Ideals form Complete Lattice

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Theorem

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

Let $\mathcal I = \struct {\mathit{Ids}\left({L}\right), \subseteq}$ 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 Semilattice 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.

$\blacksquare$


Sources