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 $$S$$, $$\mathcal{P} \left({S}\right)$$ has both one minimal element and one maximal element, that is, $$\varnothing$$ and $$S$$ itself.

It also happens that these are also the infimum and supremum of $$\mathcal{P} \left({S}\right)$$ in itself.


 * Next we note that:


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


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