# Two alleged counterexamples (about boolean algebras)

Trying to solve this question, I propose two possible counter-examples. Please help me to understand whether these cases are really counter-examples.

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 $$(\alpha_0;\beta_0)\le (\alpha_1;\beta_1) \Leftrightarrow \forall X\in\mathfrak{A}: \alpha_0(X)\le\alpha_1(X) \land \forall Y\in\mathfrak{B}: \beta_0(Y)\le\beta_1(Y).$$

Is the poset of boolean funcoids between the boolean lattice $\mathfrak{A}$ and itself also a boolean lattice, if...

1. $\mathfrak{A}$ is the boolean lattice (with $\mathord\sqcup=\mathord\cup$ and $\mathord\sqcap=\mathord\cap$) whose elements are finite unions of binary cartesian products $X\times Y$ for sets $X\in\mathscr{P}A$, $Y\in\mathscr{P}B$, where $A$ and $B$ are (fixed) infinite sets?

2. $\mathfrak{A}$ is the atomless boolean lattice from this Andreas Blass's answer? By the way, is this atomless lattice complete?

3. $\mathfrak{A}$ is the atomless boolean lattice from the comment to above mentioned Andreas Blass's answer?