Splitting Frechet filter into two proper filters

Question

Let $\Omega$ be a Frechet filter (=cofinite filter) on an infinite set.

Do there exist proper filters $\mathcal{A}$ and $\mathcal{B}$ such that $\mathcal{A}\cap\mathcal{B} = \Omega$ and $\mathcal{A}\ne\Omega$ and $\mathcal{B}\ne\Omega$?

Answer 1

Sure. Let $E=F\cup G$ where $F$ and $G$ are disjoint infinite sets. Define $$\mathcal A=\{X\subseteq E:F\setminus X\text{ is finite}\},$$$$\mathcal B=\{X\subseteq E:G\setminus X\text{ is finite}\}.$$Then $\mathcal A$ and $\mathcal B$ are proper filters, $\mathcal A\ne\mathcal B$, and $\mathcal A\cap\mathcal B=\Omega$ is the cofinite filter on $E$.