Union of Topologies on Singleton or Doubleton is Topology

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

Let $\paren{\tau_i}_{i \mathop \in I}$ be an arbitrary non-empty indexed set of topologies for a set $S$.

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

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, that the only possible topology on $S$ is the indiscrete topology.

From Union is Idempotent, 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 $x$.

Let $S$ be a set containing exactly two elements.

Suppose $S = \set{x, y}$ for certain objects $x$ and $y$.

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


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

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


 * $\tau_1 = \varnothing$
 * $\tau_2 = \set{\varnothing}$
 * $\tau_3 = \set{\set{x}}$
 * $\tau_4 = \set{\set{y}}$
 * $\tau_5 = \set{\varnothing, \set{x}}$
 * $\tau_6 = \set{\varnothing, \set{y}}$
 * $\tau_7 = \set{\set{x}, \set{y}}$
 * $\tau_8 = \set{\varnothing, \set{x}, \set{y}}$
 * $\tau_9 = \set{\set{x}, \set{x, y}}$
 * $\tau_{10} = \set{\set{y}, \set{x, y}}$
 * $\tau_{11} = \set{\set{x}, \set{y}, \set{x, y}}$
 * $\tau_{12} = \set{\set{x, y}}$
 * $\tau_{13} = \set{\varnothing, \set{x, y}}$
 * $\tau_{14} = \set{\varnothing, \set{x}, \set{x, y}}$
 * $\tau_{15} = \set{\varnothing, \set{y}, \set{x, y}}$

or
 * $\tau_{16} = \set{\varnothing, \set{x}, \set{y}, S}$.

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

Thus $\tau_1$ up to $\tau_8$ are not topologies on $S$.

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

Therefore $\tau_9$ up to $\tau_{12}$ are also not topologies on $S$.

By Indiscrete Topology is Topology, $\tau_{13}$ is a topology on $S$.

By Discrete Topology is Topology, $\tau_{16}$ is a topology on $S$.

By Particular Point Topology is Topology, both $\tau_{14}$ and $\tau_{15}$ are topologies.

Finally, it remains to show that any union of elements in $\{\tau_{13}, \tau_{14}, \tau_{15}, \tau_{16}\}$ gives a topology on $S$.

By inspection, the following statements are seen to hold:
 * $\tau_{13} \cup \tau_{14} = \tau_{14}$
 * $\tau_{13} \cup \tau_{15} = \tau_{15}$
 * $\tau_{13} \cup \tau_{16} = \tau_{16}$
 * $\tau_{14} \cup \tau_{15} = \tau_{16}$
 * $\tau_{14} \cup \tau_{16} = \tau_{16}$
 * $\tau_{15} \cup \tau_{16} = \tau_{16}$

Now let $\paren{\tau_i}_{i \mathop \in I}$ be an arbitrary non-empty indexed set of topologies for a set $S$.

By the above, if $\tau_{16}$ is one of the $\tau_i$:


 * $\displaystyle \tau = \bigcup_{i \mathop \in I} \tau_i = \tau_{16}$

So assume that $\tau_{16}$ is not one of the $\tau_i$.

If both $\tau_{14}$ and $\tau_{15}$ are in $\paren{\tau_i}_{i \mathop \in I}$, then by the above:


 * $\displaystyle \tau = \bigcup_{i \mathop \in I} \tau_i = \tau_{16}$

Now suppose that $\tau_{15}$ is in $\paren{\tau_i}_{i \mathop \in I}$ and $\tau_{14}$ is not.

Then by the above it follows that:
 * $\tau = \tau_{15}$

Now suppose that $\tau_{14}$ is in $\paren{\tau_i}_{i \mathop \in I}$ and $\tau_{15}$ is not.

Then by the above it follows that:
 * $\tau = \tau_{14}$

Finally, assume that $\tau_{14}$ and $\tau_{15}$ are not in $\paren{\tau_i}_{i \mathop \in I}$.

Then the only topologies in $\paren{\tau_i}_{i \mathop \in I}$ are $\tau_{13}$.

In the first case, we find:
 * $\tau = \tau_{13}$

Hence the result.

Also see

 * Union of Topologies is not necessarily Topology, which shows that when $\size{S} \ge 3$ this result is false.