Power Set is Complete Lattice

Theorem
Let $$S$$ be a set.

Let $$\left({\mathcal P \left({S}\right), \subseteq}\right)$$ be the relational structure defined on $$\mathcal P \left({S}\right)$$ by the relation $$\subseteq$$.

Then $$\left({\mathcal P \left({S}\right), \subseteq}\right)$$ is a complete lattice.

Proof
From Subset Relation on Power Set is Partial Ordering, we have that $$\subseteq$$ is a partial ordering.

We note in passing that for any set $$S$$:
 * From Supremum of Power Set, $$\mathcal P \left({S}\right)$$ has a supremum, that is, $$S$$ itself;
 * From Infimum of Power Set, $$\mathcal P \left({S}\right)$$ has an infimum, that is, $$\varnothing$$.

These are also the maximal and minimal elements of $$\mathcal{P} \left({S}\right)$$.

Let $$\mathbb S$$ be a subset of $$\mathcal P \left({S}\right)$$.

Then:
 * From Union Smallest:
 * $$\left({\forall X \in \mathbb S: X \subseteq T}\right) \iff \bigcup \mathbb S \subseteq T$$


 * From Intersection Largest:
 * $$\left({\forall X \in \mathbb S: T \subseteq X}\right) \iff T \subseteq \bigcap \mathbb S$$

So $$\bigcap \mathbb S$$ is the infimum and $$\bigcup \mathbb S$$ is the supremum of $$\left({\mathbb S, \subseteq}\right)$$.

Hence by definition $$\mathcal P \left({S}\right)$$ is a complete lattice.