# Bounding and dominating numbers ${\frak b}, {\frak d}$ via ultrafilters

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 ${\frak d} = {\frak d}_{\cal U}$?

• – saf Feb 12 '16 at 10:27

For any two functions $f,g$ the sets $\{n\in\omega:f(n)\leq g(n)\}$ and $\{n\in\omega:g(n)\leq f(n)\}$ cover $\omega$, so one of them must be in $\mathcal U$. Hence we have a dichotomy $f\leq_{\mathcal U}g$ or $f\leq_{\mathcal U}g$.

It follows that every unbounded family is dominating: if $B$ is an unbounded with respect to $\leq_{\mathcal U}$family, then for any $f\in\omega^\omega$ there is $g\in B$ such that $f\not\geq_{\mathcal U}g$ and hence $f\leq_{\mathcal U}g$, hence $B$ is dominating. Since every dominating set is clearly unbounded, it follows that $\frak b_{\mathcal U}=\frak d_{\mathcal U}$. However, it is consistent that $\frak b<\frak d$, so it is consistent that $\frak b_{\mathcal U}\neq\frak b$ or $\frak d_{\mathcal U}\neq\frak d$.

We can say something more: every set unbounded with respect to $\leq_{\mathcal U}$ is unbounded with respect to $\leq^*$ (because $f\not\leq_{\mathcal U}g\Rightarrow f\not\leq^*g$) and every set dominating with respect to $\leq^*$ is dominating with respect to $\leq_{\mathcal U}$ (because $f\leq^*g\Rightarrow f\leq_{\mathcal U}g$), so $\frak b\leq\frak b_{\mathcal U}=\frak d_{\mathcal U}\leq\frak d$, hence it's consistent that one of these inequalities is strict.

Of course assuming $CH$ we would have $\frak b=\frak b_{\mathcal U}=\frak d_{\mathcal U}=\frak d=\frak c$, hence the only open question left is whether we can (consistently) have $\frak b<\frak b_{\mathcal U}=\frak d_{\mathcal U}<\frak d$, which is a question I can't answer.

• Very nice and clear answer, thanks! – Dominic van der Zypen Feb 12 '16 at 9:49

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.

• Fantastic, and thanks for pointing me to the hyperreals and the references! – Dominic van der Zypen Feb 12 '16 at 20:44