(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…

read moreThis conjecture has a seemingly trivial case when $latex \mathcal{A}$ is a principal filter. When I attempted to prove this seemingly trivial case I stumbled over a looking simple but yet unsolved problem: Let $latex U$ is a set. A filter (on…

read moreI conjectured certain formula for the complete lattice generated by a strong partitioning of an element of complete lattice. Now I have found a beautiful proof of a weaker statement than this conjecture. (Well, my proof works only in the case of…

read moreIn this post I defined strong partitioning of an element of a complete lattice. For me it was seeming obvious that the complete lattice generated by the set $latex S$ where $latex S$ is a strong partitioning is equal to $latex \left\{…

read moreI proposed this open problem for the next polymath project. Now I will consider some its special simple cases.

read moreI’ve given two different definitions for partitioning an element of a complete lattice (generalizing partitioning of a set). I called them weak partitioning and strong partitioning. The problem is whether these two definitions are equivalent for all complete lattices, or if are…

read moreLet $latex \mathfrak{A}$ is a complete lattice. Let $latex a\in\mathfrak{A}$. I will call weak partitioning of $latex a$ a set $latex S\in\mathscr{P}\mathfrak{A}\setminus\{0\}$ such that $latex \bigcup{}^{\mathfrak{A}}S = a \text{ and } \forall x\in S: x\cap^{\mathfrak{A}}\bigcup{}^{\mathfrak{A}}(S\setminus\{x\}) = 0$. I will call strong partitioning…

read moreEarlier I proposed finishing writing this manuscript as a polymath project. But the manuscript contains (among other) this conjecture which can be reasonably separated into an its own detached polymath project.

read moreLet $latex U$ is a set. A filter (on $latex U$) $latex \mathcal{F}$ is by definition a non-empty set of subsets of $latex U$ such that $latex A,B\in\mathcal{F} \Leftrightarrow A\cap B\in\mathcal{F}$. Note that unlike some other authors I do not require $latex…

read moreFor filters on sets defined equivalence relation being isomorphic. Posed some open problems like this: are every two nontrivial ultrafilters isomorphic?

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