Union of Topologies on Singleton or Doubleton is Topology

Theorem
Let $S$ be a set containing either exactly one or exactly two elements.

Let $\left({\tau_i}\right)_{i \in I}$ be an arbitrary indexed set of topologies for a set $S$.

Then $\tau := \displaystyle \bigcup_{i \in I} {\tau_i}$ is also a topology for $X$.

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

Say, $S = \{x\}$ for a certain object $x$.

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


 * $\mathcal{P}(S)=\{\varnothing, \{x\}\}$

Or, in other wording:


 * $\mathcal{P}(S)=\{\varnothing, S\}$

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


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

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

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

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

Clearly the union of any number of indiscrete topologies on $S$ is the indiscrete topology.

Thus the union of any number of topologies on a set with exactly one element is a topology on that set.

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