# Maximal elements in the Rudin-Keisler ordering

Let $$\text{NPU}(\omega)$$ be the set of non-principal [ultafilters][1] 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 see that $$\leq_{RK}$$ is reflexive and transitive, but not anti-symmetric. Set $${\cal U}\simeq_{RK} {\cal V}$$ if $${\cal U}\leq_{RK}{\cal V}$$ and $${\cal V}\leq_{RK}{\cal U}$$. So $$\text{NPU}(\omega)/\simeq_{RK}$$ is a poset with the Rudin-Keisler order applied to equivalence classes.

It is known that $${\cal U}$$ is a minimal element of $$\text{NPU}(\omega)/\simeq_{RK}$$ if and only if $${\cal U}$$ is a Ramsey ultrafilter. (Ramsey ultrafilters do not necessarily exist.)

Question. Does $$\text{NPU}(\omega)/\simeq_{RK}$$ have maximal elements?

No. First, the Rudin–Keisler preorder is directed: for any ultrafilters $$\mathcal U$$ and $$\mathcal V$$ on $$\omega$$, the ultrafilter $$\mathcal U\times\mathcal V=\{X\subseteq\omega\times\omega:\{i\in\omega:\{j\in\omega:(i,j)\in X\}\in\mathcal V\}\in\mathcal U\}$$ on $$\omega\times\omega$$ is Rudin–Keisler above both $$\mathcal U$$ and $$\mathcal V$$, the maps $$f$$ being the projections.
Thus, a maximal element would in fact be a largest element $$\mathcal U$$. However, there are only $$2^\omega$$ functions $$\omega\to\omega$$, hence only $$2^\omega$$ ultrafilters $$\le_{RK}\mathcal U$$, whereas there are $$2^{2^\omega}$$ ultrafilters in total, hence there is no largest element.
• In fact, IIRC $\mathcal U\lneq_{RK}\mathcal U\times\mathcal U$ for all nonprincipal $\mathcal U$, but this requires a proof. – Emil Jeřábek Mar 14 '19 at 16:29