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April 25, 2017 / porton

A counter-example to my conjecture

I’ve found the following counter-example, to this conjecture:

Example For a set S of binary relations

\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 f such that S=\mathrm{up}\, f.

The proof is currently available at this PDF file and this wiki page.

April 11, 2017 / porton

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 \mathsf{FCD} of funcoids between some sets and a set S of principal funcoids (=binary relations)):

  1. \forall X, Y \in S : \mathrm{up} (X \sqcap^{\mathsf{FCD}} Y) \subseteq S.
  2. \forall X_0,\dots,X_n \in S : \mathrm{up} (X \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=\mathrm{up}\, f.

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

I welcome you to actively participate in the research!

Please write your comments and idea both in the wiki and as comments and trackbacks to this blog post.

March 29, 2017 / porton

A surprisingly easy proof of yesterday conjecture

I have found a surprisingly easy proof of this conjecture which I proposed yesterday.

Theorem Let S be a set of binary relations. If for every X, Y \in S we have \mathrm{up} (X \sqcap^{\mathsf{FCD}} Y) \subseteq S then there exists a funcoid f such that S = \mathrm{up}\, f.

The proof (currently available in this PDF file) is based on “Funcoids are filters” chapter of my book.

March 27, 2017 / porton

Conjecture about funcoids

Conjecture Let S be a set of binary relations. If for every X, Y \in S we have \mathrm{up} (X \sqcap^{\mathsf{FCD}} Y) \subseteq S then there exists a funcoid f such that S = \mathrm{up}\, f.

March 26, 2017 / porton

A base of a funcoid which is not a filter base

The converse of this theorem does not hold.

Counterexample: Take S = \mathrm{up}\, \mathrm{id}^{\mathsf{FCD}}_{\Omega}. We know that S is not a filter base. But it is trivial to prove that S is a base of the funcoid \mathrm{id}^{\mathsf{FCD}}_{\Omega}.

March 25, 2017 / porton

A new theorem proved

Definition A set S of binary relations is a base of a funcoid f when all elements of S are above f and \forall X \in \mathrm{up}\, f \exists T \in S : T \sqsubseteq X.

It was easy to show:

Proposition A set S of binary relations is a base of a funcoid iff it is a base of \bigsqcap^{\mathsf{FCD}} S.

Today I’ve proved the following important theorem:

Theorem If S is a filter base on the set of funcoids then S is a base of \bigsqcap^{\mathsf{FCD}} S.

The proof is currently located in this PDF file.

It is yet unknown whether the converse theorem holds, that is whether every base of a funcoid is a filter base on the set of funcoids.

January 21, 2017 / porton

I made an error in a proof

I’ve proved the following (for every funcoids f and g):

Statement \mathrm{up}\,  (f \sqcap^{\mathsf{FCD}} g) \subseteq \bigcup \{ \mathrm{up}\,  (F \sqcap^{\mathsf{FCD}} G) \mid F \in \mathrm{up}\,  f, G \in \mathrm{up}\,  g \} or equivalently: If Z\in\mathrm{up}\,  (f \sqcap^{\mathsf{FCD}} g) then there exists F \in \mathrm{up}\,  f, G \in \mathrm{up}\,  g such that Z\in\mathrm{up}\,  (F \sqcap^{\mathsf{FCD}} G).

But now I’ve noticed that the proof was with an error! So it is again a conjecture.