Lattices without nontrivial dense elements

Question

This question arose from another one of mine, Homotopy type of some lattices with top and bottom removed.

An element $d$ of a bounded lattice $L$ is called $\mathit{dense}$ if $$ \forall x\in L\ (d\land x=\bot)\Rightarrow(x=\bot) $$ holds.

It is well known that a pseudocomplemented distributive lattice is Boolean if and only if $\top$ is the only dense element: in this case dense elements are precisely those of the form $a\lor\neg a$ where $\neg$ is the pseudocomplement.

Is a characterization of general (non-distributive) lattices with the same property known? That is, which bounded lattices have the property that $\top$ is the unique dense element?

Variations: this property + the only codense element is $\bot$; not necessarily bounded lattices such that all intervals have these properties; just the finite case; etc., etc.

Also, what about those distributive lattices which are neither pseudocomplemented nor copseudocomplemented? And what about (co)pseudocomplemented non-distributive ones?

Answer 1

Let $L$ be a finite lattice. Then $\top$ is the only dense element of $L$ if and only if $\top$ is a join of atoms.

$Proof:$ The top element is always dense.

Let $L$ be a lattice such that $\top$ is a join of atoms. Let $d$<$\top$ be in $L$. Then there is an atom $x$ such that $x\nleq d$. Then $x\wedge d = \perp$ but $x\ne \perp$. Hence $d$ is not dense.

Conversely, suppose the join of the set of atoms is $d<\top$. Let $x\in L$ but $x\ne\perp$. Then there is an atom $y\le x$ and $y\le d$, so $x\wedge d\ne\perp$. Hence $d$ is dense.QED