A subset $A$ of a complete lattice $L$ is said to be join-dense if $L=\{\bigvee R|R\subseteq A\}$. An element $a\in L$ is said to be join-irreducible if $a\neq 0$ and if $a=x\vee y$ then $a=x$ or $a=y$. Is there a nice necessary and sufficient condition for when the join-irreducible elements of a complete lattice are join-dense, or is there a representation theorem for such lattices?

Of course, if $L$ is a complete lattice and $A$ is the collection of all join-irreducibles, then we may ''shrink'' the lattice $L$ to the lattice $\{\bigvee R|R\subseteq A\}$ so that the join-irreducibles are join-dense. I know that the distributive complete lattices where the join-irreducibles are join-dense are precisely the spatial coframes(the spatial coframes are the lattices isomorphic to the closed sets in some topological space). Furthermore, if $L$ is a complete lattice satisfying DCC, then the join-irreducible elements in $L$ are join-dense in $L$. However, none of these ideas characterizes the complete lattices where the join-irreducibles are join-dense.


I am not sure if I parsed your definition of join-dense correctly. If I did, then if your lattice is the dual of a continuous lattice, then the join irreducibles are join dense. In the compendium of continuous lattices it is proved that the meet-irreducibles in a continuous lattice order generate (each element is a meet of meet-irreducible elements). This is dual to what you want.

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  • $\begingroup$ Although in every continuous lattice, the meet-irreducibles generate the lattice, there are complete lattices that are not continuous but still generated by the meets of meet-irreducibles. For instance, if X is a regular space that is not locally compact and O(X) is the collection of all open sets in X, then O(X) is not a continuous lattice(see p. 44 in the compendium of continuous lattices), but each open set is a meet of meet irreducibles. $\endgroup$ – Joseph Van Name Mar 30 '12 at 20:40
  • $\begingroup$ I know it is not necessary but I think good necessary conditions are hard. $\endgroup$ – Benjamin Steinberg Mar 30 '12 at 23:51

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