TITLE: Bounding and dominating numbers ${\frak b}, {\frak d}$ via ultrafilters
QUESTION [7 upvotes]: Let $\omega^\omega$ denote the set of all functions $f:\omega\to\omega$ and suppose that ${\cal U}$ is a free ultrafilter on $\omega$. We write $f \leq_{\cal U} g$ if $$\{n\in\omega: f(n) \leq g(n)\}\in{\cal U}.$$ 
Similar to the usual bounding and dominating numbers, we define $${\frak b}_{\cal U} = \min\{|B|: B\subseteq \omega^\omega \land \forall f\in\omega^\omega \exists g\in B(g\not \leq_{\cal U} f)\},$$
and 
$${\frak d}_{\cal U} = \min\{|D|: D\subseteq \omega^\omega \land \forall f\in\omega^\omega \exists g\in D(f \leq_{\cal U} g)\}.$$
Do we have ${\frak b} = {\frak b}_{\cal U}$ and ${\frak d} = {\frak d}_{\cal U}$?

REPLY [10 votes]: The relation $\leq_{\cal U}$ is very well-studied: it is the order relation on the model of the hyperreals obtained by taking an ultrapower of $\mathbb{R}$ with respect to $\cal U$.
Letting $\mathbb{R}^*_{\cal U}$ denote this ultrapower, you are asking for the smallest size of an unbounded set in $\mathbb{R}^*_{\cal U}$ (which you call $\frak b_{\cal U}$) and the smallest size of a cofinal set in $\mathbb{R}^*_{\cal U}$ (which you call $\frak d_{\cal U}$).
As Wojowu points out, these cardinals are the same because $\leq_{\cal U}$ is a linear order on $\mathbb{R}^*_{\cal U}$. Also, this cardinal is bounded below by $\frak b$ and above by $\frak d$.
But one can say a bit more. Andreas Blass has proved (in this paper) that $\frak g \leq \frak b_{\cal U}$. Mike Canjar proved that, if you add lots of Cohen reals to a model of CH, then any uncountable regular cardinal $\leq \frak c$ is equal to $\frak b_{\cal U}$ for some free ultrafilter $\cal U$. In other words, he showed that the value of $\frak b_{\cal U}$ can depend not only on your model of set theory, but within a single model it can also depend on $\cal U$.
If you want to learn more, I suggest you look at Andreas's paper, and also his answer to this related question of Joel David Hamkins.