• The cofinality of a partially ordered set with greatest element is 1 as the set consisting only of the greatest element is cofinal (and must be contained in every other cofinal subset).
  • In particular, the cofinality of any nonzero finite ordinal, or indeed any finite directed set, is 1, since such sets have a greatest element.
• Every cofinal subset of a partially ordered set must contain all maximal elements of that set. Thus the cofinality of a finite partially ordered set is equal to the number of its maximal elements.
  • In particular, let ${\displaystyle A}$ be a set of size ${\displaystyle n,}$ and consider the set of subsets of ${\displaystyle A}$ containing no more than ${\displaystyle m}$ elements. This is partially ordered under inclusion and the subsets with ${\displaystyle m}$ elements are maximal. Thus the cofinality of this poset is ${\displaystyle n}$ choose ${\displaystyle m.}$
• A subset of the natural numbers ${\displaystyle \mathbb {N} }$ is cofinal in ${\displaystyle \mathbb {N} }$ if and only if it is infinite, and therefore the cofinality of ${\displaystyle \aleph _{0}}$ is ${\displaystyle \aleph _{0}.}$ Thus ${\displaystyle \aleph _{0}}$ is a regular cardinal.
• The cofinality of the real numbers with their usual ordering is ${\displaystyle \aleph _{0},}$ since ${\displaystyle \mathbb {N} }$ is cofinal in ${\displaystyle \mathbb {R} .}$ The usual ordering of ${\displaystyle \mathbb {R} }$ is not order isomorphic to ${\displaystyle c,}$ the cardinality of the real numbers, which has cofinality strictly greater than ${\displaystyle \aleph _{0}.}$ This demonstrates that the cofinality depends on the order; different orders on the same set may have different cofinality.

If ${\displaystyle A}$ admits a totally ordered cofinal subset, then we can find a subset ${\displaystyle B}$ that is well-ordered and cofinal in ${\displaystyle A.}$ Any subset of ${\displaystyle B}$ is also well-ordered. Two cofinal subsets of ${\displaystyle B}$ with minimal cardinality (that is, their cardinality is the cofinality of ${\displaystyle B}$) need not be order isomorphic (for example if ${\displaystyle B=\omega +\omega ,}$ then both ${\displaystyle \omega +\omega }$ and ${\displaystyle \{\omega +n:n<\omega \}}$ viewed as subsets of ${\displaystyle B}$ have the countable cardinality of the cofinality of ${\displaystyle B}$ but are not order isomorphic). But cofinal subsets of ${\displaystyle B}$ with minimal order type will be order isomorphic.

The cofinality of an ordinal ${\displaystyle \alpha }$ is the smallest ordinal ${\displaystyle \delta }$ that is the order type of a cofinal subset of ${\displaystyle \alpha .}$ The cofinality of a set of ordinals or any other well-ordered set is the cofinality of the order type of that set.

Thus for a limit ordinal ${\displaystyle \alpha ,}$ there exists a ${\displaystyle \delta }$-indexed strictly increasing sequence with limit ${\displaystyle \alpha .}$ For example, the cofinality of ${\displaystyle \omega ^{2}}$ is ${\displaystyle \omega ,}$ because the sequence ${\displaystyle \omega \cdot m}$ (where ${\displaystyle m}$ ranges over the natural numbers) tends to ${\displaystyle \omega ^{2};}$ but, more generally, any countable limit ordinal has cofinality ${\displaystyle \omega .}$ An uncountable limit ordinal may have either cofinality ${\displaystyle \omega }$ as does ${\displaystyle \omega _{\omega }}$ or an uncountable cofinality.

The cofinality of 0 is 0. The cofinality of any successor ordinal is 1. The cofinality of any nonzero limit ordinal is an infinite regular cardinal.

A regular ordinal is an ordinal that is equal to its cofinality. A singular ordinal is any ordinal that is not regular.

Every regular ordinal is the initial ordinal of a cardinal. Any limit of regular ordinals is a limit of initial ordinals and thus is also initial but need not be regular. Assuming the axiom of choice, ${\displaystyle \omega _{\alpha +1}}$ is regular for each ${\displaystyle \alpha .}$ In this case, the ordinals ${\displaystyle 0,1,\omega ,\omega _{1},}$ and ${\displaystyle \omega _{2}}$ are regular, whereas ${\displaystyle 2,3,\omega _{\omega },}$ and ${\displaystyle \omega _{\omega \cdot 2}}$ are initial ordinals that are not regular.

The cofinality of any ordinal ${\displaystyle \alpha }$ is a regular ordinal, that is, the cofinality of the cofinality of ${\displaystyle \alpha }$ is the same as the cofinality of ${\displaystyle \alpha .}$ So the cofinality operation is idempotent.

If ${\displaystyle \kappa }$ is an infinite cardinal number, then ${\displaystyle \operatorname {cf} (\kappa )}$ is the least cardinal such that there is an unbounded function from ${\displaystyle \operatorname {cf} (\kappa )}$ to ${\displaystyle \kappa ;}$ ${\displaystyle \operatorname {cf} (\kappa )}$ is also the cardinality of the smallest set of strictly smaller cardinals whose sum is ${\displaystyle \kappa ;}$ more precisely $${\displaystyle \operatorname {cf} (\kappa )=\min \left\{|I|\ :\ \kappa =\sum _{i\in I}\lambda _{i}\ \land \forall i\in I\colon \lambda _{i}<\kappa \right\}.}$$

That the set above is nonempty comes from the fact that $${\displaystyle \kappa =\bigcup _{i\in \kappa }\{i\}}$$ that is, the disjoint union of ${\displaystyle \kappa }$ singleton sets. This implies immediately that ${\displaystyle \operatorname {cf} (\kappa )\leq \kappa .}$ The cofinality of any totally ordered set is regular, so ${\displaystyle \operatorname {cf} (\kappa )=\operatorname {cf} (\operatorname {cf} (\kappa )).}$

Using König's theorem, one can prove ${\displaystyle \kappa <\kappa ^{\operatorname {cf} (\kappa )}}$ and ${\displaystyle \kappa <\operatorname {cf} \left(2^{\kappa }\right)}$ for any infinite cardinal ${\displaystyle \kappa .}$

The last inequality implies that the cofinality of the cardinality of the continuum must be uncountable. On the other hand, $${\displaystyle \aleph _{\omega }=\bigcup _{n<\omega }\aleph _{n},}$$ the ordinal number ω being the first infinite ordinal, so that the cofinality of ${\displaystyle \aleph _{\omega }}$ is card(ω) = ${\displaystyle \aleph _{0}.}$ (In particular, ${\displaystyle \aleph _{\omega }}$ is singular.) Therefore, $${\displaystyle 2^{\aleph _{0}}\neq \aleph _{\omega }.}$$

(Compare to the continuum hypothesis, which states ${\displaystyle 2^{\aleph _{0}}=\aleph _{1}.}$)

Generalizing this argument, one can prove that for a limit ordinal ${\displaystyle \delta }$ $${\displaystyle \operatorname {cf} (\aleph _{\delta })=\operatorname {cf} (\delta ).}$$

On the other hand, if the axiom of choice holds, then for a successor or zero ordinal ${\displaystyle \delta }$ $${\displaystyle \operatorname {cf} (\aleph _{\delta })=\aleph _{\delta }.}$$

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