Is the proper class a concept that relies on the axiom of regularlity?

When describing the difference between class and set, Russell's paradox always appears. But does class mean anything only about Russell's paradox? Can I assume a proper class that satisfies an axiom like the axiom of regularity?

I heard that classes and sets are defined differently by the axioms. I would like to know the existence of such an axioms.

• A genuine question that is rightly confusing and well expressed. Not sure why anyone would downvote it. So +1 to offset the ... who downvoted. – fleablood Jul 18 at 10:27
• Possible duplicate of What is the difference between a class and a set? – Vsotvep Jul 18 at 10:28
• @fleablood I would like to know what this question is asking. Unfortunately, becayse of my limitations, I haven't been able to decipher it. Something about sets and classes and regularity, but what? Since you understand it, would you mind paraphrasing it? Thanks in advance. (By the way, I'm not the ... who downvoted.} – bof Jul 18 at 11:41
• The question is what is the difference between a "set" and a "class". In beginning classes they seem like they should be the exceedingly simple concept of they are just a collection of objects. They didn't have definitions because they were an intuitive idea. Then Bertrand Russell comes along and says "Take the set of all sets that don't contain themselves. Does that set contain itself or not?" And then suddenly ZFC comes up with exceedingly abstract and obscure restrictions on what a set can technically have. en.wikipedia.org/wiki/Zermelo%E2%80%93Fraenkel_set_theory – fleablood Jul 18 at 15:47
• .....The main thing is that although a set can be any collection in general and can collect sets as elements there is a basic rule that a set may not contain itself as an element or as an element in one of its elements. So "the set of all sets that don't contain themselves" is "the set of all sets" which can not exist. Paradox solved. But now we can't just freely grab whatever we want to make a "collection". And "classes" is part of the solution. Now ZFC is not something I've studied very much so I .... can't actually answer the question. – fleablood Jul 18 at 15:52

I'm assuming we're talking about set theory without classes, such as $$\mathsf{ZF}$$.

Neither the concept of classes nor of sets is really "defined" by axioms. Classes can be seen as an informal notion describing collections of elements of a model (of set theory) that satisfy some property, whereas sets are those classes that themselves are elements of the model.

So any collection of elements that contradicts the axioms of a given set theory isn't a set (since otherwise it would not be contained in any model, as you cannot model contradictory statements).

Furthermore, there are collections of elements that could be a set, but which consistently don't have to be a set. For example, inaccessible cardinals are sets that don't contradict the axioms of $$\mathsf{ZFC}$$, but there are models of $$\mathsf{ZFC}$$ in which no inaccessible cardinals exist (all of this assuming $$\mathsf{ZFC}$$ itself is consistent, of course).

Now Russell's class $$X=\{x\mid x\notin x\}$$ of sets not containing themselves does not have a lot to do with the Axiom of Regularity, except that it is equal to the set of all sets $$V$$ under assumption of the Axiom of Regularity: it implies that $$x\notin x$$ for all sets.

Also note that it makes no sense to find out whether $$X\in X$$ or $$X\notin X$$, since $$X$$ is a proper class, and thus not an element of a model: we can therefore not quantify over collections of proper sets, or use symbols denoting proper sets in our set theoretical language. Whenever you read people talking about proper classes, they are strictly speaking not talking about objects, but about the formulas that define the proper class; they are talking about properties.

You could view Russell's class as the property of not containing yourself, $$V$$ as the property of simply existing, the class of ordinals $$\mathrm{Ord}$$ as the property of being transitively well-ordered by $$\in$$, etc.

Finally note that Russell's class can not be a set regardless of the axioms that are used, since if $$X$$ is in the domain of discourse, then by pure first order logic we could derive both $$X\in X$$ and $$X\notin X$$, giving a logical contradiction (not per se a set-theoretical one). Russell's paradox points out a flaw in the idea of Naive Set Theory that any first-order definable collection is a set. As far as I'm aware, Russell's class predates the Axiom of Regularity by about a decade (and was interestingly enough first discovered by Zermelo, one of the founders of the axiomatic approach to set theory, before it was discovered by Russell).

Now in some set theories, such as NBG, we do allow proper classes to be part of the language. However, the way to do this, is by making a partition in the domain of discourse to outline which elements of the model are sets, and which ones are proper classes.

For example, in $$\mathsf{NBG}$$, a set is defined as any object in our universe that is contained in some other object (i.e. $$x$$ is a set if $$\exists y(x\in y)$$), and alternatively a proper class is any object that is not a set (thus $$\forall y(x\notin y)$$). Furthermore, the Axioms that apply to sets in $$\mathsf{ZFC}$$ still apply to those objects that are considered sets, but don't (necessarily) apply to the proper class objects.

As an axiom, any collection that can be defined using a formula that only quantifies over sets is a class, so Russell's class is an object in $$\mathsf{NBG}$$ Set Theory. However, as classes contained in classes are always sets, (and since we have the Axiom of Regularity), it is also still the case that Russell's class is equal to $$V$$.

• You can show how Russell's Paradox does not depend on any set-theoretic axioms by observing that is equivalent to: There does not exist a widget that dabbles all those and only those widgets that do not dabble themselves. – DanielWainfleet Jul 18 at 22:29
• I've learned from your answers that "Russell's class can't be a set." and Axiom of Regularity differ from each other. However, I don't know exactly what "informal" means. In ZFC, I've seen associating a predicate $\phi(x)$ with a class $\{x:\phi(x)\}$. When a predicate isn't associated with $V$, is the class assoicated with the predicate always a set? Is it not allowed because the class is an informal concept? – istist Jul 19 at 8:20
• @UsinJung Not only do they differ from each other, they have nothing to do with each other. As I said, and as Daniel above points out, Russell's class can't be a set, because of a logical contradiction, which has in itself nothing to do with set theory. What I mean by informal concept, is that informally we usually regard a class as some kind of object. However, formally, not every class is an object, since inside a formal language, we can only talk about the objects of our model. Not all classes that can be defined by a predicate are objects of such models (e.g. $V$ and Russell's class) ... – Vsotvep Jul 19 at 8:35
• (cont'd) So when I say that $x\in C$ for some class $C=\{y\mid \phi(y)\}$, what I really mean is that $\phi(x)$ holds. The 'formula' $x\in C$ is not expressible, since $C$ is not interpretable as an object in our model. I'm not sure what you mean that a predicate is associated with $V$. – Vsotvep Jul 19 at 8:37
• Perhaps you mean a proper class that is not dependent on self reference? In that case, here are some examples: (1) the class of all sets that are singletons, (2) the class of all ordinals, (3) the class of all cardinals, (4) the class of all hierarchies $V_\alpha$ in the Von Neumann universe. – Vsotvep Jul 19 at 8:41