Indiscrete Topology is Coarsest Topology

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Theorem

Let $T = \left({S, \tau}\right)$ be an indiscrete topological space.

$\tau$ 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 $\phi$.

$\blacksquare$


Also see


Sources