# ${\frak b}$ and ${\frak d}$ defined with $\leq$ instead of $\leq^*$

Let $\omega^\omega$ denote the collection of all functions $f:\omega\to\omega$. For $f,g\in\omega$ we define

• $f\leq g$ if $f(n)\leq g(n)$ for all $n\in\omega$;
• $f\leq^* g$ if there is $N\in\omega$ such that $f(n)\leq g(n)$ for all $n\in \omega$ with $n\geq N$.

The cardinals ${\frak b}, {\frak d}$ are defined as follows:

• ${\frak b} = \min\{|S|: S \subseteq \omega^\omega \text{ and } \forall f\in\omega^\omega \exists s\in S(s \not \leq^* f)\}$,
• ${\frak d} = \min\{|S|: S \subseteq \omega^\omega \text{ and } \forall f\in\omega^\omega \exists s\in S(f \leq^* s)\}$.

Let us define ${\frak b}', {\frak d}'$ using $\leq$ instead of $\leq^*$. Can it be proved that ${\frak b}' = {\frak c}$ (which would imply ${\frak d}' = {\frak c}$)? If not: which of ${\frak b}' < {\frak b}, {\frak b}' > {\frak b}, {\frak d}' < {\frak d}, {\frak d}' > {\frak d}$ is consistent?

Using "$\le$" instead of "$\le^*$," I claim $\mathfrak{b}'=\omega$ and $\mathfrak{d}'=\mathfrak{d}$.

To see $\mathfrak{b}'=\omega$, let $f_n$ be the constant function $x\mapsto n$; then setting $S=\{f_n: n\in\omega\}$, there is clearly no function which is $\ge$ every element of $S$.

To see $\mathfrak{d}'=\mathfrak{d}$, suppose $S$ is a set of functions with $$(*)\quad\forall f\in\omega^\omega\exists s\in S(f\le^*s);$$ I'll build an $S'$, with $\vert S'\vert=\vert S\vert$, such that $$(**)\quad\forall f\in\omega^\omega\exists s\in S'(f\le s).$$

To do this we proceed as follows. For $\sigma\in\omega^{<\omega}$ and $f\in\omega^\omega$, let $f^\sigma$ be the function gotten by replacing the length-$\vert\sigma\vert$ initial segment of $f$ by $\sigma$; that is, $f^\sigma(n)=\sigma(n)$ for $n<\vert\sigma\vert$, and $f^\sigma(n)=f(n)$ otherwise.

Now, given an $S$ with the property $(*)$, let $$S'=\{f^\sigma: f\in S, \sigma\in\omega^{<\omega}\}.$$ Clearly $\vert S'\vert=\omega\times\vert S\vert=\vert S\vert$; I claim $S'$ has property $(**)$.

To see this, fix $g\in \omega^\omega$. Since $S$ has property $(*)$, we may find an $f\in S$ with $g\le^*f$. Then by definition of $\le^*$ there is a $\sigma\in\omega^\omega$ such that $g\le f^\sigma$. But this $f^\sigma$ is in $S'$.

${\frak b}'=\aleph_0$ since you can take $S$ to be the collection of constant functions.

${\frak d}' = {\frak d} + \aleph_0$ since you can take any family $S$ realizing $\frak d$ and close it under finite differences, i.e., let $S'$ consist of all functions that are eventually equal to some function in $S$.

• You don't need the "$+\aleph_0$," since $\mathfrak{d}$ is infinite. – Noah Schweber May 18 '15 at 7:38
• @NoahS true, I was in a hurry there. The argument I gave shows $\frak d\le\frak d'\le \frak d+\aleph_0$ and then an even easier one gives $\frak d\ge\aleph_0$. – Bjørn Kjos-Hanssen May 18 '15 at 7:50