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The Rudin-Blass ordering of ultrafilters. (English) Zbl 0911.04001

The Rudin-Blass order on ultrafilters over \(\omega\) is a special case of the Rudin-Keisler order: \(\mathcal U \leq_{RB} \mathcal V\) iff there is a finite-to-one function \(f: \omega \rightarrow \omega\) with \(\mathcal V \supset f^{\leftarrow}(\mathcal U)\). The strong Rudin-Blass order, \(\leq^+_{RB}\) requires that \(f\) be monotone as well.
This paper explores \(\leq_{RB}\) and \(\leq_{B}^+\). It turns out that \(\leq^+_{RB}\) sheds light on \({\mathbb R}^*\). This connection in turn is used to show that \(\leq_{RB}^+\) is tree-like, i.e., each \(\mathcal U^{\downarrow^+} = \{\mathcal V: \mathcal V \leq^+_{RB}\}\) is linear. Unlike the Rudin-Keisler order, two ultrafilters have an \(RB\) lower bound iff they have the \(RB\) upper bound; this is used to show that \(\leq_{RB} \;\neq \;\leq^+_{RB}\).
About half the paper explores the cardinal invariants of the linear orderings \((\mathcal U^{\downarrow^+},\geq^+_{RB})\) (note the reversal of the order). Note that, as with the \(RK\)-ordering, \(Q\)-points are minimal. So the interesting \(\mathcal U\)’s from the standpoint of cardinal invariants are those with no \(Q\)-points below them. The main results are that there is an ultrafilter with no \(Q\)-point \(RB\)-beneath it, and for such a \(\mathcal U\) the bounding and dominating numbers of both \((\mathcal U^{\downarrow^+}, \geq_{RB}^+)\) and \((\mathcal U^{\downarrow} = \{\mathcal V: \mathcal V \leq_{RB} \mathcal U\}, \geq_{RB})\) are equal and equal cof\((^{\omega}\omega,\leq_{\mathcal U})\).

MSC:

03E05 Other combinatorial set theory
54D40 Remainders in general topology
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