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\begin{document}

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\section*{regular cardinal}

\hypertarget{contents}{}\section*{{Contents}}\label{contents}

\noindent\hyperlink{idea}{Idea}\dotfill \pageref*{idea} \linebreak
\noindent\hyperlink{definitions}{Definitions}\dotfill \pageref*{definitions} \linebreak
\noindent\hyperlink{finite}{Finite regular cardinals?}\dotfill \pageref*{finite} \linebreak
\noindent\hyperlink{in_weak_foundations}{In weak foundations}\dotfill \pageref*{in_weak_foundations} \linebreak
\noindent\hyperlink{examples}{Examples}\dotfill \pageref*{examples} \linebreak
\noindent\hyperlink{regular_cardinals}{Regular cardinals}\dotfill \pageref*{regular_cardinals} \linebreak
\noindent\hyperlink{singular_cardinals}{Singular cardinals}\dotfill \pageref*{singular_cardinals} \linebreak


\hypertarget{idea}{}\subsection*{{Idea}}\label{idea}

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 set|uncountable]]) regular cardinals exist.

\hypertarget{definitions}{}\subsection*{{Definitions}}\label{definitions}

An [[infinite set|infinite]] [[cardinal]] $\kappa$ is a \textbf{regular cardinal} if it satisfies the following equivalent properties:

\begin{itemize}%
\item no set (in a [[material set theory]]) of cardinality $\kappa$ is the [[union]] of fewer than $\kappa$ sets of cardinality less than $\kappa$.


\item no set (in a [[structural set theory]]) of cardinality $\kappa$ is the [[disjoint union]] of fewer than $\kappa$ sets of cardinality less than $\kappa$.


\item 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 ${|P|} \lt \kappa$.


\item the [[category]] $\Set_{\lt\kappa}$ of sets of cardinality $\lt\kappa$ has all [[colimits]] (or just all [[coproducts]]) of size $\lt\kappa$.


\item the [[cofinality]] of $\kappa$ is equal to $\kappa$.



\end{itemize}
A cardinal that is not regular is called \textbf{singular}.

\hypertarget{finite}{}\subsubsection*{{Finite regular cardinals?}}\label{finite}

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:

\begin{enumerate}%
\item $Set_{\lt\kappa}$ is closed under iterated disjoint unions ($\biguplus_i A_i$).
\item $Set_{\lt\kappa}$ is closed under the nullary iterated disjoint union (the [[singleton]]).
\item $Set_{\lt\kappa}$ is closed under binary disjoint unions ($A \uplus B$).
\item $Set_{\lt\kappa}$ is closed under the nullary disjoint union (the [[empty set]]).

\end{enumerate}
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.

\hypertarget{in_weak_foundations}{}\subsubsection*{{In weak foundations}}\label{in_weak_foundations}

Thinking of a regular cardinal \emph{as} a cardinal number makes the most sense using the [[axiom of choice]]. Otherwise, we probably want to think of it as a \emph{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 \emph{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]].

\hypertarget{examples}{}\subsection*{{Examples}}\label{examples}

\hypertarget{regular_cardinals}{}\subsubsection*{{Regular cardinals}}\label{regular_cardinals}

\begin{itemize}%
\item 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.


\item The [[successor]] of any infinite cardinal, such as $\aleph_0$, is a regular cardinal. (This requires the [[axiom of choice]].) In the case of $\aleph_0$, 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 $\lambda^+$.



\end{itemize}
\hypertarget{singular_cardinals}{}\subsubsection*{{Singular cardinals}}\label{singular_cardinals}

\begin{itemize}%
\item $\aleph_\omega = \bigcup_{n\in \mathbb{N}} \aleph_n$ is singular, more or less by definition, since $\aleph_n\lt\aleph_\omega$ and ${|\mathbb{N}|} = \aleph_0 \lt\aleph_\omega$.


\item 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 cardinal|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'.


\item Assuming the consistency (with [[ZFC]]) of `there is a [[proper class]] of [[strongly compact cardinal|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.)



\end{itemize}
[[!redirects regular cardinal]] [[!redirects regular cardinals]] [[!redirects regular cardinal number]] [[!redirects regular cardinal numbers]] [[!redirects regular collection of cardinal numbers]] [[!redirects regular collections of cardinal numbers]] [[!redirects regular set of cardinal numbers]] [[!redirects regular sets of cardinal numbers]] [[!redirects regular class of cardinal numbers]] [[!redirects regular classes of cardinal numbers]]

[[!redirects singular cardinal]] [[!redirects singular cardinals]] [[!redirects singular cardinal number]] [[!redirects singular cardinal numbers]] [[!redirects singular collection of cardinal numbers]] [[!redirects singular collections of cardinal numbers]] [[!redirects singular set of cardinal numbers]] [[!redirects singular sets of cardinal numbers]] [[!redirects singular class of cardinal numbers]] [[!redirects singular classes of cardinal numbers]]



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