'm just pointing out that the result can be very different. Sep 30, 2013 at 15:35
• Yes, this is the Hajnal construction of $L(A)$ (where you take $A$ to be the transitive closure of the set $\{A\}$, of course, however in the case of $T$ it is already transitive). Sep 30, 2013 at 16:02