# ${\frak b}$ and ${\frak d}$ in the Rudin-Keisler preordering

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 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 $${\frak d}(\text{NPU}(\omega))$$?

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.
a) every maximal chain in the RK order has cardinality exactly $$\mathfrak{c}^+$$ (see Joseph van Name's answer 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).
Therefore the unbounded number is $$\mathfrak{c}^+$$.