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 non-empty 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 $x$.

$\Box$

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

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

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

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

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

$\tau_1 = \varnothing$

$\tau_2 = \{\varnothing \}$

$\tau_3 = \{\{x\}\}$

$\tau_4 = \{\{y\}\}$

$\tau_5 = \{\varnothing, \{x\}\}$

$\tau_6 = \{\varnothing, \{y\}\}$

$\tau_7 = \{\{x\}, \{y\}\}$

$\tau_8 = \{\varnothing, \{x\}, \{y\}\}$

$\tau_9 = \{\{x\}, \{x,y\}\}$

$\tau_{10} = \{\{y\}, \{x,y\}\}$

$\tau_{11} = \{\{x\}, \{y\}, \{x,y\}\}$

$\tau_{12} = \{\{x,y\}\}$

$\tau_{13} = \{\varnothing, \{x,y\}\}$

$\tau_{14} = \{\varnothing, \{x\}, \{x,y\}\}$

$\tau_{15} = \{\varnothing, \{y\}, \{x,y\}\}$

or

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

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, also $\varnothing \in \tau$, for $\tau$ to be a topology for $S$.

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

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

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

We have to prove that $\tau_{13}$ and $\tau_{14}$ are topologies.

Both proofs are similar, and quite elementary.

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

Clearly, we have the following statements:

$\tau_{13} \cup \tau_{14} = \tau_{14}$

$\tau_{13} \cup \tau_{15} = \tau_{15}$

$\tau_{13} \cup \tau_{16} = \tau_{16}$

Also, one can easily check that:

$\tau_{14} \cup \tau_{15} = \tau_{16}$

$\tau_{14} \cup \tau_{16} = \tau_{16}$

and

$\tau_{15} \cup \tau_{16} = \tau_{16}$.

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

If $\tau_{16}$ is one of the $\tau_i$, we have by the above, that:

$\tau = \bigcup_{i\in I} {\tau_i} = \tau_{16}$.

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

If both $\tau_{14}$ and $\tau_{15}$ are in $\left({\tau_i}\right)_{i\in I}$, then by the above, we find that:

$\tau = \bigcup_{i\in I} {\tau_i} = \tau_{16}$.

So now we assume that $\tau_{15}$ is in $\left({\tau_i}\right)_{i\in I}$ and $\tau_{14}$ isn't.

Then by the above we find that $\tau=\tau_{15}$.

Now assume that $\tau_{14}$ is in $\left({\tau_i}\right)_{i\in I}$ and $\tau_{15}$ isn't.

Then by the above we find that $\tau=\tau_{14}$.

Finally, let's assume that $\tau_{14}$ and $\tau_{15}$ aren't in $\left({\tau_i}\right)_{i\in I}$.

Then the only topologies in $\left({\tau_i}\right)_{i\in I}$ are $\tau_{13}$.

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

Hence, we obtain the result.

$\blacksquare$

Remark

If however, $|S| \geq 3$, the above result is false. See: Union of Topologies is Not a Topology