I’ve moved the section “Some (example) values” to my main book file (instead of the draft file addons.pdf where it was previously).
read moreI’ve calculated values of some concrete funcoids and reloids. The calculations are currently presented in the chapter 3 “Some (example) values” of addons.pdf.
read moreI have added the sections “5.25 Bases on filtrators” (some easy theory generalizing filter bases) and “16.8 Funcoid bases” (mainly a counter-example against my former conjecture) to my math book.
read moreIt is not difficult to prove (see “Counter-examples about funcoids and reloids” in the book) that $latex 1^{\mathsf{FCD}} \sqcap^{\mathsf{FCD}} (\top\setminus 1^{\mathsf{FCD}}) = \mathrm{id}^{\mathsf{FCD}}_{\Omega}$ (where $latex \Omega$ is the cofinite filter). But the result is counterintuitive: meet of two binary relations is not…
read moreI’ve noticed the following three conjectures (I expect not very difficult) for finite binary relations $latex X$ and $latex Y$ between some sets and am going to solve them: $latex X\sqcap^{\mathsf{FCD}} Y = X\sqcap Y$; $latex (\top \setminus X)\sqcap^{\mathsf{FCD}} (\top \setminus Y)…
read moreI’ve found the following counter-example, to this conjecture: Example For a set $latex S$ of binary relations $latex \forall X_0,\dots,X_n\in S:\mathrm{up}(X_0\sqcap^{\mathsf{FCD}}\dots\sqcap^{\mathsf{FCD}} X_n)\subseteq S$ does not imply that there exists funcoid $latex f$ such that $latex S=\mathrm{up}\, f$. The proof is currently available…
read moreI start the “research-in-the middle” project (an outlaw offspring of Polymath Project) introducing to your attention the following conjecture: Conjecture The following are equivalent (for every lattice $latex \mathsf{FCD}$ of funcoids between some sets and a set $latex S$ of principal funcoids…
read moreI have found a surprisingly easy proof of this conjecture which I proposed yesterday. Theorem Let $latex S$ be a set of binary relations. If for every $latex X, Y \in S$ we have $latex \mathrm{up} (X \sqcap^{\mathsf{FCD}} Y) \subseteq S$ then…
read moreConjecture Let $latex S$ be a set of binary relations. If for every $latex X, Y \in S$ we have $latex \mathrm{up} (X \sqcap^{\mathsf{FCD}} Y) \subseteq S$ then there exists a funcoid $latex f$ such that $latex S = \mathrm{up}\, f$.
read moreThe converse of this theorem does not hold. Counterexample: Take $latex S = \mathrm{up}\, \mathrm{id}^{\mathsf{FCD}}_{\Omega}$. We know that $latex S$ is not a filter base. But it is trivial to prove that $latex S$ is a base of the funcoid $latex \mathrm{id}^{\mathsf{FCD}}_{\Omega}$.
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