# Topologies on Set form Complete Lattice

## Theorem

Let $X$ be a non-empty set.

Let $\mathcal L$ be the set of topologies on $X$.

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

## Proof

Let $\mathcal K \subseteq \mathcal L$.

$\bigcap \mathcal K \in \mathcal L$

By Intersection is Largest Subset, $\bigcap \mathcal L$ is the infimum of $\mathcal K$.

Let $\tau$ be the topology generated by the sub-basis $\bigcup \mathcal K$.

Then $\tau \in \mathcal L$ and $\tau$ is the supremum of $\mathcal K$.

We have that each subset of $\mathcal L$ has a supremum and an infimum in $\mathcal L$.

Thus it follows that $\left({\mathcal L, \subseteq}\right)$ is a complete lattice.

$\blacksquare$