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 $T = \left({S, \left\{{\varnothing, S}\right\}}\right)$ be the indiscrete space on $S$.

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


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


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


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