Union of Topologies on Singleton or Doubleton is Topology

Theorem

Let $S$ be a set which is either a singleton or a doubleton.

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

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

Proof

Let $S$ be a singleton.

Let $S = \set a$.

From Topology on Singleton is Indiscrete Topology, the only possible topology on $S$ is the indiscrete topology.

From Set 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 singleton is a topology on that singleton.

$\Box$

Let $S$ be a doubleton.

Let $S = \set {a, b}$.

By Topologies on Doubleton, there are $4$ possible different topologies on $S$:

\(\ds \tau_1\)

\(=\)

\(\ds \set {\O, \set {a, b} }\)

Indiscrete Topology

\(\ds \tau_2\)

\(=\)

\(\ds \set {\O, \set a, \set {a, b} }\)

Sierpiński Topology

\(\ds \tau_3\)

\(=\)

\(\ds \set {\O, \set b, \set {a, b} }\)

Sierpiński Topology

\(\ds \tau_4\)

\(=\)

\(\ds \set {\O, \set a, \set b, \set {a, b} }\)

Discrete Topology

It remains to show that every union of elements in $\set {\tau_1, \tau_2, \tau_3, \tau_4}$ is a topology on $S$.

By inspection, the following statements are seen to hold:

$\tau_1 \cup \tau_2 = \tau_2$

$\tau_1 \cup \tau_3 = \tau_3$

$\tau_1 \cup \tau_4 = \tau_4$

$\tau_2 \cup \tau_3 = \tau_4$

$\tau_2 \cup \tau_4 = \tau_4$

$\tau_3 \cup \tau_4 = \tau_4$

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

By the above, if $\tau_4$ is one of the $\tau_i$:

$\ds \tau = \bigcup_{i \mathop \in I} \tau_i = \tau_4$

So assume that $\tau_4$ is not one of the $\tau_i$.

If both $\tau_2$ and $\tau_3$ are in $\family {\tau_i}_{i \mathop \in I}$, then by the above:

$\ds \tau = \bigcup_{i \mathop \in I} \tau_i = \tau_4$

Now suppose that $\tau_3$ is in $\family {\tau_i}_{i \mathop \in I}$ and $\tau_2$ is not.

Then by the above it follows that:

$\tau = \tau_3$

Now suppose that $\tau_2$ is in $\family {\tau_i}_{i \mathop \in I}$ and $\tau_3$ is not.

Then by the above it follows that:

$\tau = \tau_2$

Finally, assume that $\tau_2$ and $\tau_3$ are not in $\family {\tau_i}_{i \mathop \in I}$.

Then the only topology in $\family {\tau_i}_{i \mathop \in I}$ is $\tau_1$.

In this, we find:

$\tau = \tau_1$

Hence the result.

$\blacksquare$

Also see

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