After reading about Russell's paradox, I see that the set of all sets does not exist, so instead it is called a class.

What other commonly known classes exist that are not sets? I know the class of all singleton sets is not a set, because you can unite that class to get the class of all sets. This seems like the class of all sets of a fixed finite size is not a set either, correct?

What are some examples of other 'things' that can't be gathered up and put into a set?


To add on Jason's good answer, operations which are defined on all the sets can be seen as functions, which are usually treated as a collection of $n$-tuples (which are also sets)

So when the domain of a function is a class then the function itself is a proper class.

For example $\{\langle x,P(x)\rangle | x\in V\}$ is the Power Set operation, and $\{\langle x,y,x\cap y\rangle | x,y\in V\}$ is the function which takes two sets and returns their intersection.

Both of these functions are proper classes.

  • $\begingroup$ Thanks Asaf. So for your first class, I'll call it $A$, the elements of the function have the form $\{\{x\},\{x,P(x)\}\}$ for every set $x$. This means that for all $x\in V$, $x\in\cup\cup A$? So if I assumed $A$ was a set, then I would get $V=\cup\cup A$ is a set too, a contradiction? Is that how you would show $A$ is a proper class and not a set, and likewise for any relation $R$ with $\text{dom}R=V$? $\endgroup$ – user6891 Feb 11 '11 at 7:11
  • $\begingroup$ If $A$ was a set, then by the axiom of replacement you'd have that the universe is a set as well. $\endgroup$ – Asaf Karagila Feb 11 '11 at 7:12
  • $\begingroup$ Hmm, I haven't learned about the replacement axiom, but I will look into it. Without it though, is my reasoning correct, even if it's not so elegant? I really wish I could upvote, but the system won't let me. $\endgroup$ – user6891 Feb 11 '11 at 7:18
  • $\begingroup$ @Rayf: Yes, your reasoning is also correct. $\endgroup$ – Asaf Karagila Feb 11 '11 at 7:33
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    $\begingroup$ I think it's worth pointing out that while the function is a proper class, it's image can still be a set. $\endgroup$ – Jason DeVito Feb 11 '11 at 13:08

First, any description which doesn't uniquely specify the elements of a set tends to give a proper class.

For example, a group is an set together with operations satisfying a certain collection of axioms. Since these axioms don't pin down the elements of the group, one might expect the collection of all groups to not be a set.

This holds in a similar fasion for topological spaces, rings, vector spaces, fields, manifolds, metric spaces, and many of the other commonly studied objects.

This also applies to your "collection of all singletons" example, and similarly to any sort of "collection of all sets of a fixed size".

But just because a description DOES pin down the elements doesn't mean you have a set. For example, the collection of all ordinal numbers is not a set, despite the fact that if there is an order preserving bijection between two ordinals $\alpha$ and $\beta$, then one must have $\alpha = \beta$ as sets.

Similarly, the collection of all cardinals doesn't form a set.

  • $\begingroup$ Thank you Jason. So if I had a collection of all sets of fixed size $n$, for any set $x$, I could make a set with $n-1$ other sets and $x$ itself, right? Then if I took the union of this collection, I could pull out all sets $x$, and find that the class of all sets is a subset of this collection, showing this collection is not a set? Does the same conclusion hold if the sets have a fixed infinite cardinality? $\endgroup$ – user6891 Feb 11 '11 at 7:15
  • $\begingroup$ Yes - every set is an element of a $\kappa$ element set for any (possibly infinite cardinal) number $\kappa$. The rest of the argument works just as you say, with unioning giving you the whole universe (which is not a set by Russel's Paradox). $\endgroup$ – Jason DeVito Feb 11 '11 at 13:05

Since I like to represent alternative set theories, I thought I'd list some examples of proper classes in NF, a set theory that does have a universal set (and therefore provides examples besides the "you can form $V$ from it" types).

  • The Russel class, obviously.

  • $\iota$ -- the singleton function $x\mapsto\{x\}$. In NF if $\iota$ exists you can prove Cantor's paradox. Restrictions of this class to many sets, however, are sets.

  • The class of all strongly Cantorian ordinals. In NF we have a set of all ordinals because many ordinals are non-Cantorian (their power set is not necessarily bigger, and their image under $\iota$ is smaller). But the strongly Cantorian ordinals, if a set, would themselves me suceptible to the Burali-Forti paradox.

  • Where $\Omega$ is the order type of all ordinals, there are special downward closed classes from $\Omega$ that have no least member. Since NF proves every subset of ordinals has a least member, these form proper classes.

  • It is consistent with NF that there are finite proper classes of $\mathbb{N}$. (What actually happens in such models is that $\mathbb{N}$ has initial segments that contain non-standard elements.)

To add a clarification prompted by Asaf's comment below, the above are informal collections definable in NF, but not objects of the theory. In a close relative of NF called ML, there are true proper classes that are objects of the theory, having members but not members of anything. The above examples are also proper classes in this sense in ML, but the last bullet takes on a different meaning in ML:

  • The existence of a finite "proper class" subclass of $\mathbb{N}$ in NF translates in ML to the assertion that $\mathbb{N}$ is a proper class.
  • $\begingroup$ It might be worth pointing out that classes in $\sf NF$ mean the same thing as in $\sf ZF$, definable (with parameters) collections. The difference in the theories is that the comprehension schemata are limited in different ways. $\endgroup$ – Asaf Karagila Dec 21 '13 at 15:10
  • $\begingroup$ @AsafKaragila: Good point, thank you. There is a version of NF with real "can quantify over them" proper classes, and many of these are still proper classes in that theory (ML). But things get even weirder in ML. $\endgroup$ – Malice Vidrine Dec 21 '13 at 15:30

I consider the following examples canonical.

  1. The class of all cardinal numbers does not form a set.

  2. Let $T$ denote a first-order theory with at least one infinite model. Then using the upward Lownheim-Skolem theorem as well as the above observation, we see that the class of all models of $T$ can never form a set. This includes the class of groups, the class all fields, the class of all lattices, etc.

  3. Function classes into proper classes never form a set. For example, let $\mathbf{Grp}$ denote the class of all groups. Then the class of all functions $\mathbb{N} \rightarrow \mathbf{Grp}$ does not form a set. (i.e. the class of all sequences of groups does not form a set.)

Anyway, in practice, this is all much less devastating than it seems. For example, all that (2) really says is that given an adequate set-theoretic universe $M$ and a $T \in M$ such that $M$ believes that $T$ is a first-order theory, the set of all $G \in M$ such that $M$ believes that $G$ is a model of $T$ does not correspond to any particular element of $M$. Well, so be it; the set of all such $G \in M$ a perfectly well-defined subset of $M$, so who really cares that its not internalizable to an element?


In two-sorted first order set theory with proper classes (such as NBG) it is an axiom: $$ \forall x\, \exists X\, \forall z\, [ z \in x \Leftrightarrow z \in X ] $$ where lower-case letters denote type-0 objects (sets) and upper-case letters denote type-1 objects (classes).

This would mean that there are coclasses for every set: $$ \forall x\, \exists \bar X \, \forall z\, [ z \in x \Leftrightarrow z \notin \bar X\, ] $$ Thus, the elements NOT in a given set, forms a proper class.

  • $\begingroup$ I don't see how this answers the question. $\endgroup$ – Asaf Karagila Dec 21 '13 at 14:31
  • $\begingroup$ @AsafKaragila Edited to add an explicit conclusion. $\endgroup$ – Karl Damgaard Asmussen Dec 21 '13 at 14:38
  • $\begingroup$ Karl, indeed the complement of a set is never a set itself. To see that note that if $A$ and $B$ are sets then $A\cup B$ is a set. $\endgroup$ – Asaf Karagila Dec 21 '13 at 14:40
  • $\begingroup$ Karagila, Thank you, also I just realized that no set can contain a set which contains the complement of it's elements (sp?) and the complementary class would obviously contain the set itself. $\endgroup$ – Karl Damgaard Asmussen Dec 21 '13 at 14:46

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