Let $(P,\leq)$ be a poset. A subset $A\subseteq P$ is said to be cofinal in $P$ if for every $x\in P$ there is a $y\in A$ such that $x\leq y$. A function $f\colon X\to P$ is said to be cofinal if $f(X)$ is cofinal in $P$. The least cardinality of a cofinal set of $P$ is called the cofinality of $P$. Equivalently, the cofinality of $P$ is the least ordinal $\alpha$ such that there is a cofinal function $f\colon\alpha\to P$. The cofinality of $P$ is written $\operatorname{cf}(P)$, or $\operatorname{cof}(P)$.

If $(T,\leq)$ is a totally ordered set, then it must contain a well-ordered cofinal subset which is order-isomorphic to $\operatorname{cf}(T)$. Or, put another way, there is a cofinal function $f\colon\operatorname{cf}(T)\to T$ with the property that $f(x)<f(y)$ whenever $x<y$.

For any ordinal $\beta$ we must have $\operatorname{cf}(\beta)\leq\beta$, because the identity map on $\beta$ is cofinal. In particular, this is true for cardinals, so any cardinal $\kappa$ either satisfies $\operatorname{cf}(\kappa)=\kappa$, in which case it is said to be regular, or it satisfies $\operatorname{cf}(\kappa)<\kappa$, in which case it is said to be singular.

The cofinality of any totally ordered set is necessarily a regular cardinal.

$0$ and $1$ are regular cardinals. All other finite cardinals have cofinality $1$ and are therefore singular.

It is easy to see that $\operatorname{cf}(\aleph_{0})=\aleph_{0}$, so $\aleph_{0}$ is regular.

$\aleph_{1}$ is regular, because the union of countably many countable sets is countable. More generally, all infinite successor cardinals are regular.

The smallest infinite singular cardinal is $\aleph_{{\omega}}$. In fact, the function $f\colon\omega\to\aleph_{{\omega}}$ given by $f(n)=\omega_{n}$ is cofinal, so $\operatorname{cf}(\aleph_{\omega})=\aleph_{0}$. More generally, for any nonzero limit ordinal $\delta$, the function $f\colon\delta\to\aleph_{\delta}$ given by $f(\alpha)=\omega_{\alpha}$ is cofinal, and this can be used to show that $\operatorname{cf}(\aleph_{\delta})=\operatorname{cf}(\delta)$.

Let $\kappa$ be an infinite cardinal. It can be shown that $\operatorname{cf}(\kappa)$ is the least cardinal $\mu$ such that $\kappa$ is the sum of $\mu$ cardinals each of which is less than $\kappa$. This fact together with König’s theorem tells us that $\kappa<\kappa^{{\operatorname{cf}(\kappa)}}$. Replacing $\kappa$ by $2^{\kappa}$ in this inequality we can further deduce that $\kappa<\operatorname{cf}(2^{\kappa})$. In particular, $\operatorname{cf}(2^{{\aleph_{0}}})>\aleph_{0}$, from which it follows that $2^{{\aleph_{0}}}\neq\aleph_{\omega}$ (this being the smallest uncountable aleph which is provably not the cardinality of the continuum).