On Fraenkel-Mostowski choiceless set theory

Question

I have been trying to solve an exercise from Kunen (1980) on Fraenkel and Mostowski's construction of a choiceless model of set theory. I have a couple of questions:

1. The model is constructed from an infinite set $U$ of "urelements", or "atoms", and it lacks a well-ordering of $U$ in it. If I understand correctly, not only does it lack a well-ordering of $U$ but it also lacks any linear orders on it. Am I correct?

2. In Kunen, instead of talking about urelements or atoms, one assumes ZF - foundation and working in that theory starts with a set $U$ that is infinite and whose elements are of the form $x = \{x\}$. It is another exercise in the book to show that the existence of such $U$ is consistent, but how can one do so? I know how to interpret the theory ZF - foundation + "such $U$ exists" by using graphs, but I would like to know how to obtain a (transitive) model that interprets $\in$ as $\in$. I find it hard to think about the latter possibility because for example under the presence of antifoundation (which is equiconsistent with the rest of the axioms) such $U$ has to be a singleton.

Answer 1

1. I believe that's right (without specifying the exercise, I can't know for sure; but this is a feature of the simplest permutation model, so it's probably true). Note, however, that other models can be constructed where $U$ can be linearly ordered, but not well-ordered.

2. I don't understand what you are asking when you say you "would like to know how to obtain a transitive model that interprets $\in$ as $\in$" - you can't have a transitive model where Foundation fails, at all! So what sort of thing are you looking for, here?

Answer 2

1. Fraenkel's original model was one where the set of atoms is amorphous. Namely every set of atoms is finite or cofinite. It's a fun exercise to show that an amorphous set cannot be linearly ordered. In fact, you can show more. You can show that if $A$ is amorphous, then every function from $A$ into a set which can be linearly ordered has a finite range.

   Mostowski constructed, some 20 years later, a model where the axiom of choice fails but every set can be linearly ordered.

   (As a side note, the reason these are called Fraenkel-Mostowski models is that Fraenkel had some mistakes in his ideas and Mostowski fixed them; Specker improved the presentation too, which is why in some places these are called Fraenkel-Mostowski-Specker models.)

2. Anti-Foundation-Axioms are numerous. For example Boffa's axiom (BAFA) implies that there is a proper class of Quine's atoms (sets of the form $x=\{x\}$). So you don't have to stick to accessible pointed graphs. But it's easier to look at the construction which appears in Jech, which gives you a class model with atoms, rather than the failure of foundation. In this construction we slightly modify the power set operation, and we begin with a set of $\in$-independent sets as our atoms and proceed as usual.