Continuous lattice

Remarks

Since $x \ll y$ implies $x \le y$, the condition says that $y$ is not larger than the supremum of the elements way below it. Example: $L = [0, 1]$ with the usual order, where $x \ll y$ if and only if $x = 0$ or $x < y$. For $y > 0$ the supremum of $[0, y)$ is $y$, and for $y = 0$ the set $\set{0}$ has supremum $0$. Example: every algebraic lattice is continuous, because each compact $c \le y$ satisfies $c \ll y$. Non-example: the lattice of closed subsets of $[0, 1]$ under inclusion is complete. But $\varnothing$ is the only element way below $[0, 1]$, so $[0, 1]$ is not the supremum of the elements way below it. A complete lattice is continuous if and only if $\bigwedge_{j \in J} \bigvee D_j = \bigvee_{f} \bigwedge_{j \in J} f(j)$ for every family $(D_j)_{j \in J}$ of directed subsets, where $f$ runs over the choice functions with $f(j) \in D_j$.