A regular cardinal is is a cardinal number that is ‘closed under union’. The category of sets bounded by a regular cardinal has several nice properties, making it a universe that is handy for some purposes but falls short of being a Grothendieck universe. Unlike Grothendieck universes (which are based on inaccessible cardinals? rather than regular cardinals), it is easy to prove that (even uncountable) regular cardinals exist.
An infinite cardinal $\kappa$ is a regular cardinal if it satisfies the following equivalent properties:
no set (in a material set theory) of cardinality $\kappa$ is the union of fewer than $\kappa$ sets of cardinality less than $\kappa$.
no set (in a structural set theory) of cardinality $\kappa$ is the disjoint union of fewer than $\kappa$ sets of cardinality less than $\kappa$.
given a function $P \to X$ (regarded as a family of sets $\{P_x\}_{x\in X}$) such that ${|X|} \lt \kappa$ and ${|P_x|} \lt \kappa$ for all $x \in X$, then ${|P|} \lt \kappa$.
the category $\Set_{\lt\kappa}$ of sets of cardinality $\lt\kappa$ has all colimits (or just all coproducts) of size $\lt\kappa$.
the cofinality of $\kappa$ is equal to $\kappa$.
A cardinal that is not regular is called singular.
Traditionally, one requires regular cardinals to be infinite. This clause may be removed, in which case $0$, $1$, and $2$ are all regular cardinals.
Other modifications of the definition which are equivalent for infinite cardinals may include some of $0$, $1$, and $2$ but not all. For instance, if we regard an indexed disjoint union $\sum_{i\in\lambda} \alpha_i$ as a binary operation taking as input $\lambda$ and a $\lambda$-indexed family $\alpha$, then closure under this binary operation as in the above definition also entails closure under the ternary version $\sum_{i\in\lambda} \sum_{j\in \mu_i} \alpha_{i,j}$, and so on. The unary version is simply the identity operation, whereas the nullary version will always output the singleton set $1$. (This can be seen by thinking in terms of trees of uniform finite height, or remembering that a dependent sum includes a binary cartesian product as a special case, so a nullary dependent sum should at least include a nullary product.) Thus, from this perspective, $2$ is a regular cardinal, but $0$ and $1$ are not. In applications for which this perspective is the relevant one, such as familial regularity and exactness, one may more precisely be interested in an arity class rather than a regular cardinal.
We may rule out all three finite regular cardinals by additionally generalising from indexed disjoint unions to finitary disjoint unions.
Then in terms of $Set_{\lt\kappa}$, the (potential) conditions on a (possibly finite) regular cardinal are as follows:
These are all variations on the theme of closure under disjoint unions.
Clauses (2–4) hold of all infinite cardinals, while clauses (2&3) together force $\kappa$ to be greater than any finite cardinal. However, if we require only clauses (1&2), then $2$ is a regular cardinal.
Thinking of a regular cardinal as a cardinal number makes the most sense using the axiom of choice. Otherwise, we probably want to think of it as a collection of cardinals, or equivalently think of it as the category $Set_{\lt\kappa}$.
From this perspective, a regular cardinal is a full subcategory of $Set$ that is closed under taking quotient objects and satisfies the condition on $Set_{\lt\kappa}$ above. We can then recover $\kappa$ as the smallest cardinal number greater than every cardinal in $Set_{\lt\kappa}$, if we accept the axiom of choice.
Note that if we require only conditions (1&2) on $Set_{\lt\kappa}$, then (even classically), $\{1\}$ is an acceptable (and finite) regular collection of cardinals, even though it is not actually of the form $Set_{\lt\kappa}$ for any cardinal number $\kappa$.
In the absence of the axiom of choice, it is not clear that there exist arbitrarily large regular cardinals. Thus in weaker foundations, regular cardinals (or “regular sets of cardinals”) can be regarded as a large cardinal property.
At least if “regular cardinal” has its classical meaning of a particular ordinal, then the statement that there exist arbitrarily large regular cardinals is independent of ZF; in fact it is consistent with ZF that all uncountable cardinals are singular. A foundational axiom which is related to the existence of regular cardinals (but considers them as sets with various closure properties, rather than cardinal numbers) is the regular extension axiom.
The first (infinite) regular cardinal is $\aleph_0 = {|\mathbb{N}|}$, because a set with cardinality less than $\aleph_0$ is a finite set, and a finite union of finite sets is still a finite set.
The successor of any infinite cardinal, such as $\aleph_1$, is a regular cardinal. (This requires the axiom of choice.) In the case of $\aleph_1$, this means that a countable union of countable sets is countable. Note that this implies that there exist arbitarily large regular cardinals: for any cardinal $\lambda$ there is a greater regular cardinal, namely $\lambda^+$.
$\aleph_\omega = \bigcup_{n\in \mathbb{N}} \aleph_n$ is singular, more or less by definition, since $\aleph_n\lt\aleph_\omega$ and ${|\mathbb{N}|} = \aleph_0 \lt\aleph_\omega$.
More generally, any limit cardinal that can be “written down by hand” should be singular, since if it were regular then it would be weakly inaccessible?, and the existence of weakly inaccessible cardinals cannot be proven in ZFC (if $ZFC$ is consistent). We say ‘should’ rather than ‘must’, since there are exceptions, but they are sort of cheating: one (definable with Choice) is ‘the smallest limit cardinal that is regular if and only if some weakly innaccessible cardinal exists’.
Assuming the consistency (with ZFC) of ‘there is a proper class of strongly compact cardinals?’, it is consistent with $ZF$ that every uncountable cardinal is singular (and in fact every infinite well-orderable cardinal has cofinality $\aleph_0$), a result due to Moti Gitik?. (Of course this conclusion is inconsistent with $ZFC$, in which many uncountable cardinals, starting with $\aleph_1$, are regular.)