meet continuous
Let $L$ be a meet semilattice. We say that $L$ is meet continuous if
 1.

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 nondecreasing 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).

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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:
$$\overline{lim}{x}_{i}=\underset{j\in I}{\bigwedge}\{\underset{j\le i}{\bigvee}{x}_{i}\}\mathit{\hspace{1em}\hspace{1em}}\text{and}\mathit{\hspace{1em}\hspace{1em}}\underset{\xaf}{lim}{x}_{i}=\underset{j\in I}{\bigvee}\{\underset{i\le j}{\bigwedge}{x}_{i}\}$$ If there is an $x\in L$ such that $\overline{lim}{x}_{i}=x=\underset{\xaf}{lim}{x}_{i}$, 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^{}.
References
 1 G. Birkhoff, Lattice Theory, 3rd Edition, Volume 25, AMS, Providence (1967).
 2 G. Gierz, K. H. Hofmann, K. Keimel, J. D. Lawson, M. W. Mislove, D. S. Scott, Continuous Lattices and Domains, Cambridge University Press, Cambridge (2003).
 3 G. Grätzer, General Lattice Theory, 2nd Edition, Birkhäuser (1998).
Title  meet continuous 
Canonical name  MeetContinuous 
Date of creation  20130322 16:36:41 
Last modified on  20130322 16:36:41 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  12 
Author  CWoo (3771) 
Entry type  Definition 
Classification  msc 06A12 
Classification  msc 06B35 
Synonym  order convergence 
Related topic  CriteriaForAPosetToBeACompleteLattice 
Related topic  JoinInfiniteDistributive 
Defines  join continuous 
Defines  order converges 