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 $I$.



Is the intersection of Boolean sublattices a Boolean sublattice?
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 $L$.


