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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 $I$.

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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 $L$.

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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.

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  • $\begingroup$ 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. $\endgroup$ – Jon Doyle Apr 4 at 12:34
  • $\begingroup$ @JonDoyle right, I don't know either in the incomplete case $\endgroup$ – Bjørn Kjos-Hanssen Apr 5 at 3:14

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