The cardinal characteristic $\mathfrak d$

Question

I am reading a chapter in a book of Andreas Blass which is called: "Combinatorial Cardinal Characteristics of the Continuum". In there, the cardinal characteristic $\mathfrak d$ is defined as folows:

Definition: A familly $\mathcal D \subseteq \omega^\omega$ is dominating, if for each $f \in \omega^\omega$, there is $g \in \mathcal D$ with $f \leq^*g$. The dominating number $\mathfrak d$, is the smallest cardinality of any dominating family, $\mathfrak d = min \{|\mathcal D|; \mathcal D \space is \space dominating\} $

Remark: $f \leq^*g$ means $f \leq g$ for all $\omega$, exapt maybe a finite set of points.

My question is: can I say that,$cf(\mathfrak d)$ $=\mathfrak d$.

The proof I was thinking of is this: Suppose that, $cf(\mathfrak d) < \mathfrak d$. Let $D$ be a dominating set with $|D|= \mathfrak d$. We assumed that $cf(\mathfrak d) < \mathfrak d$, so, there is a cofinal subset $A$ of $D$, with $|A|<|D|$. But, we know that $D$ is dominating and $A$ is not bounded in $D$ so, this implies that $A$ is dominating. contradicting the fact that $|D| = min \{|\mathcal D|; \mathcal D \space is \space dominating\}$

So far, I haven't encountered this claim so I guess it is not true (otherwise it probably would have been stated). So I guess some of the arguments I have stated are wrong. But, I can't see where. any help?

Thank you! Shir

Answer 1

I don't think your reasoning works, because you seem to be conflating two different orders and thus two different "bounded".

When you're speaking of a cofinal subset of $\mathfrak d$, the order you're speaking of is the standard well-order on $\mathfrak d$ viewed as an ordinal.

A priori, this has nothing to do with the dominance ordering $\le^*$ of the functions $\omega\to\omega$. You have no reason to assume that just because a function comes later than another in the enumeration of $D$ induced by $|D|=\mathfrak d$, it will also dominate the other function.

Answer 2

It is consistent to have $\operatorname{cf}(\mathfrak d)<\mathfrak d$. However we do know that $\operatorname{cf}(\mathfrak d)\geq\mathfrak b$.

To see a proof you can check Handbook of Set Theory, the chapter by Andreas Blass (chapter 6, "Combinatorial Cardinal Characteristics of the Continuum") has the following theorem of $\sf ZFC$.

$2.4$ Theorem. $\aleph_1\leq\operatorname{cf}(\mathfrak b)=\mathfrak b\leq\operatorname{cf}(\mathfrak d)\leq\mathfrak d\leq\mathfrak c$.

And the next theorem stating that any three uncountable cardinals satisfying the above inequality are consistently the values of $\frak b,d,c$. Therefore taking, for example $\mathfrak b=\aleph_1$ and $\mathfrak d=\aleph_{\omega_1}$ and $\mathfrak c=\aleph_{\omega_1+1}$ is a counterexample.