3

Let ZFCU be ZFC modified in the usual way to allow for urelements but without an axiom stating that there is a set of all urelements. Let the principle of Schematic Dependent Choice (SDC) be:

$\forall x\exists y\phi(x, y) \to \forall x\exists f(\mbox{dom}(f) = \omega \wedge f(0) = x \wedge \forall n\in\omega: \phi(f(n), f(n+1)))$

Question: does SDC follow from ZFCU?

hmakholm left over Monica
  • 286,031

2 Answers2

4

The answer is, as Noah suggests, negative.

Suppose that $\frak M$ is a model of $\sf ZFCU$ where $U$ is not a set. Consider the model generated by taking all the permutations of $U$ which are sets (namely, permutations of a subset of $U$, and therefore the identity elsewhere). And consider $\cal F$ to be the filter generated by $\langle\operatorname{fix}(E)\mid E\text{ finite set of atoms}\rangle$, where $\operatorname{fix}(X)$ is the pointwise stabilizer of a set. Let $\frak N$ be the permutation model defined from these, and I claim that:

  1. $\frak N\models\sf ZFCU$.
  2. $\frak N\not\models\sf SDC$.

To see (1) holds, first prove that every set of atoms in $\frak N$ is finite; that every transitive set in $\frak N$ contains only finitely many atoms; by induction on the rank of a set show that if $x$ is a transitive set and $E=x\cap U$, then $\operatorname{fix}(E)=\operatorname{fix}(x)$. In particular a classical theorem about permutation models is that if the pointwise stabilizer of a set is in the filter, then the set can be well-ordered.

To see that (2) holds, or rather that $\sf SDC$ fails, consider the statement $\phi(x,y)$ to be "If $x$ is a set of atoms, then $y$ is a set of atoms and $x\subsetneq y$". Of course every set of atoms can be extended to a larger set of atoms; but if $f$ exists as in the schema, then $\bigcup f(n)$ exists by replacement and union, but it has to be an infinite set of atoms.

Here is a tentative (better) example. I don't have time to work out all the details yet.

Start with a model of $\sf ZFCU$ where $U$ is not a set, and there is a bijection between $U$ and the ordinals. Let $F$ denote a definable class function from $U$ onto $\omega$ such that every fiber is a proper class. This is doable from the bijection with the ordinals, for example (which is really the only reason to require that bijection).

Now consider all the set permutations of $U$ which preserve $F$. Namely $\pi\colon U\to U$, which is the identity outside some set $d$, and $F(a)=F(\pi(a))$ for all $a\in d$.

Let $\cal F$ be the filter generated by $\{\operatorname{fix}(E)\mid E\text{ is any set of urelements and }F[E]\text{ is bounded in }\omega\}$. Let $\frak N$ be the permutation model.

  1. $\frak N\models\sf ZFCU$.
  2. $F$ is definable in $\frak N$.
  3. There is no $f\colon\omega\to U$ such that $F(f(n))=n$.

Now consider the formula $\phi(x,y)$ to be "If $x$ is a finite partial inverse of $F$ [read: $F(x(n))=n$ for $n\in\operatorname{dom}(x)$], then so is $y$, and $x\subsetneq y$". This is an obvious counterexample to $\sf SDC$ as in the previous case.

Asaf Karagila
  • 393,674
  • Thanks, Asaf! As I mentioned to Noah in the comments to his answer, I'm particularly interested in the case where the sets are injectable into the urelements (and thus where there is an infinite sets of them). Do you have a sense how the proof might go in that case? –  Jun 24 '15 at 15:28
  • Suppose that there was a [class] function $f\colon U\to\omega$ which was surjective, but without an inverse. Then you could consider finite partial inverses ordered by inclusion. It's easy to make these sort of models without choice; but it's harder if you want to preserve the axiom of choice. I'm not sure how a proof like that would go, but I believe it's possible. – Asaf Karagila Jun 24 '15 at 15:32
  • I added an idea for an example based on my comment. I think it should work, but I don't have time to sit through the details right now. Please let me know if you check it out, and if it works. – Asaf Karagila Jun 24 '15 at 15:44
3

If I understand ZFCU correctly, then the answer should be "no": there seems to be nothing ruling out a model $M$ of ZFCU in which the atoms do not form a set, there is a linear order $L$ on the class of atoms with no $L$-greatest element, and there is no infinite set of atoms. In such a case, SDC would fail for the formula $\phi(x, y)\equiv$"$x$ is not an atom or $y>_Lx$."

In fact, here's an outline of how to build such a model. Start with a countable model of ZFCU in which the atoms form a proper class which is linearly ordered by some definable relation $L$, of (external) order type $\mathbb{Q}$ - so $L$ is externally homogeneous. Now let $G$ be the group of $L$-automorphisms of the class of atoms, and let $\mathcal{F}$ be the filter of finite supports as usual; the permutation model constructed from $M$ via $G, \mathcal{F}$ should have the desired properties.

Note that if the atoms form a set, then (1) we would have to have an infinite set of atoms, of course, and (2) if we demanded instead "no injection from $\omega$ to atoms," this would violate choice. But since the atoms don't form a set, neither objection applies.

Noah Schweber
  • 245,398
  • Thanks! I guessed it doesn't follow, but it wasn't clear how to construct a counter model. I'll need to think more about your suggestion, as I don't see it immediately. Would it be much harder to get a counter model if we assume that there is an infinite set of urelements? Indeed, I'm actually interested in the case where there's a definable 1-1 function from the sets into the urelements. –  Jun 24 '15 at 13:36
  • BTW: if we had a set of urelements, then it can be proved. We assign ranks in the usual way, giving urelements rank 0. A simple recursion then shows that there is a set of things of rank $\alpha$ for each $\alpha$. Starting with any $x$ we can then find a rank closed under $\phi$ and the conclusion follows from Choice. –  Jun 24 '15 at 13:44
  • @GME I think that the question you mention in the comment - about whether we can do this in case there's an embedding of (pure? all?) sets into the atoms - is likely to be hard: I suspect the answer is no, SDC is still violable, but I don't see how to do it off the top of my head. – Noah Schweber Jun 24 '15 at 13:47
  • (All). Cool; thanks! –  Jun 24 '15 at 13:57
  • Hm, actually I think I see how to do it. Start with a model $M$ of ZFCU in which the atoms $A$ form a proper class, and in fact are the disjoint union of two proper classes, $A_0$ and $A_1$. On $A_0$ we have a linear order $L$ of (external) order-type $\mathbb{Q}$; we also have a class bijection $F$ from $M$ to $A_1$. Now, any $L$-automorphism $p: A_0\rightarrow A_0$ induces a bijection $\hat{p}: A_1\rightarrow A_1$ in the obvious way. Let $G$ be the group of all maps $f: A_0\sqcup A_1\rightarrow A_0\sqcup A_1$ of the form $p\sqcup \hat{p}$, and take the permutation model with finite supports. – Noah Schweber Jun 24 '15 at 14:03
  • As written, I think this is lacking, but something like it might work. – Noah Schweber Jun 24 '15 at 14:05
  • If I understand you correctly, $L$ would need to be given by some concrete formula in the language of set theory. I have trouble imagining how such an $L$ can distinguish well enough between the atoms to order them linearly without somehow forcing an infinite set of atoms to exist. Could you add some hints as to how it would work? – hmakholm left over Monica Jun 24 '15 at 14:59
  • You don't need a linear order. Just look at finite injective sequences of atoms ordered by extensions. – Asaf Karagila Jun 24 '15 at 15:06
  • @AsafKaragila: Ah, that I get. Or finite sets of atoms ordered by inclusion. – hmakholm left over Monica Jun 24 '15 at 15:12
  • @Henning: Yes. I realized that when I was writing the [new] answer! :-) – Asaf Karagila Jun 24 '15 at 15:15