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?
The class of all classes not containing themselves
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0New 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. – 2016-06-23
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1@Timothy: Engineers??? Huh? – 2016-06-23
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0@karagila: set theoretical engineers digging up new foundations where the old ones won't do.. – 2016-06-24
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1(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. – 2016-06-28
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0@AsafKaragila Quine might be [beside himself](http://idioms.thefreedictionary.com/beside+oneself) with indignation ;-D – 2016-06-29
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0@asaf: well, Quine might not find wit appealing; but joking aside, I have heard the term 'machinery' bandied about in algebraic topology, if not 'engineering'. – 2016-07-03
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0and the term 'heavy lifting'! – 2016-07-04
2 Answers
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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0Once 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? – 2012-03-04
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0@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. – 2012-03-04
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0@Mozibur: I'm no expert in set theory, but as I understand it, that's basically how [NF set theory](http://en.wikipedia.org/wiki/New_Foundations) is built up. – 2012-03-04
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1@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. – 2012-03-04
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0@Mozibur: Sure, you can do that. Then you can throw in another large class axiom so there's a class that can hold all kinds of $n$-classes. And then again, make another layer of classes that can hold those large classes. Of course, you're really just doing set theory at this point, and it's much more straightforward to throw in a large cardinal axiom and talk about large and small sets, rather than fiddle with higher-order logic. – 2012-03-04
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0@Karagila: Thanks. The reason why I asked this question, is that I find the notion of there being just one receptacle for collections - ie sets, natural, but two - sets & classes unnatural. One can surely then say if two, well, why not more. – 2012-03-04
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1@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). – 2012-03-04
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0@Karagila: I don't see what the problem is with having objects which are definable by recursion for every ordinal, can you explain this. – 2012-03-04
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1@Mozibur: Well, usually the meta-language is very "natural" so you only have natural numbers in the meta-language (luckily this is enough to define ZF and *within* ZF have the transfinite recursion). Since classes are syntactic objects they live in the meta-language. If your meta-language and theory are not strong enough, you cannot define such things. If you use ZF for your meta-theory then I think you can do that, but you still run into problem since if you define "the class of all $\alpha$-classes$" for every $\alpha$ then the collection of all those classes will be too big to exist, again. – 2012-03-04
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0@Karagila: In that case NBG seems a more natural framework for set theory than ZFC! Ok, if your collection is too large to exist, then that seems an opportunity to introduce an axiom to assert its existence. Is there any connection between what we're discussing here and large cardinal axioms? And are these axioms available only in ZFC, given what you said about NBG? – 2012-03-04
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0@Mozibur: That depends on what you want to do. In most of analysis and non-categorical algebra you don't even go much beyond the continuum so classes are a bit excessive. In categories there are indeed many people preferring NBG over ZFC for some cases, but you quickly run into the limitations of this theory as well, and it's easier to work with large cardinals as pretty soon (2-inaccessible) you get *at least* the universes axiom of Tarksi-Grothendieck. Furthermore, NBG is a conservative extension of ZFC, so if your proof only uses sets it means it was provable in ZFC to begin with. – 2012-03-04
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0@AsafKaragila, is it really true that every set is a class under your definition? How about a set that exist, but which cannot be defined by a formula in the language of set theory? – 2013-09-17
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0@user18921: Remember that these formulas can have parameters. $x=\{y\mid y\in x\}$ is a valid formula in the language of set theory with $x$ as a parameter. – 2013-09-17
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0Okay, but what if $x$ isn't definable? Looking at it model-theoretically, if ZFC is consistent, then there exist uncountable models, which will necessarily have elements that aren't definable within the language. This is actually something I've been unsure about for quite a while, so it will be good have it cleared up. – 2013-09-17
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0@user18921: Again, in the definition of a class we allow the formula to have arbitrary parameters from the model. Since $x$ is an element of the model it can be used as a parameter instead of $p$ in the formula $y\in p$. – 2013-09-17
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0I see. So basically, given a model $V$ of ZFC, we have subsets of $V$, some of which are subclasses, some of which are internal subsets. – 2013-09-17
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0Well, since you like being extremely nitpicky, the answer is not exactly. $M\models x=\{y\mid y\in x\}$, but $V$ might think that $x\neq\{y\mid M\models y\in x\}$. Because (1) $V$ might know a lot more about the set $x$; and (2) even if it doesn't, you have to remember that the objects of $M$ don't have to be actual subsets of $M$. Although you can indeed replace them with subsets and $\in^M$ with $\subsetneqq$. Even then, though, you're not guaranteed that the subsets of $M$ are its elements. But if $M$ is a model of $\sf ZFC$ then it has classes, some of which are isomorphic to its elements. – 2013-09-17
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0Right. So in summary: every class of $M$ is a subset of $M$; the converse needn't hold; and furthermore, some of the classes of $M$ are isomorphic to elements of $M$ (we call these improper classes) while others are not (by definition, the proper classes). Have I understood correctly? – 2013-09-17
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0@user18921: Yes, it seems correct enough to my sober (but so very tired) eyes. – 2013-09-17
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0Okay, thanks.$\!\;$ – 2013-09-17
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0Asaf, I have a couple of questions, I hope you don't mind. Firstly, would it be fair to say that the notion of a class make sense for any first-order structure, not just a model of set theory? For example, does it make sense to speak of the classes of $(\mathbb{N},0,S,+,\times)$ ? I'm thinking yes. Secondly, would it be fair to say that, since MK allows quantification over proper classes, thus some of its so called "classes" aren't really classes at all? – 2013-09-19
Von Neumann–Bernays–Gödel set theory is consistent and it's a theorem of Von Neumann–Bernays–Gödel set theory that there is no class of all classes that don't contain themselves. For any predicate describable in that theory, you can prove that there is a class of all sets satisfying that predicate but not necessarily prove that there is a class of all classes satisfying that predicate. You can describe the statement that no class contains itself and prove it but that's doesn't let you assert the existence of the class of all classes that don't contain themselves. In fact, in Von Neumann–Bernays–Gödel set theory, no class contains any proper class.
Source: https://en.wikipedia.org/wiki/Von_Neumann%E2%80%93Bernays%E2%80%93G%C3%B6del_set_theory