# Is the Rudin-Keisler ordering a continuous relation?

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?

• What is $V$ in the first sentence? Is "For every open subset $V$ of $Y$" missing somewhere? – YCor Jan 10 at 16:47
• Thanks for your comment, and apologies for my mistake. $V$ is supposed to mean an open subset of $Y$, as you suggest. I corrected it. – Dominic van der Zypen Jan 10 at 16:55

## 1 Answer

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 $$\mathcal U$$.