A construction on boolean lattices is itself a boolean lattice? (Solved)

Importance: Medium ✭✭
Author(s): Porton, Victor
Subject: Algebra
Recomm. for undergrads: no
Posted by: porton
on: October 12th, 2015
Solved by: Porton, Victor

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 math.SE answer.

See Algebraic General Topology for definitions of used concepts.

It is mostly solved:

Theorem   Pointfree funcoids between a complete boolean lattice and an atomistic boolean lattice always form a boolean lattice.
Theorem   Pointfree funcoids between a non-atomic boolean lattice and itself always do not form a boolean lattice.

It is not a complete answer, but the most important cases are considered. So I mark this question as solved. Further consideration however is welcome.

Bibliography

*First appeared as this math.SE question.


* indicates original appearance(s) of problem.

A special case proved

I have proved a weird (due its asymmetry) special case of this conjecture:

Theorem. The set of pointfree funcoids between a complete boolean lattice and an atomistic boolean lattice is itself a boolean lattice.

See my blog.

Victor Porton - http://www.mathematics21.org

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