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In ZF classes are used informally to resolve Russells Paradox, that is the collection of all sets that do not contain themselves does not form a set but a proper class. But doesn't the same paradox manifest itself when discussing the class of all classes that do not contain themselves?

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New foundations is a set theory that was probably formulated by engineers having meetings to try and find a resolution to derived contradictions. I think an expert in new foundations should give a big long answer that changes people's method of thinking entirely to show why a contradiction can't be derived. – Timothy Jun 23 at 1:18
@Timothy: Engineers??? Huh? – Asaf Karagila Jun 23 at 4:46
@karagila: set theoretical engineers digging up new foundations where the old ones won't do.. – Mozibur Ullah Jun 24 at 16:29
(1) pinging to "karagila" will notify no one. Try "Asaf", which is my actual name. (2) Also no, Quine would turn in his grave if he would know people call him a set theoretic engineer. – Asaf Karagila Jun 28 at 20:02
@AsafKaragila Quine might be beside himself with indignation ;-D – Joffan Jun 29 at 23:29
up vote 13 down vote accepted

Classes in ZF are merely collections defined by a formula, that is $A=\{x\mid \varphi(x)\}$ for some formula $\varphi$.

It is obvious from this that every set is a class. However proper classes are not sets (as that would induce paradoxes). This means, in turn, that classes are not elements of other classes.

Thus discussion on "the classes of all classes that do not contain themselves" is essentially talking about sets again, which we already resolved.

Of course if you allow classes, and allow classes of classes (also known as hyper-classes or 2-classes) then the same logic applies you have have another level of a collection which you can define but is not an object of your universe.

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Once you allow the notion of 2-classes, then I assume you can define n-classes for any n a natural number=finite ordinal. Does this mean this construction can be carried through at limit ordinals? – Mozibur Ullah Mar 4 '12 at 20:05
@Mozibur: I don't really know. I suppose you can. Simply by saying that $\omega$-classes are classes whose elements are $n$-classes for unbounded $n$. Then you'll have the problem in $\omega+1$-classes. The problem is that even if you allow classes for every $\alpha$ then you still get stuck with objects which are definable by recursion for every ordinal. Be forewarned that what said in this comment might just as well be a load of manure. I'll see my advisor tomorrow and ask him, then I'll have a better answer to give you here about this question. – Asaf Karagila Mar 4 '12 at 20:10
@Mozibur: I'm no expert in set theory, but as I understand it, that's basically how NF set theory is built up. – Ilmari Karonen Mar 4 '12 at 20:26
@Ilmari: I'd think there is a lot of fine points to the New Foundation theory which $\alpha$-classes do not necessarily agree upon. In fact, it would seem to me that NF is "the other way around" which resolves the paradoxes by allowing only "uncomplicated formulas" to define classes (and thus sets). However, I don't know a lot about NF so I cannot really answer that. – Asaf Karagila Mar 4 '12 at 20:33
@Mozibur: Indeed, which is why in set theory you cannot really have the two. In the NBG set theory you can have classes, but not 2-classes and you cannot have collections of classes (unless those were sets to begin with). – Asaf Karagila Mar 4 '12 at 21:00

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