Union of Topologies is not necessarily Topology

Theorem
Let $\tau_1$ and $\tau_2$ be topologies on a set $S$.

Then $\tau_1 \cup \tau_2$ is not necessarily also a topology on $S$.

Proof
Let $S := \set {0, 1, 2}$ be a set.

Let:
 * $\tau_1 := \set {\O, \set 0, \set 1, \set {0, 1}, S}$
 * $\tau_2 := \set {\O, \set 0, \set 2, \set {0, 2}, S}$

be topologies on $S$.

Then:
 * $\tau := \tau_1 \cup \tau_2 = \set {\O, \set 0, \set 1, \set 2, \set {0, 1} \set {0, 2}, S}$

For $\tau$ to be a topology the union of any number of elements of $\tau$ should also be in $\tau$.

But:
 * $\set 1 \cup \set 2 = \set {1, 2} \not \in \tau$

Therefore $\tau$ is not a topology on $S$.

Hence the result.

Also see

 * Union of Topologies on Singleton or Doubleton is Topology where it is shown that this result does not hold if $\size S \le 2$.