I'm confused about the notion of the cofinality of a cardinal. Since I think the source of the confusion is the Von Neumann cardinal assignment, my first question is:

Question 0. Is there an article or book that purposely distinguishes between a cardinal and its associated initial ordinal?

Anyway, for the duration of this question, lets adopt this distinction. Thus we have two order-isomorphic proper classes, $\mathrm{Ord}$ and $\mathrm{Card}$. And although we have retained the identification $[0,\alpha)=\alpha$ for $\alpha \in \mathrm{Ord}$, we have abandoned it for cardinals. Additionally, given cardinal numbers $\mu$ and $\nu$, lets write $[\mu,\nu)$ for the set of all cardinals $\kappa$ with $\mu \leq \kappa < \nu$.

Furthermore, let $x \mapsto \underline{x}$ denote the unique order isomorphism $\mathrm{Ord}\rightarrow \mathrm{Card}$, so for example: if $\omega$ is the least infinite ordinal, then $\underline{\omega}$ is the least infinite cardinal. Also, let $\eta : \mathrm{Card} \rightarrow \mathrm{Ord}$ denote the proper class function that maps every cardinal number to its initial ordinal. And finally, for every subset $A$ of a well-ordered set, lets write $\mathrm{ord}(A)$ for the unique ordinal that is order-isomorphic to $A$. So in general, we have that $\mathrm{ord}(A) \in \mathrm{Ord}$.

Now my understanding is that the cofinality $\mathrm{cf}(\alpha)$ of an ordinal $\alpha$ is defined as the least ordinal $\beta$ such that there exists a cofinal subset of $\alpha$, call it $B$, such that $\mathrm{ord}(B)=\beta$.

Question 1. Under these definitions, how does one define the cofinality of a cardinal number $\kappa$? Is it:

1. The ordinal $\mathrm{cf}\,\mathrm{ord}[0,\kappa)$
2. The cardinal $|\mathrm{cf}\,\mathrm{ord}[0,\kappa)|$
3. The ordinal $\mathrm{cf}\,\eta({\kappa})$
4. The cardinal $|\mathrm{cf}\,\eta({\kappa})|$
5. Something else???

I honestly can't work it out.

• The cofinality of an ordinal is always an initial ordinal, hence, may be regarded as a cardinal. The definition of the cofinality of a cardinal is, in your notation, $\operatorname{cf} \eta (\alpha)$. May 4, 2013 at 7:49
• @ZhenLin, thank you!!! May 4, 2013 at 7:50
• How are options 1 and 3 different? (I assume the $\alpha$ in 3 and 4 are meant to be $\kappa$) May 4, 2013 at 7:50
• @AndresCaicedo In his notation, $[0, \aleph_1)$ denotes $\{ 0, 1, \ldots, \aleph_0 \}$, which is countable. May 4, 2013 at 7:51
• Ah, I see, thanks! OK. Then option 3 is what you want as definition. And option does not always give the correct answer. May 4, 2013 at 7:53

One can formulate two forms of cofinalities:

1. Ordinal cofinality, which is the least order type of an unbounded set.
2. Cardinal cofinality, which is the least cardinality of a partition that all its parts are smaller then the original cardinal.

Some facts that are useful to know:

• Both versions have the property that the resulting ordinal is always a regular cardinal. That is, the cofinality of the cofinality is the original cofinality.

• If $\delta$ is an initial ordinal, i.e. a cardinal, then both cofinalities are equal.

We are not interested not usually interested in the cofinality of $\text{ord}[0,\kappa)$, because it will usually not coincide with either the cofinalities above. For example, $\text{ord}[0,\omega_1)=\omega+1$, and its cofinality as an ordinal is $1$. That is very uninteresting, and it doesn't tell us much. Whereas knowing that $\omega_1$ is a regular cardinal tells us plenty.

And as for your question $0$. If the same context uses both ordinal and cardinal arithmetics, then there is benefit to using $\aleph_\alpha$ instead of $\omega_\alpha$, to discern the cardinal and ordinal arithmetics. But in many cases people use $\kappa,\lambda,\mu,\nu$ for cardinals and $\alpha,\beta,\gamma,\delta$ for ordinals so it's possible to understand from the context of the letters, Shelah is a prime example and many of his papers begin with this. (Of course they always say whether or not a letter denotes a cardinal or an ordinal.)

• With regards to the two notions of cofinality, is there a third? Namely, the least cardinality of an unbounded subset? And which, if either, of 1 and 2 does this coincide with? Also.... I think dot point 2 is meant to read: "If $\delta$ is an initial ordinal." And thank you for your answer. May 5, 2013 at 14:35
• Thank you for pointing out the typo. The third notion is really just the first. Because the least order type is always a cardinal and it is trivially the least cardinality of an unbounded set. May 5, 2013 at 14:43
• Gotcha. And I suppose that for all ordinals (indeed, for all partially ordered sets), definition 1 is always less than or equal to definition 2? May 5, 2013 at 16:46
• Hm. For ordinals the two are equal, so yes. I'm not sure what happens in the case of partial orders. But in the case of partial orders unbounded sets need not be well-ordered, even if the set is directed. So talking about order-types is slightly awkward. In that case we talk about cardinality. May 5, 2013 at 17:07
• Are you sure they're equal for ordinals? It seems to me (and I may have completely misunderstood) that $\mathrm{cf_1}(\omega+1)=1,$ while $\mathrm{cf_2}(\omega+1)=\aleph_0$. May 5, 2013 at 21:13

Here is an ordinal-free way of defining cofinality of a cardinal.

The cofinality of a cardinal $\kappa$ is the least cardinal $\lambda$ for which there exists a set $X$ of cardinality $\lambda$ such that every member of $X$ is a set of cardinality $< \kappa$ and $\coprod_{x \in X} x$ has cardinality equal to $\kappa$.

A slightly cleaner definition is available if you permit cardinal arithmetic:

The cofinality of a cardinal $\kappa$ is the least cardinal $\lambda$ for which there exists a set $X$ of cardinality $\lambda$ and a function $f : X \to [0, \kappa)$ such that $\sum_{x \in X} f (x) = \kappa$.

Of course, these both compute the same answer as the standard definition (exercise!), but have the advantage of making sense in contexts where the well-ordering principle is not available.

• Actually, I think the previous comment I left here may have been a bit silly. Please disregard it. May 4, 2013 at 8:37
• Just to confirm: by $[0,\kappa)$ you're meaning the set of all cardinals $\nu$ such that $\nu < \kappa$? May 5, 2013 at 16:52
• Yes. But if $[0, \kappa)$ is the set of all ordinals of cardinality less than $\kappa$, then the answer is the same, provided the well-ordering principle is available. May 5, 2013 at 17:10