meet continuous

Let $L$ be a meet semilattice. We say that $L$ is meet continuous if

1. for any monotone net $D=\{x_i\mid i\in I\}$ in $L$, its supremum $\bigvee D$ exists, and

2. for any $a\in L$ and any monotone net $\{x_i\mid i\in I\}$,

   $$a\wedge \bigvee \{x_i\mid i\in I\}=\bigvee \{a\wedge x_i\mid i\in I\}.$$

A monotone net $\{x_i\mid i\in I\}$ is a net $x:I\to L$ such that $x$ is a non-decreasing function; that is, for any $i\le j$ in $I$, $x_i\le x_j$ in $L$.

Note that we could have replaced the first condition by saying simply that $D\subseteq L$ is a directed set. (A monotone net is a directed set, and a directed set is a trivially a monotone net, by considering the identity function as the net). It’s not hard to see that if $D$ is a directed subset of $L$, then $a\wedge D:=\{a\wedge x\mid x\in D\}$ is also directed, so that the right hand side of the second condition makes sense.

Dually, a join semilattice $L$ is join continuous if its dual (as a meet semilattice) is meet continuous. In other words, for any antitone net $D=\{x_i\mid i\in I\}$, its infimum $\bigwedge D$ exists and that

$$a\vee \bigwedge \{x_i\mid i\in I\}=\bigwedge \{a\vee x_i\mid i\in I\}.$$

An antitone net is just a net $x:I\to L$ such that for $i\le j$ in $I$, $x_j\le x_i$ in $L$.

Remarks.

• A meet continuous lattice is a complete lattice, since a poset such that finite joins and directed joins exist is a complete lattice (see the link below for a proof of this).

• Let a lattice $L$ be both meet continuous and join continuous. Let $\{x_i\mid i\in I\}$ be any net in $L$. We define the following:

  $$\overline{lim}{x}_i=\underset{j\in I}{\bigwedge }\{\underset{j\le i}{\bigvee }{x}_i\}\mathit{\hspace{1em}\hspace{1em}}\text{ and }\mathit{\hspace{1em}\hspace{1em}}\underset{¯}{lim}{x}_i=\underset{j\in I}{\bigvee }\{\underset{i\le j}{\bigwedge }{x}_i\}$$

  If there is an $x\in L$ such that $\overline{lim}{x}_i=x=\underset{¯}{lim}{x}_i$, then we say that the net $\{x_i\}$ order converges to $x$, and we write $x_i\to x$, or $x=lim{x}_i$. Now, define a subset $C\subseteq L$ to be closed (in $L$) if for any net $\{x_i\}$ in $C$ such that $x_i\to x$ implies that $x\in C$, and open if its set complement is closed, then $L$ becomes a topological lattice. With respect to this topology, meet $\wedge$ and join $\vee$ are easily seen to be continuous.