Indiscrete Topology is Coarsest Topology

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Let $T = \struct {S, \tau}$ be an indiscrete topological space.

$\tau$ is the coarsest topology on $S$.

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


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

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

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

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


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