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!

1. porton / Apr 11 2017 21:35 I present an attempted proof in the wiki.

The idea behind this attempted proof is to reduce behavior of funcoids $\langle f\rangle$ with better known behavior of filters $\langle f\rangle x$ for an arbitrary ultrafilter $x$ (I remind that knowing $\langle f\rangle x$ for all ultrafilters $x$ on the domain, it’s possible to restore funcoid $f$) and then to replace $\langle X_0 \sqcap^{\mathsf{FCD}} \ldots \sqcap^{\mathsf{FCD}} X_n\rangle x$ with $\langle X_0 \rangle x \sqcap \dots \sqcap \langle X_n \rangle x$.

2. porton / Apr 13 2017 11:59 At https://conference.portonvictor.org/wiki/Funcoid_bases/Another_reduce_to_ultrafilters I introduce a proof attempt of the statement:

If $\forall X_0, \ldots, X_n \in S : \mathrm{up} (X_0 \sqcap^{\mathsf{FCD}} \ldots \sqcap^{\mathsf{FCD}} X_n) \subseteq S$ (for every natural $n$), then there exists a funcoid $f$ such that $S = \mathrm{up}\, f$.

3. porton / Apr 13 2017 23:27 I’ve published some easy basic results related to the conjecture at https://conference.portonvictor.org/wiki/Funcoid_bases/Basic_results

First I define $S'=S\cap\Gamma$. Second, I prove $\bigwedge^{\mathsf{FCD}} S' = \bigwedge^{\mathsf{FCD}} S$.

4. porton / Apr 13 2017 23:41 At https://conference.portonvictor.org/wiki/Funcoid_bases/Proving_existence_of_funcoid_through_lattice_Gamma I tried to prove that $S$ is a an up of a funcoid (under another conjecture conditions). My attempted proof uses the lattice $\Gamma$ from the chapter “Funcoids are filters” of my book

5. porton / Apr 16 2017 00:19 I propose also the following two conditions (possibly) equivalent to the conditions mentioned in the original conjecture:

4. $\forall X,Y\in S': \mathrm{up}(X\sqcap Y)\subseteq S'$;
5. $\forall X_0,\dots X_n\in S': \mathrm{up}(X_0\sqcap\dots\sqcap X_n)\subseteq S'$ (for every natural $n$).
6. porton / Apr 16 2017 00:26 The two above conditions 4 and 5 are each equivalent to $S'$ being a filter on the boolean lattice $\Gamma$.

7. porton / Apr 16 2017 00:41 It is easy to show that $S'$ being a filter is not enough for the (other) conditions of the conjecture to hold (for a counter-example consider $S\subseteq\Gamma$ and thus $S=S'$).

Probably the following is equivalent to the conditions of the conjecture: $S'$ is a filter on $\Gamma$ and $S$ is an upper set.

8. porton / Apr 17 2017 00:53 Added condition “4” defined above to the main wiki page. It is quite obvious that $1\Rightarrow 4$ and $3\Rightarrow 4$.

9. porton / Apr 17 2017 02:10 Should we also add to “4” the requirement for $S$ to be filter-closed? (see my book for a definition of being filter-closed).

10. porton / Apr 18 2017 19:09 The condition “ $S'$ is a filter on the lattice $\Gamma$ and $S$ is an upper set” is not enough for existence of $f$ such that $S=\mathrm{up}\, f$. See https://conference.portonvictor.org/wiki/Funcoid_bases/Failed_condition in the wiki. So the condition “4” is removed from consideration.

11. porton / Apr 18 2017 19:25 Can the same counter-example as in https://conference.portonvictor.org/wiki/Funcoid_bases/Failed_condition (the topic of the previous comment) be applied to some implications between conditions 1, 2, 3?

12. porton / Apr 18 2017 22:29 The conjecture was declined with a counter-example https://conference.portonvictor.org/wiki/Funcoid_bases/Disproof

It yet remains the question whether the condition “1” implies “2”.

13. porton / Apr 18 2017 22:50 The proof at https://conference.portonvictor.org/wiki/Funcoid_bases/Disproof was with an error, but the proof idea was right. Now it contains the corrected proof.

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