Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Recently I heard a joke:

If an object exists, mathematicians call it a set and study it. But if an object does not exist, mathematicians call it a proper class and study it anyway.

I wonder, if there is any mathematical or philosophical meaning in it?

share|cite|improve this question
a possibly relevant definition:… – Dan Rust Aug 16 '13 at 22:43
This very much reminds me of the field with one element; there are strong indications that it is worthwhile to study, so one looks for a different setting in which to define fields that allows for this sort of field while not breaking anything. – Karl Kronenfeld Aug 16 '13 at 22:52
One should maybe also comment that "doesn't exists" doesn't necessarily mean you can view it as a class. The cases in which an object doesn't make sense, according to the underlying logic for example, will usually be rules out not to be suitable to be talked about at all. – NikolajK Aug 16 '13 at 23:29
@NickKidman Isn't the existence of, for example, the class of all ordinals ruled out by the same logical rules in the same way, when an assumption of its existence leads to a plain contradiction? How can we distinguish between cases "does not exist at all, so don't mention it anymore" and "well, formally does not exist, but kinda 'exists' as a proper class"? – Լ.Ƭ. Aug 16 '13 at 23:54
@Լ.Ƭ.: The point is that a proper class is still definable with a first-order formula. – Asaf Karagila Aug 17 '13 at 1:23
up vote 22 down vote accepted

Let me start by saying that yes. There is some mathematical meaning to this joke.

Sets, as you may know, are the objects of interest in set theory. For example $\sf ZFC$, which is probably the "default" set theory in the eyes of many. One of the most beautiful parts of modern set theory is that we can use it as a foundation for mathematics. That is, we can, with only the $\in$ relation at our disposal, build and describe pretty much all the constructions in mathematics within set theory. Okay, that's inaccurate, but if we limit ourselves to classical mathematics, or things like basic analysis and so on, then the answer is positive. Yes, we can do that just with $\sf ZFC$. I am not going to go into details on how we can do that, but let's assume that we agree on that for now.

If so, we can treat the mathematical universe, the collection of all objects in mathematics as a universe of sets which adheres to the axioms of $\sf ZFC$. Meaning all our objects are sets.

So what does it mean to exist? If $\varphi(x)$ is a formula in the language of set theory, then we can ask whether or not $\exists x\varphi$ is a true sentence in the universe. But when is it true? It is true when there is some object in our universe which satisfies the formula $\varphi$. In this case, we say that there exists such $x$.

But wait, object in the universe? We already agreed that objects in the universe are sets. So if there exists an object we can say that there exists a set. So when we have a certain property $\psi$, we can define from it the following formula $$\varphi(x):=\forall y(y\in x\leftrightarrow\psi(y))$$

Now we can ask, does there exists $x$ which satisfies $\varphi$? If there is one, then $x$ is a set, because existing is synonymous with being an object in the universe, which itself is synonymous with being a set.

However, the paradoxes of set theory tell us that not every such $\varphi$ can be realized in the universe. If $\psi(y)$ is simply $y\notin y$, then $\lnot(\exists x\varphi(x))$. There is no set which includes all the sets which are not members of themselves. But we can still ask meaningful questions about the collection of objects which do not include themselves. Is this collection closed under unions? intersections? taking power sets?

In order to handle these sort of constructs we define the notion of a class. Classes are just collections which are defined by a formula. Sometimes those are sets, but sometimes not. If a class is not a set, we say it is a proper class.

Examples of proper classes are the collection of all sets; all non-empty sets; all groups; all functions; all the singletons; and so on and so forth. Each of these can be described using a formula. And indeed, in many ways, a correct way to look at classes is to consider them as syntactical objects. When we talk about classes we really just talk about formulas.

The joke, if so, says that when mathematician try to analyze something, if it exists then it's a set -- because it's an object of our universe; but if it doesn't exist then it's a proper class.

But the joke, of course, being just a joke, hides a deep and important point. For a collection to be a class it needs to be definable by a formula. So while existence is synonymous with being a set, non-existence is not synonymous with being a proper class.

For example, if we denote by $V$ the proper class of all sets, then $\{V\}$ is neither a set nor a proper class. It doesn't exist. But wait, what does it mean? We just wrote it. So how can it not exist? Well, there are a lot of fine points in a formal proof, but the shortest way to argue is to prove that elements of classes are sets. Therefore, if $\{V\}$ is a class, we have that $V$ is a set, but we already know it's not.

Therefore the collection $\{V\}$ is not definable by a formula.

Another example is that we can prove that there is no function from $\Bbb N$ onto $\Bbb R$. This doesn't mean that such function is a proper class. No, it simply cannot be defined.

To read more:

  1. Sets and classes
  2. Difference between a class and a set
  3. Why bother proving that a class is a set?
  4. difference between class, set , family and collection
share|cite|improve this answer
Would it be correct (at least formally) to replace a phrase "... So, functions with given properties do not exist. QED" in a proof with "... So, function with given properties are proper classes. QED"? – Լ.Ƭ. Aug 17 '13 at 0:29
No. non-existence is not synonymous with "being a proper class". – Asaf Karagila Aug 17 '13 at 0:30
Could you include in your answer some details about what's the difference between them? – Լ.Ƭ. Aug 17 '13 at 0:33
There. I rewrote the answer, and I hope it will be better. – Asaf Karagila Aug 17 '13 at 1:11
@Mariano: Grothendieck universes, for example. – Asaf Karagila Aug 17 '13 at 1:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.