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

## 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 $x = \bigvee S_x$.