Topology forms Complete Lattice

Theorem
Let $\struct {X, \tau}$ be a topological space.

Then $\struct {\tau, \subseteq}$ is a complete lattice.

Proof
To show that $\struct {\tau, \subseteq}$ is a complete lattice, we must show that every subset of $\tau$ has a supremum and an infimum.

Let $S \subseteq \tau$.

By the definition of a topology:
 * $\ds \bigcup S \in \tau$

By Union is Smallest Superset, $\ds \bigcup S$ is the supremum of $S$.

Let $I$ be the interior of $\ds \bigcap S$, where by Intersection of Empty Set $\ds \bigcap \O$ is conventionally taken to be $X$.

Then by the definition of interior and Intersection is Largest Subset:
 * $I \in \tau$

and
 * $I \subseteq U$

for each $U \in S$.

Let $V \in \tau$ with $V \subseteq U$ for each $U \in S$.

By Intersection is Largest Subset:
 * $\ds V \subseteq \bigcap S$.

Then by the definition of interior:
 * $V \subseteq I$

Thus $I$ is the infimum of $S$.

So each subset of $\tau$ has a supremum and an infimum.

Thus, by definition, $\struct {\tau, \subseteq}$ is a complete lattice.