Is a finite lattice uniquely atomistic iff it is boolean?

Question

A finite lattice $(L,\wedge,\vee)$ is atomistic if every nonzero element is a join of atoms.

Let $A = \{a_1, \dots , a_n \} \subset L$ be the subset of atoms, then $L$ is called uniquely atomistic if in addition: $$\forall I , J \subset \{1,2, \dots , n\} , \bigvee_{i \in I}a_i = \bigvee_{j \in J}a_j \Rightarrow I=J$$

A finite lattice is boolean if it is equivalent to the subsets lattice of $ \{ 1,2, \dots , n\}$, for some $n$.

Question: Is a finite lattice uniquely atomistic if and only if it is boolean?

Answer 1

Yes, this is true. If $L$ is uniquely atomistic with atoms $A$, that means that there is a bijection $f:P(A)\to L$ given by $f(S)=\bigvee S$. I claim this map is an isomorphism. It is clear that if $S\subseteq T$ then $f(S)\leq f(T)$, so it suffices to show the converse. Suppose $f(S)\leq f(T)$. It follows from the definition of $f$ that $f$ preserves joins, so $f(S\cup T)=f(S)\vee f(T)=f(T)$. Since $f$ is injective, this implies $S\cup T=T$, so $S\subseteq T$.