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Consider NFU set theory as presented in this:

On page 15 of that pdf it states that the following is an axiom: The set $\{X\colon X=X\}$ exists.

Let $V = \{X\colon X=X\}$.

$V$ is an element of $V$ as $V=V$.

How is this not circular?

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Have you read the relevant Wikipedia article? – Asaf Karagila Nov 30 '11 at 22:50
Thanks for the link. The wiki article says V exists as an entailment of the Comprehension axiom whereas the Holmes book is an axiom. {X:X=X} is an element of {X:X=X}. How is that different from something like defining the set y by the equation y={y}? – atat Dec 1 '11 at 0:25
It is possible, without the axiom of foundation (also known as regularity) to have $x=\{x\}$. – Asaf Karagila Dec 1 '11 at 0:33
Do you also view the fact that $0\cdot 0=0$ in arithmetic to be circular? – JDH Dec 1 '11 at 1:26
@JDH No mainly because that isn't the definition of zero. In NFU, V is defined to be {X:X=X}. As Asaf pointed out, this would only be "circular" if there were an axiom of foundation which would prohibit such things as V∈V. – atat Dec 6 '11 at 16:27

While we are very much used to this, and it seems very natural to us, the axiom of foundation which asserts that $\in$ is well founded (every set has a $\in$-minimal element) is independent of the rest of the axioms.

It is a very interesting construction to see how one can have a model of set theory in which there is some $x=\{x\}$. It can get stretched even further, $P(x)\in x$ for some $x$ (where $P(x)$ denotes the power set).

Going through the .pdf that you linked I can observe several things:

  1. There is no axiom of foundations in NFU (somewhat ironic, isn't it?). Therefore there is no real contradiction in $V\in V$.

  2. In chapter 7 (page 39) it is proved that $\in$ is not a set-relation of the universe, however $\subseteq$ is. Meaning that we care more about subsets than we do about elements.

  3. There is no "axiom of power set" either, so $P(V)$ need not exist as a set, which is just fine with the above, although as I said earlier, it is quite possible to have $P(x)\in x$.

  4. Comprehension itself is bounded. Unlike ZFC, however, the bound is not by taking a subset of some pre-existing set, but rather bounding the formula we use to a stratified formula - the definition of which appearing on page 40 of the .pdf file.

I'd guess that the further you get in this book, the more observations you can collect. However skimming through the first 50 would seem enough to gather the above and see that there is no real apparent paradox in NFU. It is only your intuition which was heavily laced with ZFC which stands in your way.

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Here's why I asked. I think I can write down a stratified formula F(X) that is satisfied iff X is a structure (in the math logic sense). Then S={X:F(X)} exists by the stratified comprehension theorem; ie, the aggregate of all structures exists and is a set. Then I want to define some sort of product of all elements of S, related to another question I posted a little while back, in such a way that every structure can be immersed via 1-1 homomorphism from that structure to the product. This would then be the product of all structures. That seems circular, given that the productis a structure. – atat Dec 1 '11 at 1:06
@atat: Remember that a structure is a universe with an interpretation of relations, constants and symbols. As my second remark says, $\{(x,y)\mid x\in y\}$ is not a subset of the universe, so you cannot encode the membership relation as a subset of the universe. Furthermore, the product is not always a structure, let alone a model of a theory. – Asaf Karagila Dec 1 '11 at 11:37
I'm looking for a way to endow a product of elements of S with structure to make it true that all elements of S can be homomorphically injected into the product. I've been having difficulty with doing that though it seems like it should be easy. Thanks for all your feedback. – atat Dec 6 '11 at 16:31

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