Topology forms Complete Lattice

From ProofWiki
Jump to navigation Jump to search

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:

$\displaystyle \bigcup S \in \tau$

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

Let $I$ be the interior of $\displaystyle \bigcap S$, where by Intersection of Empty Set $\displaystyle \bigcap \O$ 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:

$\displaystyle V \subseteq \bigcap S$.

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

$\blacksquare$