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 $$\phi$$.