# Is the intersection of Boolean sublattices a Boolean sublattice?

Let $$L$$ be a boolean lattice, $$A$$ and $$B$$ sublattices of $$L$$ that are themselves boolean lattices, and suppose that $$I = A \cap B$$ is nonempty.

Is $$I$$ a boolean sublattice of $$L$$? Is it a homomorphic image or retract of $$L$$? If so, is there an explicit characterization of its upper and lower bounds?

$$I$$ is clearly a distributive sublattice of $$L$$, $$A$$, and $$B$$, and is bounded above in $$L$$ by $$1_A \sqcap 1_B$$ and below by $$0_A \sqcup 0_B$$.

For example, consider the boolean lattice consisting of the powerset of $$\{1,2,3\}$$ with intersection as meet and union as join. Let $$A$$ be the boolean sublattice containing $$\{1,2\}$$, $$\{1\}$$, $$\{2\}$$, and $$\{\}$$, and $$B$$ the boolean sublattice containing $$\{1,2,3\}$$, $$\{1\}$$, $$\{2,3\}$$, and $$\{\}$$. Then $$I$$ is the boolean sublattice containing $$\{1\}$$ and $$\{\}$$, but not containing the upper bound $$1_A \sqcap 1_B = \{1,2\}$$.

More generally, if $$A$$ and $$B$$ are intersecting intervals of $$L$$, then $$I$$ is a boolean sublattice and a retraction of $$L$$ to $$I$$ takes an element $$x$$ of $$L$$ to the element $$(x \sqcap 1_I) \sqcup 0_I$$ of $$I$$.

Let $$L$$ be the lattice of all subsets of $$\{1,2,...\}$$. Let $$X=\{2^k3^l:k,l\in\omega\}$$ and $$Y=\{2^k5^l:k,l\in\omega\}$$. Let $$A=\{a\subset X:$$ $$a$$ is finite or $$X\backslash a$$ is finite$$\}$$, and $$B=\{a\subset Y:$$ $$a$$ is finite or $$Y\backslash a$$ is finite$$\}$$. Then $$A$$ and $$B$$ are Boolean sublattices of $$L$$, but $$A\cap B=\{a\subset X:$$ $$a$$ is finite and $$a\subset\{2^k:k\in\omega\}\}$$ is not a Boolean sublattice of $$L$$.
If $$A\cap B$$ contains a greatest element $$y$$ and another element $$x$$ then $$\neg x := y\setminus x\quad \in A\cap B$$ is a "complement" of $$x$$ within $$A\cap B$$.
So in that sense, $$A\cap B$$ will always be a Boolean sublattice.
• The existence of such a greatest element $y$ is guaranteed if $A$ and $B$ are complete lattices, but does the conclusion hold if they need not be complete? Let $A = Pfc(2Z)$ denote all finite and cofinite subsets of the multiples of 2. Then $A$ is a bounded boolean sublattice of the powerset $P(Z)$, but is not complete because the intersection of all cofinite subsets of $2Z$ that lack only elements of $4Z$ is the set $2Z - 4Z$, which is not finite or cofinite. If $B=Pfc(3Z)$, then $I = Pfc(6Z)$, which is a bounded boolean sublattice. – Jon Doyle Apr 4 '19 at 12:34