We can define a very simple cardinal characteristic in the following way. Recall the relation $\leq^*$ on $\omega^\omega$ defined by $x\leq^* y$ iff $x(i)\leq y(i)$ for all but finitely many $i$. For $x,y\in\omega^\omega$, say that $x$ and $y$ are comparable, denoted by $x\parallel y$, if either $x\leq^* y$ or $y\leq^* x$.

Let $\omega^{*\omega}$ denote the set of sequences of natural numbers that diverge to $\infty$. Define

$\mathfrak{cp}=\min\{|C|:C\subseteq\omega^{*\omega},\ (\forall x\in\omega^{*\omega})(\exists y\in C)\, x\parallel y\}$.

It is not hard to see that $\mathfrak{b}\leq\mathfrak{cp}\leq\mathfrak{d}$. On the other hand, since Cohen forcing can add a real in $\omega^{*\omega}$ incompatible with all ground model reals in $\omega^{*\omega}$, we can prove that $\mathrm{cov}(\mathcal{M})\leq\mathfrak{cp}$.

Now, to the questions:

1) Is $\mathfrak{cp}=\mathfrak{d}$?

2) What happens to $\mathfrak{cp}$ in Miller's model?

Perhaps this cardinal invariant is something trivial, but I haven't been able to figure it out so far.


1 Answer 1


I claim that $\mathfrak{c}\mathfrak{p}= \mathfrak{d}$.

For any $x\in \omega^{*\omega}$ define its "inverse" $x'$ by $x'(n) = \min \{k\mid\forall j\ge k : x(j)>n\}$. If $x$ grows very fast, then $x'$ grows very slowly.

In particular we have $x\le ^* y \Rightarrow y'\le^* x'$, and $x\le^* x''$.

Define $x^+$ as follows: $x^+(2n)=x(n)$, and $x^+(2n+1)=x'(n)$. So if $x$ grows very fast, then $x^+$ grows very fast on the even numbers, but very slowly on the odd numbers.

Now assume that $C$ is a witness for $\mathfrak c\mathfrak p\le \lambda$. Let $D$ be the closure of $C$ under some natural operations (such as $x\mapsto x'$), then the set $D$ will witness $\mathfrak d\le \lambda$.

Proof: Let $x\in \omega^{*\omega}$ be strictly increasing. Find $y$ in $C$ such that $x^+ \|y$.

Case 1: $x^+\le^* y$. Then let $x(n)=x^+(2n) \le y(2n)$ for almost all $n$, so the function $n\mapsto y(2n)$ dominates $x$.

Case 2: $y \le^* x^+$. Then for almost all $n$ we have $y(2n+1)\le x^+(2n+1)=x'(n)$. Consider the function $z(n)=y(2n+1)$: We have $z\le^* x'$, so $x\le x''\le^* z'$. So all we need to do here is to ensure that $D$ is closed under composition with $n\mapsto 2n+1$ and under $z\mapsto z'$.

  • 1
    $\begingroup$ This is great! One correction. Actually we have $x''\leq x$: according to your definition, $x'(x(i))>i$ and, since $x'$ is monotone increasing, $x'(k)>i$ for all $k\geq x(i)$, so $x''(i)\leq x(i)$. However, in the case when $x$ is increasing and $x>0$, we get $x''=x$. The reason is because $x'(x(i)-1)\leq i$. So your proof still works. $\endgroup$
    – dragoon
    Jul 4, 2019 at 9:37

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