Existence of maximal boolean-algebra sublattice (preserving top and bottom) of finite distributive lattice

Question

If I regard a modal logic as some sort of many-valued logic, a "modal operator" projecting into a classical propositional logic context could sometimes be useful. Such an operator would provide a projection into a (maximal?) Boolean-algebra which is a sublattice of the given lattice $L$. I'm only interested in distributive lattices in the context of many-valued logic.

I ask myself whether such an operator could always be defined. Such an operator $P$ should satisfy $P(0)=0$ and $P(1)=1$ in addition to the properties of a closure operator $P(P(x))=P(x)$, $x \leq P(x)$ and $x \leq y \rightarrow P(x) \leq P(y)$. In addition, it would be nice if $P$ restricted to any Boolean-algebra sublattice of $L$ containing $0$ and $1$ would be the identity.

I wonder whether the additional condition can always be satisfied, and whether it would characterize the operator uniquely. To simplify the question, let's restrict ourselves to finite distributive lattices, and just ask whether there exists a (unique) maximal Boolean-algebra sublattice of $L$ containing $0$ and $1$.

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Edit The following statement from the initial question is incorrect:

... an operator similar to double-negation is sometimes useful. Such an operator provides a projection into a Boolean-algebra which is a sublattice of the given lattice $L$.

The wikipedia article on Heyting algebras explains that the regular elements constitute a Boolean algebra, but that they are in general not a sublattice, because the join operation can be different.

Answer 1

For the simplified version, in the last paragraph of your question, the answer is affirmative: Every finite distributive lattice $L$ (with $0$ and $1$) has a unique largest Boolean algebra sublattice. The proof I have in mind uses (possibly overkill) the structure theorem for such lattices: $L$ is, up to isomorphism, the lattice of downward-closed subsets of some finite partially ordered set $P$, with ordinary union and intersection as the lattice operations (and $0=\varnothing$ and $1=P$). Now suppose $B$ is a sublattice of $L$ containing $0$ and $1$, and $B$ is also a Boolean algebra. Consider any element $x$ of $B$ and its Boolean complement $\neg x$ in $B$. Their meet is $0$ and their join is $1$; that is, if we regard them as downward-closed subsets of $P$, their intersection is empty and their union is $P$, so they are complementary subsets of $P$. That is, the negation operation in $B$ must be ordinary set-theoretic complementation of subsets of $P$. Note that, for the complement of a downward-closed set $x$ to also be downward-closed, $x$ must be upward-closed. In a finite poset $P$, the subsets that are simultaneously downward-closed and upward-closed are exactly the unions of connected components of the Hasse diagram of $P$. (This diagram is defined as the graph with vertex set $P$, where two elements are joined iff one is below the other with nothing strictly between them in the ordering of $P$.) So $B$ has to consist of some such unions of connected components. But the collection of all such unions is clearly closed under all the Boolean operations, so this collection is the (unique) largest Boolean sublattice of $L$.

Answer 2

There is no need to assume that $L$ is finite.

An element $x$ of a bounded lattice is complemented iff there is an element $y$ such that $x\vee y=1$ and $x\wedge y=0$. It is easy to check that, in a distributive lattice $L$, the set of all complemented elements forms a sublattice $C(L)$. Since the lattice $C(L)$ is distributive and complemented, it is a Boolean algebra. Clearly, any other Boolean subalgebra of $L$ is under $C(L)$, since Boolean subalgebras of $L$ consist only of complemented elements.

Answer 3

The operator from the question cannot always be defined. As counter-example, take the following set-algebra on $\mathbb N$: Define a partial ordering where $m > n$ if and only if $m$ and $n$ are even, $m/2$ is even, $n/2$ is odd and $|m-n|=2$. Now take the algebra of finite or cofinite subsets of $\mathbb N$ which are also upper sets with respect to this partial ordering.

If we take an upper set with a finite number of even numbers and no odd numbers, then $P$ can only project it to the set of all even numbers. (Because of $x \leq P(x)$, but of course some minor details are missing here.) But the set of all even numbers is neither finite nor cofinite. This argument can be made precise to show that $P$ can't be defined.

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This counter-example is based on this answer and this comment/answer to questions I had while trying to better understand Priestley spaces.