Indiscrete Topology is Topology

Theorem
Let $S$ be a set.

Let $\vartheta$ be the indiscrete topology on $S$.
 * $\vartheta$ is a topology on $S$.

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

Then by definition $\vartheta = \mathcal P \left({S}\right)$, that is, is the power set of $S$.

We confirm the criteria for $T$ to be a topology:
 * $(1): \quad$ Trivially, by definition, $\varnothing \in \vartheta$ and $S \in \vartheta$.
 * $(2): \quad \varnothing \cup \varnothing = \varnothing \in \vartheta$, $\varnothing \cup S = S \in \vartheta$ and $S \cup S = S \in \vartheta$ from Union with Null and Union is Idempotent.
 * $(3): \quad \varnothing \cap \varnothing = \varnothing \in \vartheta$, $\varnothing \cap S = \varnothing \in \vartheta$ and $S \cap S = S \in \vartheta$ from Intersection with Null and Intersection is Idempotent.