Cofinality of the bounding number and the dominating number Let's define a relation $\leq^*$ on $\omega^{\omega}$ by $$f\leq^* g \iff \exists m\in\omega\ \forall n>m\ f(n)\leq g(n),\ \ f,g\in\omega^{\omega}.$$
Then let $$\mathfrak{b}=\min\{|\mathcal{F}|:\ \mathcal{F}\subseteq\omega^{\omega}\ \wedge\ \neg\exists\ g\in\omega^{\omega}\ \ \forall f\in\omega^{\omega}\ f\leq^* g\},$$ $$\mathfrak{d}=\min\{|\mathcal{F}|:\ \mathcal{F}\subseteq\omega^{\omega}\ \wedge\ \forall\ g\in\omega^{\omega}\ \ \exists f\in\omega^{\omega}\ g\leq^* f\}.$$
It has to proved that $i)\ \mathfrak{b}$ is regular and $cf(\mathfrak{b})>\omega$,  $\ ii)\ cf(\mathfrak{d})\geq\mathfrak{b}$.
Before I started doing this exercise I didn't know about dominating and bounding numbers  - I've just read about them a couple minutes ago. I've learned that, hmm, they exist.
All I know so far is that, as it can be easily noticed, $\mathfrak{b}\leq\mathfrak{d}$. But I don't know what's next... Should I think about the sets that satisfy the above-stated contraints and notice any cofinal subsets have the same cardinality? It seems intuitive (though is probably false for the dominating number). Or perhaps there's a different ingenious way to do it? Please, guys, help.
 A: HINT: You’ll want to show that if $\mathscr{F}\subseteq{^\omega\omega}$ is countable, there is a $\in{^\omega\omega}$ such that $f<^*g$ for each $f\in\mathscr{F}$; just construct $g(n)$ by recursion on $n$.
To show that $\mathfrak{b}$ is regular, let $\mathscr{F}$ be an unbounded family of cardinality $\mathfrak{b}$, and suppose that $\operatorname{cf}\mathfrak{b}=\kappa<\mathfrak{b}$. Then there are $\mathscr{F}_\xi\subseteq\mathscr{F}$ for $\xi<\kappa$, such that $\mathscr{F}=\bigcup_{\xi<\kappa}\mathscr{F}_\xi$, and $|\mathscr{F}_\xi|<\mathfrak{b}$ for each $\xi<\kappa$. But $|\mathscr{F}_\xi|<\mathfrak{b}$ means that $\mathscr{F}_\xi$ is bounded; use this to produce an unbounded family of cardinality less than $\mathfrak{b}$.
To show that $\operatorname{cf}\mathfrak{d}\ge\mathfrak{b}$, let $\kappa=\operatorname{cf}\mathfrak{d}$, and let $\mathscr{F}=\bigcup_{\xi<\kappa}\mathscr{F}_\xi$ be a dominating family of cardinality $\mathfrak{d}$ such that $|\mathscr{F}_\xi|<\mathfrak{d}$ for each $\xi<\kappa$. None of the families $\mathscr{F}_\xi$ is dominating, so for each $\xi<\kappa$ there is a $g_\xi\in{^\omega\omega}$ such that $g_\xi\not<^*f$ for each $f\in\mathscr{F}_\xi$. If $\kappa<\mathfrak{b}$, then $\{g_\xi:\xi<\kappa\}$ is bounded, say by $g$. Now use the fact that $\mathscr{F}$ is a dominating family to get a contradiction.
