Topology forms Complete Lattice

Theorem
Let $(X, \tau)$ be a topological space.

Then $(\tau, \subseteq)$ is a complete lattice.

Proof
To show that $(\tau, \subseteq)$ 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, $\bigcup S \in \tau$.

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

Let $I$ be the interior of $\bigcap S$, where $\bigcap \varnothing$ is conventionally taken to be $X$.

Then by the definition of interior and Intersection Largest, $I \in \tau$ and $I \subseteq U$ for each $U \in S$.

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

By Intersection Largest, $V \subseteq \bigcap S$.

Then by the definition of interior, $V \subseteq I$.

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

As each subset of $\tau$ has a supremum and an infimum, $(\tau, \subseteq)$ is a complete lattice.