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](https://en.wikipedia.org/wiki/Ultrafilter#Ultrafilters_on_.E2.84.98.28.CF.89.29) that ${\cal U}$ is a minimal element of $\text{NPU}(\omega)/\simeq\_{RK}$ if and only if ${\cal U}$ is a [Ramsey](https://en.wikipedia.org/wiki/Ultrafilter#Ultrafilters_on_.E2.84.98.28.CF.89.29) ultrafilter. (Ramsey ultrafilters do not necessarily exist.)


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



Maximal elements in the Rudin-Keisler ordering
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.


