filters


Inner reloid through the lattice Gamma ★★

Author(s): Porton

\begin{conjecture} $(\mathsf{RLD})_{\operatorname{in}} f = \bigcap^{\mathsf{RLD}} \operatorname{up}^{\Gamma (\operatorname{Src} f ; \operatorname{Dst} f)} f$ for every funcoid $f$. \end{conjecture}

Counter-example: $(\mathsf{RLD})_{\operatorname{in}} f \sqsubset \bigcap^{\mathsf{RLD}} \operatorname{up}^{\Gamma (\operatorname{Src} f ; \operatorname{Dst} f)} f$ for the funcoid $f = (=)|_\mathbb{R}$ is proved in \href[this online article]{http://www.mathematics21.org/binaries/funcoids-are-filters.pdf}.

Keywords: filters; funcoids; inner reloid; reloids

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

Co-separability of filter objects ★★

Author(s): Porton

\begin{conjecture} Let $a$ and $b$ are filters on a set $U$ and $a\cap b = \{U\}$. Then $$\exists A\in a,B\in b: (\forall X\in a: A\subseteq X \wedge \forall Y\in b: B\subseteq Y \wedge A \cup B = U).$$ \end{conjecture}

\href[See here for some equivalent reformulations of this problem.]{http://portonmath.wordpress.com/2009/11/29/co-separability/}

This problem (in fact, a little more general version of a \href[problem equivalent to this problem]{http://portonmath.wordpress.com/2009/11/29/co-separability/}) was solved by the problem author. See \href[here]{http://filters.wikidot.com/primary-filtrator-is-filtered} for the solution.

Maybe this problem should be moved to "second-tier" because its solution is simple.

Keywords: filters

Pseudodifference of filter objects ★★

Author(s): Porton

Let $U$ is a set. A filter $\mathcal{F}$ (on $U$) is 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 $\emptyset \notin \mathcal{F}$.

I will call \emph{the set of filter objects} the set of filters ordered reverse to set theoretic inclusion of filters, with principal filters equated to the corresponding sets. \href[See here for the formal definition of filter objects]{http://www.mathematics21.org/binaries/filters.pdf}. I will denote $(\operatorname{up} a)$ the filter corresponding to a filter object $a$. I will denote the set of filter objects (on $U$) as $\mathfrak{F}$.

I will denote $(\operatorname{atoms} a)$ the set of atomic lattice elements under a given lattice element $a$. If $a$ is a filter object, then $(\operatorname{atoms} a)$ is essentially the set of ultrafilters over $a$.

\begin{problem} Which of the following expressions are pairwise equal for all $a, b \in \mathfrak{F}$ for each set $U$? (If some are not equal, provide counter-examples.) \begin{enumerate} \item $\bigcap^{\mathfrak{F}} \left\{ z \in \mathfrak{F} | a \subseteq b \cup^{\mathfrak{F}} z \right\}$;

\item $\bigcup^{\mathfrak{F}} \left\{ z \in \mathfrak{F} | z \subseteq a \wedge z \cap^{\mathfrak{F}} b = \emptyset \right\}$;

\item $\bigcup^{\mathfrak{F}} (\operatorname{atoms} a \setminus \operatorname{atoms} b)$;

\item $\bigcup^{\mathfrak{F}} \left\{ a \cap^{\mathfrak{F}} (U\setminus B) | B \in \operatorname{up} b \right\}$. \end{enumerate} \end{problem}

Keywords: filters; pseudodifference

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