If $(Q,\leq)$ is any preordered set (that is, $\leq$ is a reflexive and transitive, but not necessarily anti-symmetric relation), then we say that $S\subseteq Q$ is


1. *unbounded* if for all $q\in Q$ there is $s\in S$ such that $s \not \leq q$, and
2. *dominating* if for all $q\in Q$ there is $d \in S$ such that $q\leq d$.


We let the *unbounded number* ${\frak b}(Q)$ and the *dominating number* ${\frak d}(Q)$ be the smallest cardinality of an unbounded, respectively dominating subset of $Q$.


Let $\text{NPU}(\omega)$ be the set of non-principal [ultafilters](https://en.wikipedia.org/wiki/Ultrafilter) on $\omega$. The *Rudin-Keisler preorder* on $\text{NPU}(\omega)$ is defined by
$${\cal U} \leq\_{RK} {\cal V} :\Leftrightarrow (\exists f:\omega\to\omega)(\forall U\in{\cal U}) f^{-1}(U)\in {\cal V} .$$ It is easy to verify that $\leq\_{RK}$ is a preorder.


**Question.** What are ${\frak b}(\text{NPU}(\omega))$ and ${\frak d}(\text{NPU}(\omega))$?



${\frak b}$ and ${\frak d}$ in the Rudin-Keisler preordering
Since every ultrafilter has at most $\mathfrak{c}$ predecessors in the Rudin-Keisler ordering, it follows the the relevant dominating number is $2^{\mathfrak{c}}$, the cardinality of the entire collection.


On the other hand, we have:


a) every maximal chain in the RK order has cardinality exactly $\mathfrak{c}^+$ (see Joseph van Name's [answer](https://mathoverflow.net/a/283899/18128) to one of your previous questions), and


b) every set of ultrafilters of size at most $\mathfrak{c}$ has an upper bound in the the RK-order (by his answer [here](https://mathoverflow.net/a/131122/18128)).


Therefore the unbounded number is $\mathfrak{c}^+$.


