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?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.