Continuous Lattice Subframe of Algebraic Lattice is Algebraic Lattice

Theorem
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a bounded below algebraic lattice.

Let $P = \struct {T, \precsim}$ be a continuous lattice subframe of $L$.

Then $P$ is algebraic lattice.

Proof
By definition of algebraic ordered set:
 * $L$ is up-complete.

By Up-Complete Lower Bounded Join Semilattice is Complete:
 * $L$ is a complete lattice.

By definition:
 * $P$ is closure system of $L$.

By Image of Operator Generated by Closure System is Set of Closure System
 * $\map {\operatorname {operator} } P \sqbrk S = T$

By Closure Operator Preserves Directed Suprema iff Image of Closure Operator Inherits Directed Suprema
 * $\map {\operatorname {operator} } P$ preserves directed suprema.

Thus by Image of Directed Suprema Preserving Closure Operator is Algebraic Lattice:
 * $P$ is an algebraic lattice.