Let $\mathfrak{A}$ and $\mathfrak{B}$ be (fixed) boolean lattices (with lattice operations denoted $\sqcup$ and $\sqcap$, bottom element $\bot$ and top element $\top$).

I call a boolean funcoid a pair $(\alpha;\beta)$ of functions $\alpha:\mathfrak{A}\rightarrow\mathfrak{B}$, $\beta:\mathfrak{B}\rightarrow\mathfrak{A}$ such that (for every $X\in\mathfrak{A}$, $Y\in\mathfrak{B}$) $$Y\sqcap^{\mathfrak{B}}\alpha(X)\ne\bot^{\mathfrak{B}} \Leftrightarrow X\sqcap^{\mathfrak{A}}\beta(Y)\ne\bot^{\mathfrak{A}}.$$

(Boolean funcoids are a special case of pointfree funcoids as defined in my free ebook.)

Order boolean funcoids by the formula $$(\alpha_0;\beta_0)\le (\alpha_1;\beta_1) \Leftrightarrow \forall X\in\mathfrak{A}: \alpha_0(X)\le\alpha_1(X) \land \forall Y\in\mathfrak{B}: \beta_0(Y)\le\beta_1(Y).$$

Conjecture The set of boolean funcoids with above defined order is itself a boolean lattice.

If this conjecture does not hold in general, does it hold for: a. atomic boolean lattices? b. atomistic boolean lattices? c. complete boolean lattices?

For the special case when $\mathfrak{A}$ and $\mathfrak{B}$ are complete atomic boolean lattices, the conjecture easily follows from this my answer to my own question.

  • $\begingroup$ Why do you use such cumbersome notation? $\endgroup$ – J.-E. Pin Sep 22 '15 at 19:22
  • $\begingroup$ Can we construct a counterexample such that $\mathfrak{B}$ is a finite boolean algebra or even a two-element boolean algebra? $\endgroup$ – porton Mar 14 '16 at 21:46
  • $\begingroup$ I've proved a special case of my conjecture: The set of boolean funcoids between a complete boolean lattice and an atomistic boolean lattice is itself a boolean lattice. Very weird condition (it requires different properties of the first and the second algebras). It is now available in my addons.pdf file but will be integrated into my book in the future. See mathematics21.org/algebraic-general-topology.html $\endgroup$ – porton Mar 15 '16 at 22:24
  • $\begingroup$ mathematics21.org/binaries/addons.pdf - chapter "Boolean funcoids" $\endgroup$ – porton Mar 15 '16 at 22:27
  • $\begingroup$ I have proved that the poset of boolean funcoids between non-atomic boolean lattices is always not a boolean lattice portonmath.wordpress.com/2016/03/22/… $\endgroup$ – porton Mar 22 '16 at 20:39

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