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

  • $\begingroup$ Ping him. $\endgroup$ – Git Gud Jan 12 '14 at 17:00
  • $\begingroup$ oh.. I didn't know he was here... sure but how do I do that? $\endgroup$ – topsi Jan 12 '14 at 17:03
  • $\begingroup$ Type: @Andreas Blass Can you please help me out with this? $\endgroup$ – Git Gud Jan 12 '14 at 17:05
  • $\begingroup$ @GitGud: Stop messing with the young ones. Pinging someone arbitrarily doesn't work. $\endgroup$ – Asaf Karagila Jan 12 '14 at 17:06
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    $\begingroup$ The problem with your proof is that, even though $|A|<|D|$, you do not know that a function $f$ is dominated by a function in $A$, because the well-ordering you used to list $D$ does not necessarily relate to the ordering $<^*$. $\endgroup$ – Andrés E. Caicedo Jan 12 '14 at 17:12

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.

  • $\begingroup$ I see.. yes, this was the confusion I made. Thank you! $\endgroup$ – topsi Jan 12 '14 at 17:18

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.

  • $\begingroup$ Ok i will go over these theorems. Thanks! $\endgroup$ – topsi Jan 12 '14 at 17:17
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    $\begingroup$ Note that the second theorem uses forcing. So if you're unfamiliar with the technique, it might be difficult to fully understand. But then again, pretty much all the consistency results about cardinal characteristics require forcing (except "Assume $\sf CH$, then everything is equal to $\aleph_1$", and so equal)... So trying to understand a consistency result without knowing forcing is a difficult task to begin with. $\endgroup$ – Asaf Karagila Jan 12 '14 at 17:19
  • $\begingroup$ I see. ok thanks for pointing that out for me. It's true. I am not familier with forcing. But I have also noticed that, He talks about forcing only later in that chapter. Do you know of any, short introduction for forcing that will be sufficient for this theorems? Thank you! $\endgroup$ – topsi Jan 12 '14 at 17:23
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    $\begingroup$ @Shir, there is a nice proof of the particular case $\mathfrak b=\aleph_1$ and $\mathfrak d=\kappa$ for any $\kappa$ of uncountable cofinality, in Kunen's new book on set theory. $\endgroup$ – Camilo Arosemena-Serrato Jan 12 '14 at 22:55

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