Trivial Topological Space is Indiscrete

Theorem
Let $S$ be a singleton.

The only possible topology on $S$ is the indiscrete topology.

Proof
Let $S$ be a set containing exactly one element.

Suppose $S = \set x$ for some object $x$.

Then the power set of $S$ is the set:


 * $\powerset S = \set {\O, \set x}$

That is:


 * $\powerset S = \set {\O, S}$

Let $\tau$ be a topology on $S$.

$\tau$ is a subset of $\powerset S$, $\tau$ must equal one of the following sets:


 * $\tau_1 = \O$
 * $\tau_2 = \set \O$
 * $\tau_3 = \set S$

or
 * $\tau_4 = \set {\O, S}$.

By definition of a topology, $S \in \tau$.

Thus $\tau \ne \tau_1$ and $\tau \ne \tau_2$.

By Empty Set is Element of Topology, also $\O \in \tau$.

Therefore $\tau \ne \tau_3$.

By Indiscrete Topology is Topology, $\tau_4$ is a topology on $S$.

So if $S$ is a set containing exactly one element, the only possible topology on $S$ is the indiscrete topology.