Ideals form Complete Lattice

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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.

$\blacksquare$

Sources

• Mizar article YELLOW_2:48