Ideals form Algebraic Lattice

Theorem
Let $L = \struct {S, \vee, \preceq}$ be a bounded below join semilattice.

Let $I = \struct {\map {\operatorname{Ids} } L, \precsim}$ be an inclusion ordered set

where
 * $\map {\operatorname{Ids} } L$ denotes the set of all ideals in $L$
 * $\mathord \precsim = \mathord \subseteq \cap \paren {\map {\operatorname{Ids} } L \times \map {\operatorname{Ids} } L}$

Then $I$ is an algebraic lattice.

Proof
By definition of subset:
 * $\map {\operatorname{Ids} } L \subseteq \powerset S$

where $\powerset S$ denotes the power set of $S$.

Define:
 * $P = \struct {\powerset S, \precsim'}$

where:
 * $\mathord \precsim' = \mathord\subseteq \cap \paren {\powerset S \times \powerset S}$

By Ideals are Continuous Lattice Subframe of Power Set:
 * $I$ is an continuous lattice subframe of $P$.

By Lattice of Power Set is Algebraic:
 * $P$ is an algebraic lattice.

Thus by Continuous Lattice Subframe of Algebraic Lattice is Algebraic Lattice:
 * $I$ is an algebraic lattice.