# Must $\mathfrak{sd} = \mathfrak{d}$?

This should be fun.

Let $\mathfrak{sd}$ be the least cardinal $\kappa$ such that there exists $\langle s_i : i < \kappa \rangle$ satisfying the following.

(1) Each $s_i: \omega^{\omega} \to \omega^{\omega}$.

(2) For every $n$, knowing $n$ bits of input gives you $n+1$ bits of output for each $s_i$ - i.e., for every $x, y \in \omega^{\omega}$, if $x \upharpoonright n = y \upharpoonright n$ then $s_i(x) \upharpoonright (n+1) = s_i(y) \upharpoonright (n+1)$.

(3) For every $x \in \omega^{\omega}$ there is some $i < \kappa$ such that $s_i(x)$ dominates $x$, i.e., $|\{n< \omega: s_i(x)(n) \leq x(n)\}| < \omega$.

Must $\mathfrak{sd} = \mathfrak{d} =$ least size of a family $F \subseteq \omega^{\omega}$ such that $(\forall x \in \omega^{\omega})(\exists y \in F)(|\{n < \omega: y(n) \leq x(n)\}| < \omega)$?

Here's some history.

• I am sorry. Hopefully all the typos are fixed now. – hot_queen May 15 '15 at 23:56

Temporarily consider one fixed "strategy" $s_i$, consider one fixed natural number $n$, consider one sequence $t$ of $n$ natural numbers, and consider the set $M(i,n,t)$ of all the functions $x:\omega\to\omega$ such that $x(k)=t(k)$ for all $k<n$ and $x(k)\leq s_i(x)$ for all $k\geq n$. I claim that the functions $x\in M(i,n,t)$ are all majorized (everywhere, not just eventually) by a single function $f_{i,n,t}$. Namely, let $f_{i,n,t}$ agree with $t$ at the first $n$ arguments $k$, and then, for larger $k$ define $f_{i,n,t}(k)$ inductively as follows. Consider all the finite sequences of length $k-1$ that are majorized by the initial segment of $f_{i,n,t}$ of length $k-1$; apply $s_i$ to all those finite sequences, and let $f_{i,n,t}(k)$ be larger than all the (finitely many) outputs produced by $s_i$.
Now un-fix $i$, $n$, and $t$. If the functions $s_i$ are as in your definition of $\mathfrak{sd}$, then every $x:\omega\to\omega$ is in $M(i,n,t)$ for some $i<\kappa$, some $n\in\omega$ and some $t\in{}^n\omega$. So the functions $f_{i,n,t}$ constitute a dominating family. Since there are only countably many $n$"s and $t$'s, this dominating family has size only $\mathfrak{sd}$. So $\mathfrak d\leq\mathfrak{sd}$.
The reverse inequality is trivial as you can take all the $s_i$'s to be constant.
• In "apply $s_i$ to all those finite sequences, I was tacitly thinking of $s_i$, by virtue of your requirement 2, as taking finite sequences of length $k-1$ as inputs and producing sequences of length $k$ as outputs, and in particular producing as outputs the last components of those sequences. – Andreas Blass May 16 '15 at 2:10
• Although I'm responsible for a few multi-character names of cardinal characteristics, I would advise against combinations like $\mathfrak{sd}$ that can be easily read as the product of $\mathfrak{s}$ and $\mathfrak{d}$ (thereby making the question really easy.) – Andreas Blass May 16 '15 at 2:12
• $~$Thanks! $~~$ – hot_queen May 18 '15 at 20:00