# A simple cardinal characteristic associated with $\omega^\omega$

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.

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 $$z\mapsto z'$$.

• 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. Jul 4, 2019 at 9:37