Ideals form Complete Lattice

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

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

where $\map {\operatorname{Ids} } L$ denotes the set of all ideals in $L$.

Then $\II$ is complete lattice.

Proof
Let $X \subseteq \map {\operatorname{Ids} } L$.

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

By Intersection is Largest Subset/General Result:
 * $\ds \forall A \in \map {\operatorname{Ids} } L: \paren {\forall I \in X: A \subseteq I} \iff A \subseteq \bigcap X$

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

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