Is the intersection of Boolean sublattices a Boolean sublattice?

Question

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 $I$.

Answer 1

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 $L$.

Answer 2

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.