Let $\kappa\geq \aleph_0$ be a cardinal, and suppose that ${\cal U}$ is a non-principal ultrafilter on $\kappa$. We regard ${\cal U}$ as a poset $({\cal U}, \subseteq)$.

Suppose that there are posets $P, Q$ such that ${\cal U} \cong P\times Q$. Does this imply one of $P, Q$ consists of one point only?


No. Every nonprincipal ultrafilter $U$, considered as a partial under $\subseteq$, is a nontrivial product order. To see this, suppose that $U$ is a nonprincipal ultrafilter on $\kappa$. Partition $\kappa=A\sqcup B$ into two sets with $A\in U$ and $B$ nonempty. Every $X\in U$ can be written as $X=(X\cap A)\sqcup (X\cap B)$, and furthermore, $X\subseteq Y$ just in case $(X\cap A)\subseteq (Y\cap A)$ and $(X\cap B)\subseteq (Y\cap B)$. Let $P=U\upharpoonright A=\{ X\subseteq A\mid X\in U\}$ and $Q=P(B)=\{X\mid X\subseteq B\}$. These are both nontrivial and $\langle U,\subseteq\rangle$ is isomorphic to the product order $\langle P,\subseteq\rangle\times\langle Q,\subseteq\rangle$ by the map $X\mapsto (X\cap A,X\cap B)$.

Indeed, you don't even need the ultrafilter to be non-principal, provided $\kappa\geq 3$. The reason is that if $\kappa\geq 3$, then you can partition $\kappa=A\sqcup B$ where $A\in U$, $B$ is nonempty and $A$ has at least two points. In this case, both $P$ and $Q$ again will have at least two elements each, and the rest of the argument is as before.

(Meanwhile, if $\kappa=1$ or $\kappa=2$, then $U$ has only one or two elements, respectively, and so it is not a nontrivial product.)


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.