Indiscrete Topology is Coarsest Topology

Theorem
Let $S$ be a set.

Let $\tau$ be the indiscrete topology on $S$.
 * $\tau$ is the coarsest topology on $S$.

Hence it is comparable with all other topologies on $S$.

Proof
Let $\phi$ be any topology on $S$.

Then by definition of topology, $\varnothing \in \phi$ and $S \in \phi$

Hence by definition of subset, $\tau \subseteq \phi$.

Hence by definition of coarser topology, $\tau$ is coarser than $\phi$.