Properties of Indiscrete Topology

Theorem
Let $S$ be a set.

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

Then:


 * $\vartheta$ is indeed a topology on $S$;
 * $\vartheta$ is the coarsest topology on $S$.

Proof
Let $\vartheta$ be the indiscrete topology on $S$.

Then by definition $\vartheta = \left\{{\varnothing, S}\right\}$.


 * $\vartheta$ is a topology on $S$:


 * 1. Trivially, $\varnothing \in \vartheta$ and $S \in \vartheta$.
 * 2. $\varnothing \cup \varnothing = \varnothing \in \vartheta$, $\varnothing \cup S = S \in \vartheta$ and $S \cup S = S \in \vartheta$ from Union with Null and Union is Idempotent.
 * 3. $\varnothing \cap \varnothing = \varnothing \in \vartheta$, $\varnothing \cap S = \varnothing \in \vartheta$ and $S \cap S = S \in \vartheta$ from Intersection with Null and Intersection is Idempotent.


 * $\vartheta$ is the coarsest topology on $S$:

Let $\phi$ be any topology on $S$.

Then by definition of topology, $\varnothing \in \phi$ and $S \in \phi$

Hence by definition of subset, $\vartheta \subseteq \phi$.

Hence by definition of coarser topology, $\vartheta$ is coarser than $\phi$.