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.



A simple cardinal characteristic associated with $\omega^\omega$
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 $z\mapsto z'$.


