Filters on Posets and Generalizations: Submitted to another math journal
I submitted to “Topology” math journal by email the manuscript Filters on Posets and Generalizations. In the email I asked them to confirm receipt of the email as soon as they receive it. Until now there were no response from “Topology” math journal. So I count them unresponsive and submitted the same work to an […]
Filters on Posets and Generalizations – preprint
I submitted a preprint of the “Filters on Posets and Generalizations” article for peer-review and publication in Topology journal. The current version of this article is located at this URL.
Co-separability of filter objects – solved
I solved a problem earlier formulated in this blog post. A solution (of a slightly more general problem) can be found at this wiki page.
Chain-meet-closed sets on complete lattices
Let $latex \mathfrak{A}$ is a complete lattice. I will call a filter base a nonempty subset $latex T$ of $latex \mathfrak{A}$ such that $latex \forall a,b\in T\exists c\in T: (c\le a\wedge c\le b)$. I will call a chain (on $latex \mathfrak{A}$) a linearly ordered subset of $latex \mathfrak{A}$. Now as a part my research of […]
Open problem: co-separability of filter objects
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] This conjecture can be equivalently reformulated in terms of filter objects: Conjecture Let […]
Open problem: Pseudodifference of filters
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 $latex {U}&fg=000000$ such that $latex {A, B \in \mathcal{F} \Leftrightarrow A \cap B \in \mathcal{F}}&fg=000000$. Note that unlike some other authors I do not require $latex {\emptyset \notin \mathcal{F}}&fg=000000$. I will call the set of […]
Exposition: Complementive filters are complete lattice
(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 $latex {U}&fg=000000$ such that $latex {A, B \in \mathcal{F} \Leftrightarrow A \cap B […]
Filter objects
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 $latex \varnothing\notin\mathcal{F}$. For greater clarity I will use filter objects instead of filters. Below […]
Principal filters are center – solved
I 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 $latex {U}&fg=000000$ is the center of the lattice of all filters on $latex […]
Are principal filters the center of the lattice of filters?
This 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 $latex U$) $latex \mathcal{F}$ is by definition a non-empty set of subsets of […]