What are some examples of classes that are not sets?

Solution 1:

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.

Solution 2:

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.

Solution 3:

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.

Solution 4:

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?