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Inner reloid through the lattice Gamma ★★

Author(s): Porton

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

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 this online article.

Keywords: filters; funcoids; inner reloid; reloids

Posted by porton
updated September 25th, 2014
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Chain-meet-closed sets ★★

Author(s): Porton

Let $ \mathfrak{A} $ is a complete lattice. I will call a 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) $.

Definition   A subset $ S $ of a complete lattice $ \mathfrak{A} $ is chain-meet-closed iff for every non-empty chain $ T\in\mathscr{P}S $ we have $ \bigcap T\in S $.
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 $.

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

Posted by porton
updated December 12th, 2009
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Co-separability of filter objects ★★

Author(s): Porton

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

See here for some equivalent reformulations of this problem.

This problem (in fact, a little more general version of a problem equivalent to this problem) was solved by the problem author. See here for the solution.

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

Keywords: filters

Posted by porton
updated September 10th, 2010
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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 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. See here for the formal definition of filter objects. 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 $.

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.)
    \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\} $.

Keywords: filters; pseudodifference

Posted by porton
updated May 19th, 2012
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