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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 $\mathcal{B}\ne\Omega$?

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1 Answer 1

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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 $E$.

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