Trivial Topological Space is Indiscrete

Theorem
Let $S$ be a singleton.

The only possible topology on $S$ is the trivial topological space.

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

Suppose $S = \set{x}$ for a certain object $x$.

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


 * $\map {\mathcal P} S = \set{\varnothing, \set{x}}$

That is:


 * $\map {\mathcal P} S = \set{\varnothing, S}$

Since all topologies $\tau$ on $S$ are subsets of $\tau \subseteq \map {\mathcal P} S$, one of the following must hold:


 * $\tau_1 = \varnothing$
 * $\tau_2 = \set{\varnothing}$
 * $\tau_3 = \set{S}$

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

By definition of a topology, $S$ must be an element of the topology.

Thus $\tau_1$ and $\tau_2$ are not topologies on $S$.

By Empty Set is Element of Topology, also $\varnothing \in \tau$, for $\tau$ to be a topology for $S$.

Therefore $\tau_3$ is also not a topology.

Finally, 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.

This topology is the trivial topological space on $x$.