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What are some differences between pure and impure set theory?

For example, this paper references the result that ZFC with urelements is categorical if you assume that the urelements form a set. ZFC, however, is not known to categorical.

Likewise, there are important metatheoretic differences between New Foundations (NF) and NF with urelements. For example, there is a relative consistency proof for NF with urelements, but none is known for NF.

Are there any other major differences like these between set theories and their impure counterparts?

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One big difference is that models of impure set theories admit nontrivial automorphisms. Consider $\text{ZFU}$ for instance. Take a set $U$ of urelements and a permutation $\pi : U \to U$ of these urelements. Then $\pi$ extends to an automorphism of the (impure) set theoretic universe by $$\pi(x) = \{ \pi(y)\, :\, y \in x \}$$ for all sets $x$. This is a well-defined recursive definition because by the axiom of foundation you'll always eventually hit the empty set or an urelement.

This kind of construction allows you form models of $\text{ZFU}+(\neg \text{AC})$. Take a group $G$ of automorphisms of $U$, which extend to automorphisms of the universe as above. Then under a suitable definition of 'symmetric', the hereditarily symmetric sets form a model of $\text{ZFU}$ in which the axiom of choice fails.

This is impossible in $\text{ZF}$ because in this setting the only automorphism of the universe is trivial. In fact, even if you admit choice in the construction given above, even though choice fails in the symmetric submodel, it still holds in the class of pure sets of the submodel.


Further reading: The Axiom of Choice by T.J. Jech.

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Thank you in particular for the reference to Jech. From a quick scan, I can already tell that the book is full of incredibly interesting material. The stuff on ZFU (ZFA in Jech's terminology) is especially interesting. –  Dennis Jan 3 '13 at 8:16
    
@Clive: Actually, any first-order theory with an infinite model also has a model that has a non-trivial automorphism. First, construct an uncountable model, and then observe that there must be an element of the model that is not definable without parameters. Then apply the argument here. The point is that a well-founded model of ZF has no automorphisms, but not every model of ZF is well-founded. (Or perhaps the point is that not every automorphism is definable.) –  Zhen Lin Jan 3 '13 at 8:35
    
So, if I'm following you @ZhenLin, what you're saying is that it is impossible to replicate the proof in ZF not because the only automorphism of the universe is trivial, but because there are no automorphisms of ZF (assuming a well-wounded model)? –  Dennis Jan 3 '13 at 9:00
    
I'm not sure what you're asking. The construction Clive gives works in any model of ZFU and in particular goes through in ZF. The only difference is that the set of urelements in ZF is always empty, so this only ever produces a trivial automorphism. (When I say "no automorphism", I mean "every automorphism is trivial", which is the same thing for practical non-pedantic purposes.) –  Zhen Lin Jan 3 '13 at 9:16
    
@ZhenLin: I guess I meant definable automorphism. –  Clive Newstead Jan 3 '13 at 9:24
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As well as Jech's excellent book, you might also be interested in looking at one of Michael Potter's two set theory books, particularly the later version Set Theory and Its Philosophy. Here he principally develops $\mathsf{ZU}$, a version of what has become known as Scott-Potter set theory with urelemente. Potter gives reasons too for why it is natural to start with, and continue to work with, theories with urelemente.

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Another great suggestion Peter. I'm a big fan of this book. It was my introduction to Set Theory when I took Foundations of Mathematics with Hartry Field. Incidentally he also assigned your Godel book, the first time I read that. –  Dennis Jan 3 '13 at 9:02
    
Off topic, but when do you expect Tom Forster's rumored proof of the relative consistency of NF to be made public? –  Dennis Jan 3 '13 at 9:05
    
@Dennis: It's not Forster's proof (but hopefully it'll be announced in the coming months). –  Clive Newstead Jan 3 '13 at 9:25
    
@CliveNewstead Ah, that's right, it was Randall Holmes. I got the post announcing the NF conference confused with the post announcing the consistency proof. –  Dennis Jan 3 '13 at 19:37
    
@Dennis: Where's this announcement? I'm not sure how discrete everyone is meant to be about the proof at the moment (which is why I said "not Forster's" rather than "Randall Holmes's"!). –  Clive Newstead Jan 3 '13 at 20:34
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