most recent 30 from http://mathoverflow.net 2013-05-19T12:51:13Z http://mathoverflow.net/feeds/question/92519 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/92519/when-are-the-join-irreducibles-in-a-complete-lattice-join-dense Joseph Van Name 2012-03-29T02:16:38Z 2012-12-25T23:57:04Z

<p>A subset $A$ of a complete lattice $L$ is said to be join-dense if <code>$L=\{\bigvee R|R\subseteq A\}$</code>. 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?</p> <p>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 <code>$\{\bigvee R|R\subseteq A\}$</code> 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.</p>

http://mathoverflow.net/questions/92519/when-are-the-join-irreducibles-in-a-complete-lattice-join-dense/92586#92586 Benjamin Steinberg 2012-03-29T17:15:51Z 2012-03-29T17:15:51Z

<p>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. </p>