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 $$\mathcal{P} \left({S}\right)$$.

Next we note that:


 * $$\forall S_1, S_2 \in \mathcal P \left({S}\right) : S_1 \subseteq \left({S_1 \cup S_2}\right) \and S_2 \subseteq \left({S_1 \cup S_2}\right)$$ from Subset of Union;


 * $$\forall S_1, S_2 \in \mathcal P \left({S}\right) : \left({S_1 \cap S_2}\right) \subseteq S_1 \and \left({S_1 \cap S_2}\right) \subseteq S_2$$ from Intersection Subset.

It is then straightforward to prove that $$S_1 \cap S_2$$ is the infimum and $$S_1 \cup S_2$$ the supremum of $$\left\{{S_1, S_2}\right\} \subseteq \mathcal P \left({S}\right)$$.

Thus the conditions for $$\left({\mathcal P \left({S}\right), \subseteq}\right)$$ being a lattice are fulfilled.


 * Now we need to show that $$\left({\mathcal P \left({S}\right), \subseteq}\right)$$ is a complete lattice.