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

**Example** For a set of binary relations

does not imply that there exists funcoid such that .

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

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 of funcoids between some sets and a set of principal funcoids (=binary relations)):

- .
- (for every natural ).
- There exists a funcoid such that .

and 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.

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

**Theorem** Let be a set of binary relations. If for every we have then there exists a funcoid such that .

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

**Conjecture** Let be a set of binary relations. If for every we have then there exists a funcoid such that .

The converse of this theorem does not hold.

Counterexample: Take . We know that is not a filter base. But it is trivial to prove that is a base of the funcoid .

**Definition** A set of binary relations is a *base* of a funcoid when all elements of are above and .

It was easy to show:

**Proposition** A set of binary relations is a base of a funcoid iff it is a base of .

Today I’ve proved the following important theorem:

**Theorem** If is a filter base on the set of funcoids then is a base of .

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.

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

**Statement** or equivalently: If then there exists , such that .

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

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