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Difference between revisions of "Funcoid bases"

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# $\forall X_0,\dots,X_n \in S : \operatorname{up} (X_0 \sqcap^{\mathsf{FCD}} \dots \sqcap^{\mathsf{FCD}} X_n) \subseteq S$ (for every natural $n$).
 
# $\forall X_0,\dots,X_n \in S : \operatorname{up} (X_0 \sqcap^{\mathsf{FCD}} \dots \sqcap^{\mathsf{FCD}} X_n) \subseteq S$ (for every natural $n$).
 
# There exists a funcoid $f\in\mathsf{FCD}$ such that $S=\operatorname{up} f$.
 
# There exists a funcoid $f\in\mathsf{FCD}$ such that $S=\operatorname{up} f$.
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Another possibly equivalent condition:
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4. $S$ is an upper set and $S\cap\Gamma$ is a filter on the boolean lattice $\Gamma$ (defined in the chapter "Funcoids are filters" of the book).
  
 
$3\Rightarrow 2$ and $2\Rightarrow 1$ are obvious.
 
$3\Rightarrow 2$ and $2\Rightarrow 1$ are obvious.

Revision as of 21:44, 16 April 2017

This page presents Victor Porton's conjectures about funcoid bases and related stuff as first defined in this draft document.

Please read Algebraic general topology book before attempt to participate in this research.

The main conjecture about funcoid bases is the following:

Conjecture The following are equivalent (for every lattice $\mathsf{FCD}$ of funcoids between some sets and a set $S$ of principal funcoids (=binary relations)):

  1. $\forall X, Y \in S : \operatorname{up} (X \sqcap^{\mathsf{FCD}} Y) \subseteq S$.
  2. $\forall X_0,\dots,X_n \in S : \operatorname{up} (X_0 \sqcap^{\mathsf{FCD}} \dots \sqcap^{\mathsf{FCD}} X_n) \subseteq S$ (for every natural $n$).
  3. There exists a funcoid $f\in\mathsf{FCD}$ such that $S=\operatorname{up} f$.

Another possibly equivalent condition:

4. $S$ is an upper set and $S\cap\Gamma$ is a filter on the boolean lattice $\Gamma$ (defined in the chapter "Funcoids are filters" of the book).

$3\Rightarrow 2$ and $2\Rightarrow 1$ are obvious.

Please write your research ideas at this wiki and as comments and trackbacks to this blog post.

Funcoid bases/Basic results

Proposed ways to attack this conjecture:

$1\Rightarrow 2$:

$2\Rightarrow 3$: