Indiscrete Topology is Topology

Theorem

Let $S$ be a set.

Let $\tau$ be the indiscrete topology on $S$.

$\tau$ is a topology on $S$.

Proof

Let $T = \struct {S, \set {\O, S} }$ be the indiscrete space on $S$.

Confirming the criteria for $T$ to be a topology:

$(1): \quad$ Trivially, by definition, $\O \in \tau$ and $S \in \tau$.

$(2): \quad \O \cup \O = \O \in \tau$, $\O \cup S = S \in \tau$ and $S \cup S = S \in \tau$ from Union with Empty Set and Set Union is Idempotent.

$(3): \quad \O \cap \O = \O \in \tau$, $\O \cap S = \O \in \tau$ and $S \cap S = S \in \tau$ from Intersection with Empty Set and Set Intersection is Idempotent.

$\blacksquare$

Sources

• 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 $\text {II}$: Counterexamples: $4$. Indiscrete Topology
• 2011: John M. Lee: Introduction to Topological Manifolds (2nd ed.) ... (previous) ... (next): $\S 2$: Topological Spaces: Topologies