New sections in my math book

I 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.

The mystery of meet of funcoids solved?

It 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 a binary relation. After proving this I always felt that there is some […]

Three (seemingly not so difficult) new conjectures

I’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: $latex X\sqcap^{\mathsf{FCD}} Y = X\sqcap Y$; $latex (\top \setminus X)\sqcap^{\mathsf{FCD}} (\top \setminus Y) = (\top \setminus X)\sqcap (\top \setminus Y)$; $latex (\top \setminus X)\sqcap^{\mathsf{FCD}} Y = […]

A counter-example to my conjecture

I’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 at this PDF file and this wiki page.

A new research project (a conjecture about funcoids)

I 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 (=binary relations)): $latex \forall X, Y \in S : \mathrm{up} (X \sqcap^{\mathsf{FCD}} Y) […]

A surprisingly easy proof of yesterday conjecture

I 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 there exists a funcoid $latex f$ such that $latex S = \mathrm{up}\, f$. […]

Conjecture about funcoids

Conjecture 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 $latex S = \mathrm{up}\, f$.

A base of a funcoid which is not a filter base

The 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 $latex \mathrm{id}^{\mathsf{FCD}}_{\Omega}$.