Ideals form Algebraic Lattice

Theorem
Let $L = \left({S, \vee, \preceq}\right)$ be a bounded below join semilattice.

Let $I = \left({\mathit{Ids}\left({L}\right), \precsim}\right)$ be an inclusion ordered set

where
 * $\mathit{Ids}\left({L}\right)$ denotes the set of all ideals in $L$,
 * $\mathord\precsim = \mathord\subseteq \cap \left({\mathit{Ids}\left({L}\right) \times \mathit{Ids}\left({L}\right)}\right)$

Then $I$ is algebraic lattice.

Proof
By definition of subset:
 * $\mathit{Ids}\left({L}\right) \subseteq \mathcal P\left({S}\right)$

where $\mathcal P\left({S}\right)$ denotes the power set of $S$.

Define $P = \left({\mathcal P\left({S}\right), \precsim'}\right)$

where $\mathord\precsim' = \mathord\subseteq \cap \left({\mathcal P\left({S}\right) \times \mathcal P\left({S}\right)}\right)$

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 algebraic lattice.

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