Topology forms Complete Lattice

Theorem
Let $\left({X, \tau}\right)$ be a topological space.

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

Proof
To show that $(\left({\tau, \subseteq}\right)$ 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:
 * $\displaystyle \bigcup S \in \tau$

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

Let $I$ be the interior of $\displaystyle \bigcap S$, where $\displaystyle \bigcap \varnothing$ 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:
 * $\displaystyle V \subseteq \bigcap S$.

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, $\left({\tau, \subseteq}\right)$ is a complete lattice.