While reading the paper LATTICES WITH UNIQUE COMPLEMENTS by R. P. DILWORTH, I get to know that any number of weak additional restrictions are sufficient for a lattice with unique complement to be a boolean algebra, including properties like modular, etc.
But I'm wondering whether the restriction like "finite" be sufficient enough. However, it's so difficult for me to prove or give a counterexample.
Is a finite lattice where each element has exactly one complement distributive? Why or why not
discrete mathematicslattice-orders
Related Solutions
Suppose that $x$ and $y$ are two complements of $a$. We have:
$x = x \vee 0 = x \vee (a\wedge y) = (x\vee a)\wedge (x \vee y) = 1\wedge (x \vee y) = x\vee y$.
Similarly, $y = x\vee y$, so $x=y$.
A lattice in which each element has at most one complement may have elements with no complement at all.
It is rather easy to come up with non-distributive lattices with that property, even if we require that there is at least one pair of elements which are complements of one another.
For example, take a bounded, non-distributive lattice, in which no element except $0$ and $1$ has a complement.
For a less trivial example, one in which there are other pairs of complements, take the following example:
Here $\langle 0,1\rangle$ and $\langle a,b\rangle$ are the only pairs of complements, and the lattice is not distributive, since it has $M_3$ as a sublattice (see here).
If you're looking for a non-distributive lattice in which every element has a unique complement, that is not easy to point out.
I believe that there is no finite such lattice, but basically this answer tells that it exists (even non-modular).
Actually every lattice is a sublattice of a lattice with unique complements (Dilworth's paper on this).
Best Answer
It is immediate that every finite lattice is complete and atomic, i.e., every element is above some atom.
So the following result yields that a finite uniquely complemented lattice is Boolean.
Notice that it's not even asked in the hypothesis that $\mathbf L$ is atomistic (which is stronger than being atomic). We will see by the end of the proof that that will follow from the hypothesis.
Proof: Let us denote by $\mathcal At(\mathbf L)$ the set of atoms of $\mathbf L$. To $S \subseteq \mathcal At(\mathbf L)$, let $$\bigvee S = \bigvee \{ x : x \in S \}$$ and $$\bigwedge S' = \bigwedge \{ x' : x \in S \}.$$ It follows that $$\bigvee (S \cup T) = \bigvee S \vee \bigvee T$$ and $$\bigwedge(S \cup T)' = \bigwedge S' \wedge \bigwedge T'.$$ For each atom $x$ of $\mathbf L$, we have that $x' \prec 1$ ($x'$ is covered by $1$, that is, $x'<1$ and if $x'\leq y \leq 1$ then $y=x'$ or $y=1$). Indeed, if $x' \leq y \leq 1$, then either $x \leq y$ or $x \nleq y$; in the former case, $1 = x \vee x' \leq y$, whence $y=1$; in the later, $x \wedge y = 0$, and $x \vee y \geq x \vee x' = 1$, whence $y = x'$.
It follows that for $x \neq y$ in $\mathcal At(\mathbf L)$ we have $x \leq y'$, for otherwise $x \wedge y' = 0$ and $x \vee y' = 1$, yielding $x=y$ by the unique complementation. Hence, if $S,T \subseteq \mathcal At(\mathbf L)$ are such that $S \cap T = \varnothing$, then $$\bigvee S \leq \bigwedge T'.$$ Thus, denoting by $S^c$ the complement of $S$ in $\mathcal At(\mathbf L)$,
\begin{align} \bigvee S \wedge \bigvee S^c &\leq \bigwedge(S^c)' \wedge \bigwedge(S^{cc})'\\ &= \bigwedge(S^c)' \wedge \bigwedge S'\\ &= \bigwedge(S^c \cup S)'\\ &= \bigwedge(\mathcal At(\mathbf L))' \\ &= 0 \tag{$\dagger$} \end{align} Thus, $\mathcal At(\mathbf L) \cap {\downarrow}\bigvee S = S$ and so $\bigvee S \neq \bigvee T$, whenever $S \neq T$, and therefore the poset whose elements are the family $\{ \bigvee S : S \subseteq \mathcal At(\mathbf L) \}$, with the order inherited from $\mathbf L$ is isomorphic to the power-set $\wp(\mathcal At(\mathbf L))$, which is clearly a Boolean algebra.
It remains to show that $x = \bigvee S$ for some $S \subseteq \mathcal At(\mathbf L)$ and each $x \in L$. Let $$S_x = \{ a \in \mathcal At(\mathbf L) : a \leq x \}.$$ We will show that $x = \bigvee S_x$ (i.e., $\mathbf L$ is atomistic).
It is clear that the only atoms below $x \wedge \bigvee S_x^c$ are those which are in $S_x \cap S_x^c = \varnothing$, and so $x \wedge \bigvee S_x^c = 0$. On the other hand \begin{align} x \vee \bigvee S_x^c &\geq \bigvee S_x \vee \bigvee S_x^c\\ &= \bigvee (S_x \cup S_x^c)\\ &= \bigvee \mathcal At(\mathbf L)\\ &= 1. \end{align} Thus $x$ is the (unique) complement of $\bigvee S_x^c$. From ($\dagger$) and $$\bigvee S_x \vee \bigvee S_x^c = \bigvee (S_x \cup S_x^c) = \bigvee \mathcal At(\mathbf L) = 1,$$ it follows that $x = \bigvee S_x$.