# Finite pre-orders embeddable in the Rudin-Keisler ordering

$$\DeclareMathOperator{\NPU}{\operatorname{NPU}}\DeclareMathOperator{\RK}{\,\mathrm{RK}}$$A pre-ordered set is a pair $$(P, \leq)$$ where $$P$$ is a set and $$\leq\subseteq P\times P$$ is a reflexive and transitive relation.

Let $$\NPU(\omega)$$ be the set of non-principal ultrafilters on $$\omega$$. The Rudin-Keisler pre-order on $$\NPU(\omega)$$ 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 see that $$\leq_{\RK}$$ is reflexive and transitive, but not anti-symmetric.

Question. Given a finite pre-ordered set $$(P,\leq)$$, is there a finite subset $$S\subseteq\NPU(\omega)$$ such that $$(P,\leq)$$ is isomorphic to $$S$$ with the pre-order inherited from $$\text{NPU}(\omega)$$?

• There are incomparable ultrafilters, and since the product operation gives you upper bounds, one might hope to find a sufficiently independent family of ultrafilters, whose products make a copy of a finite Boolean algebra in the RK order. This would embed any partial order into the RK order. For pre-orders, you can easily handle clusters of equivalent nodes, since any nontrivial permutation of $\omega$ makes an isomorphic ultrafilter, which is RK equivalent but not equal. So my strategy is: find a strong antichain of ultrafilters, meaning incomparable products are RK incomparable. – Joel David Hamkins Oct 31 at 21:41
• In particular, I claim that to embed every finite preorder it suffices to embed every finite partial order. – Joel David Hamkins Oct 31 at 21:49
• I agree with @Joel's strategy, and I'll add that under CH (or weaker assumptions like cov(category)=c), there are lots of non-isomorphic selective ultrafilters, and these constitute what Joel calls a strong antichain. – Andreas Blass Oct 31 at 22:10
• I wouldn't be surprised if the existence of an infinite strong antichain could be proved in ZFC, perhaps by an independent-sets argument. – Andreas Blass Oct 31 at 22:13
• @AndreasBlass Could you post an answer explaining why nonisomorphic selective ultrafilters necessarily form a strong antichain? – Joel David Hamkins Nov 1 at 9:41