complete lattice

Chain-meet-closed sets ★★

Author(s): Porton

Let $\mathfrak{A}$ is a complete lattice. I will call a \emph{filter base} a nonempty subset $T$ of $\mathfrak{A}$ such that $\forall a,b\in T\exists c\in T: (c\le a\wedge c\le b)$.

\begin{definition} A subset $S$ of a complete lattice $\mathfrak{A}$ is \emph{chain-meet-closed} iff for every non-empty \Def[chain]{Total_order#Chains} $T\in\mathscr{P}S$ we have $\bigcap T\in S$. \end{definition}

\begin{conjecture} A subset $S$ of a complete lattice $\mathfrak{A}$ is chain-meet-closed iff for every filter base $T\in\mathscr{P}S$ we have $\bigcap T\in S$. \end{conjecture}

Keywords: chain; complete lattice; filter bases; filters; linear order; total order

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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