Ideals form Complete Lattice

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

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

where $\mathit{Ids}\left({L}\right)$ denotes the set of all ideals in $L$.

Then $\mathcal I$ is complete lattice.

Proof
Let $X \subseteq \mathit{Ids}\left({L}\right)$.

By Intersection of Ideals is Ideal/Set of Sets:
 * $\bigcap X$ is an ideal.

By Intersection is Largest Subset/General Result:
 * $\forall A \in \mathit{Ids}\left({L}\right):\left({\forall I \in X: A \subseteq I}\right) \iff A \subseteq \bigcap X$

Thus by definition:
 * $X$ admits an infimum.

Thus by dual to Lattice is Complete iff it Admits All Suprema:
 * $\mathcal I$ is complete lattice.