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.


  1. 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. 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).
  3. 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.

  4. Should we also add to “4” the requirement for S to be filter-closed? (see my book for a definition of being filter-closed).

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