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


1 Answer 1


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 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 $\mathfrak{c}^+$.


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