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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 $x\le y$ . 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 $\cf{P}$ , or $\cof{P}$ .
If $(T,\leq)$ is a totally ordered set, then it must contain a well-ordered cofinal subset which is order-isomorphic to $\cf{T}$ . Or, put another way, there is a cofinal function $f\colon\cf{T}\to T$ with the property that $f(x)<f(y)$ whenever $x<y$ .
For any ordinal $\beta$ we must have $\cf{\beta}\leq\beta$ , because the identity map on $\beta$ is cofinal. In particular, this is true for cardinals, so any cardinal $\kappa$ either satisfies $\cf{\kappa}=\kappa$ , in which case it is said to be regular, or it satisfies $\cf{\kappa}<\kappa$ , 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 $\cf{\aleph_0}=\aleph_0$ , 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 $\cf{\aleph_\omega}=\aleph_0$ . 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 $\cf{\aleph_\delta}=\cf{\delta}$ .
Let $\kappa$ be an infinite cardinal. It can be shown that $\cf{\kappa}$ 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 $\kappa<\kappa^{\cf{\kappa}}$ . Replacing $\kappa$ by $2^\kappa$ in this inequality we can further deduce that $\kappa<\cf{2^\kappa}$ . In particular, $\cf{2^{\aleph_0}}>\aleph_0$ , 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).
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