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 '16 at 01:18

1@Timothy: Engineers??? Huh? – Asaf Karagila Jun 23 '16 at 04:46

@karagila: set theoretical engineers digging up new foundations where the old ones won't do.. – Mozibur Ullah Jun 24 '16 at 16:29

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. – Asaf Karagila Jun 28 '16 at 20:02

@AsafKaragila Quine might be [beside himself](http://idioms.thefreedictionary.com/beside+oneself) with indignation ;D – Joffan Jun 29 '16 at 23:29

@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'. – Mozibur Ullah Jul 03 '16 at 22:50

and the term 'heavy lifting'! – Mozibur Ullah Jul 04 '16 at 08:46

I used to think some people created NBG because they thought it was a true model of set theory. I guess I thought they thought that surely, there is a real existing object called the class of all sets, and that there are strictly more classes than there are sets. It turns out that NBG was intended as a short cut for determining when a statement describable in ZFC is a theorem of ZFC because it has been proven that a statement describable in ZFC is a theorem of ZFC if and only if it is a theorem of NBG. There is a new theory called New Foundations which proves that there is a universal set but – Timothy Mar 17 '20 at 02:55

not a set of all sets that don't contain themselves. However, there is another set that has some connection with the formalization of the property of not containing itself. As soon as you introduce classes in NBG, you suddenly realize that that would mean the class of all sets doesn't contain everything that doesn't contain itself anymore. Once you describe a class, you realize that you hadn't actually described everything that exists when you talk about the class of all sets. We might as well call all those objects sets rather than classes. We find that none of them contain all of the ones – Timothy Mar 17 '20 at 03:03

that don't contain themselves and only those ones. There are no proper classes. When you think you conceived of a proper class, it's because you're conceiving of a set of only some of the sets and not all of them. – Timothy Mar 17 '20 at 03:06

@Timothy: There are 'larger' set theories than NBG; for example, look at how the large cardinal hierarchy is used to construct a hierarchy of sets, classes, and so on; it's used as a stop gap measure in category theory by asking for the existence of an inaccessible cardinal (which can't be proved from ZFC), or alternatively, using Grothendieck universes. – Mozibur Ullah Mar 17 '20 at 13:43

@MoziburUllah I guess NBG makes sense if it's treated like a conservative extension of ZFC, that is, as a short cut for finding theorems of ZFC. I know anything is possible including all sorts of wierd theories. – Timothy Mar 17 '20 at 19:21

@Timothy: ZFC was one of the first axiomatised set theory that won general acceptance, that's its importance; this does not mean there aren't others; after all, we have lots of number systems; that we don't know more about other set theories is simply because of the hegemonic presence of ZFC; set pluralism  to my mind  makes a great deal of sense. – Mozibur Ullah Mar 17 '20 at 19:41
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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 hyperclasses or 2classes) 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.
Asaf Karagila
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Once you allow the notion of 2classes, then I assume you can define nclasses for any n a natural number=finite ordinal. Does this mean this construction can be carried through at limit ordinals? – Mozibur Ullah Mar 04 '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 04 '12 at 20:10

@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. – Ilmari Karonen Mar 04 '12 at 20:26

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. – Asaf Karagila Mar 04 '12 at 20:33

@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 higherorder logic. – Mar 04 '12 at 20:50

@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. – Mozibur Ullah Mar 04 '12 at 20:57

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 2classes and you cannot have collections of classes (unless those were sets to begin with). – Asaf Karagila Mar 04 '12 at 21:00

@Karagila: I don't see what the problem is with having objects which are definable by recursion for every ordinal, can you explain this. – Mozibur Ullah Mar 04 '12 at 21:10

1@Mozibur: Well, usually the metalanguage is very "natural" so you only have natural numbers in the metalanguage (luckily this is enough to define ZF and *within* ZF have the transfinite recursion). Since classes are syntactic objects they live in the metalanguage. If your metalanguage and theory are not strong enough, you cannot define such things. If you use ZF for your metatheory 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. – Asaf Karagila Mar 04 '12 at 21:15

@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? – Mozibur Ullah Mar 04 '12 at 21:51

@Mozibur: That depends on what you want to do. In most of analysis and noncategorical 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 (2inaccessible) you get *at least* the universes axiom of TarksiGrothendieck. 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. – Asaf Karagila Mar 04 '12 at 22:07

@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? – goblin GONE Sep 17 '13 at 17:10

@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. – Asaf Karagila Sep 17 '13 at 17:17

Okay, but what if $x$ isn't definable? Looking at it modeltheoretically, 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. – goblin GONE Sep 17 '13 at 18:08

@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 $y\in p$. – Asaf Karagila Sep 17 '13 at 18:15

I 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. – goblin GONE Sep 17 '13 at 18:32

Well, 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. – Asaf Karagila Sep 17 '13 at 18:38

Right. 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? – goblin GONE Sep 17 '13 at 20:14

@user18921: Yes, it seems correct enough to my sober (but so very tired) eyes. – Asaf Karagila Sep 17 '13 at 20:15


Asaf, 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 firstorder 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? – goblin GONE Sep 19 '13 at 04:30

Are you talking about something like the fact that some people say "The class of all sets is proper" to informally mean "There is no set of all sets" which is a theorem of ZF, and explaining how that's why we cannot say something like "For all classes" when we're informally referring to a statement describable in ZF? I guess so because it seems to have solved the author's problem. – Timothy Mar 17 '20 at 03:11