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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.



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  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 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

    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 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 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 (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

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

  13. porton / Apr 18 2017 22:50

    The proof at was with an error, but the proof idea was right. Now it contains the corrected proof.

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