Let $latex U$ 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 moreI have proved this conjecture: Theorem 1 If $latex {\mathfrak{F}}&fg=000000$ is the set of filter objects on a set $latex {U}&fg=000000$ then $latex {U}&fg=000000$ is the center of the lattice $latex {\mathfrak{F}}&fg=000000$. (Or equivalently: The set of principal filters on a set…

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 moreThere were much talking about writing math research articles collaboratively but no real action. I present probably the first real example of a research manuscript ready to be written collaboratively. I wrote the draft Filters on Posets and Generalizations which I present…

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 more