Ideals are Continuous Lattice Subframe of Power Set

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 $L$,
 * $\mathord\precsim = \mathord\subseteq \cap \left({\mathit{Ids}\left({L}\right) \times \mathit{Ids}\left({L}\right)}\right)$

Let $P = \left({\mathcal P\left({S}\right), \precsim'}\right)$ be an inclusion ordered set

where
 * $\mathcal P\left({S}\right)$ denotes the power set of $S$,
 * $\mathord\precsim' = \mathord\subseteq \cap \left({\mathcal{P}\left({S}\right) \times \mathcal{P}\left({S}\right)}\right)$

Then $I$ is continuous lattice subframe of $P$.

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

Then
 * $\mathord\precsim = \mathord\precsim' \cap \left({\mathit{Ids}\left({L}\right) \times \mathit{Ids}\left({L}\right)}\right)$

Hence $I$ is ordered subset of $P$.

Infima Inheriting
Let $A$ be a subset of $\mathit{Ids}\left({L}\right)$ such that
 * $A$ admits an infimum in $P$.

By the proof of Power Set is Complete Lattice:
 * $\inf_P A = \bigcap A$

By Intersection of Ideals is Ideal:
 * $\inf_P A \in \mathit{Ids}\left({L}\right)$

Thus by Infimum in Ordered Subset:
 * $A$ admits an infimum in $I$ and $\inf_I A = \inf_P A$

Hence $I$ inherits infima.

Directed Suprema Inheriting
Let $D$ be a directed subset of $\mathit{Ids}\left({L}\right)$ such that
 * $D$ admits a supremum in $P$.

By the proof of Power Set is Complete Lattice:
 * $\sup_P D = \bigcup D$

We will prove that
 * $\bigcup D$ is an ideal in $L$.

Directed
Let $x, y \in \bigcup D$.

By definition of union:
 * $\exists I_1 \in D: x \in I_1$

and
 * $\exists I_2 \in D: y \in I_2$

By definition of directed:
 * $\exists I \in D: I_1 \precsim I \land I_2 \precsim I$

By definition of $\precsim$:
 * $I_1 \subseteq I$ and $I_2 \subseteq I$

By definition of subset:
 * $x, y \in I$

By definition of directed:
 * $\exists z \in I: x \preceq z \land y \preceq z$

Thus by definition of union:
 * $\exists z \in \bigcup D: x \preceq z \land y \preceq z$