Join of oblique products ★★

Author(s): Porton

\begin{conjecture} $\left( \mathcal{A} \ltimes \mathcal{B} \right) \cup \left( \mathcal{A} \rtimes \mathcal{B} \right) = \mathcal{A} \times^{\mathsf{\ensuremath{\operatorname{RLD}}}} \mathcal{B}$ for every filter objects $\mathcal{A}$, $\mathcal{B}$. \end{conjecture}

Keywords: filter; oblique product; reloidal product

Do filters complementive to a given filter form a complete lattice? ★★

Author(s): Porton

Let $U$ is a set. A filter (on $U$) $\mathcal{F}$ is by definition a non-empty set of subsets of $U$ such that $A,B\in\mathcal{F} \Leftrightarrow A\cap B\in\mathcal{F}$. Note that unlike some other authors I do not require $\varnothing\notin\mathcal{F}$. I will denote $\mathscr{F}$ the lattice of all filters (on $U$) ordered by set inclusion.

Let $\mathcal{A}\in\mathscr{F}$ is some (fixed) filter. Let $D=\{\mathcal{X}\in\mathscr{F} | \mathcal{X}\supseteq \mathcal{A}\}$. Obviously $D$ is a bounded lattice.

I will call complementive such filters $\mathcal{C}$ that:

  1. $\mathcal{C}\in D$;
  2. $\mathcal{C}$ is a complemented element of the lattice $D$.

\begin{conjecture} The set of complementive filters ordered by inclusion is a complete lattice. \end{conjecture}

Keywords: complete lattice; filter

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