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.
$\blacksquare$