Indiscrete Topology is Coarsest Topology

Theorem

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 $\phi$.

$\blacksquare$

1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $3$: Continuity generalized: topological spaces: $3.1$: Topological Spaces: Example $3.1.6$
1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $1$: General Introduction
1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $4$. Indiscrete Topology: $1$
1989: Ephraim J. Borowski and Jonathan M. Borwein: Dictionary of Mathematics ... (previous) ... (next): coarser (of a topology)
2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): Chapter $\S 2.3$: The normed space $\map {CL} {X, Y}$. Strong and weak operator topologies on $\map {CL} {X, Y}$