Cofinality

Properties

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.

Cofinality of ordinals and other well-ordered sets

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.

Regular and singular ordinals

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.

Cofinality of cardinals

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

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 ${\displaystyle \mathrm{cf}(\kappa )\le \kappa .}$ The cofinality of any totally ordered set is regular, so ${\displaystyle \mathrm{cf}(\kappa )=\mathrm{cf}(\mathrm{cf}(\kappa )).}$

Using Kőnig's theorem, one can prove ${\displaystyle \kappa <{\kappa }^{\mathrm{cf}(\kappa )}}$ and ${\displaystyle \kappa <\mathrm{cf}\left(2^{\kappa }\right)}$ 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, $${\displaystyle {\mathrm{\aleph }}_{\omega }=\bigcup _{n<\omega }{\mathrm{\aleph }}_n,}$$ the ordinal number ω being the first infinite ordinal, so that the cofinality of ${\displaystyle {\mathrm{\aleph }}_{\omega }}$ is card(ω) = ${\displaystyle {\mathrm{\aleph }}_0.}$ (In particular, ${\displaystyle {\mathrm{\aleph }}_{\omega }}$ is singular.) Therefore, $${\displaystyle 2^{{\mathrm{\aleph }}_0}\ne {\mathrm{\aleph }}_{\omega }.}$$

(Compare to the continuum hypothesis, which states ${\displaystyle 2^{{\mathrm{\aleph }}_0}={\mathrm{\aleph }}_1.}$)

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

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