Discrete Topology is Finest Topology
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Theorem
Let $T = \struct {S, \tau}$ be a discrete topological space.
- $\tau$ is the finest topology on $S$.
Proof
Let $\phi$ be any topology on $S$.
Let $U \in \phi$.
Then, by the definition of topology, $U \subseteq S$.
Then, by the definition of discrete topological space, $U \in \tau$.
Hence by definition of subset, $\phi \subseteq \tau$.
Hence by definition of finer topology, $\tau$ is finer than $\phi$.
$\blacksquare$
Also see
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $3$: Continuity generalized: topological spaces: $3.1$: Topological Spaces
- 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: $1 \text { - } 3$. Discrete 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 $\map {CL} {X, Y}$