Conjecture Let $latex a$ and $latex b$ are filters on a set $latex U$. Then $latex a\cap b = \{U\} \Rightarrow \\ \exists A,B\in\mathcal{P}U: (\forall X\in a: A\subseteq X \wedge \forall Y\in b: B\subseteq Y \wedge A \cup B = U).$ [corrected]…

read moreLet $latex {U}&fg=000000$ is a set. A filter $latex {\mathcal{F}}&fg=000000$ (on $latex {U}&fg=000000$) is a non-empty set of subsets of $latex {U}&fg=000000$ such that $latex {A, B \in \mathcal{F} \Leftrightarrow A \cap B \in \mathcal{F}}&fg=000000$. Note that unlike some other authors I…

read more(In a past version of this article I erroneously concluded that our main conjecture follows from join-closedness of $latex {Z (D \mathcal{A})}&fg=000000$.) Let $latex {U}&fg=000000$ is a set. A filter $latex {\mathcal{F}}&fg=000000$ (on $latex {U}&fg=000000$) is a non-empty set of subsets of…

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