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.

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    $\begingroup$ In fact, IIRC $\mathcal U\lneq_{RK}\mathcal U\times\mathcal U$ for all nonprincipal $\mathcal U$, but this requires a proof. $\endgroup$ – Emil Jeřábek Mar 14 '19 at 16:29

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