Is the Rudin-Keisler ordering a continuous relation?

Question

If $X, Y$ are topological, and $R\subseteq X\times Y$ we say that $R$ is continuous (from $X$ to $Y$) if for every $V\subseteq Y$ with $V$ open, we have $$R^{-1}(V) = \{u\in U: \exists v\in V:(u,v)\in R\}$$ is open in $X$.

Let $\text{NPU}(\omega)$ be the set of non-principal ultafilters on $\omega$. The Rudin-Keisler preorder on $\text{NPU}(\omega)$ is defined by $${\cal U} \leq_{RK} {\cal V} :\Leftrightarrow (\exists f:\omega\to\omega)(\forall U\in{\cal U}) f^{-1}(U)\in {\cal V} $$ for ${\cal U}, {\cal V}\in \text{NPU}(\omega)$. It is easy to see that $\leq_{RK}$ is reflexive and transitive, but not anti-symmetric.

Let $[\omega]^\omega$ denote the collection of infinite subsets of $\omega$. For $A\in[\omega]^\omega$, let $$u(A) = \{{\cal U}\in \text{NPU}(\omega): A\in {\cal U}\}.$$ On $\text{NPU}(\omega)$ we consider the topology generated by the the collection $\{u(A): A\in[\omega]^\omega\}$.

Question. Is $\leq_{\text{RK}}$ a continuous relation on $\text{NPU}(\omega)$ with this topology?

Answer 1

I assume that, in your definition of continuity of relations, the unspecified $V$ is intended to be an open subset of $Y$. With this assumption, the answer to your question is affirmative. If $V$ is any nonempty open subset of NPU$(\omega)$ then $(\leq_{RK})^{-1}(V)$ is all of NPU$(\omega)$.

To prove it, first notice that it suffices to consider the case where $V=u(A)$ for some infinite, co-infinite subset of $\omega$. Then, given any ultrafilter $\mathcal U$, fix some infinite, co-infinite $B\in\mathcal U$, and let $g:\omega\to\omega$ be a bijection sending $B$ onto $A$. Then $g(\mathcal U)\in u(A)$ and $g(\mathcal U)$ is RK-equivalent (i.e., isomorphic) to $\mathcal U$.