Union of Topologies is not necessarily Topology

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

Then $\tau := \displaystyle \bigcup_{i \mathop \in I} {\tau_i}$ is not necessarily also a topology for $S$.

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

Let:
 * $\tau_1 := \left\{{\varnothing, \left\{{0}\right\}, \left\{{1}\right\}, \left\{{0, 1}\right\}, S}\right\}$
 * $\tau_2 := \left\{{\varnothing, \left\{{0}\right\}, \left\{{2}\right\}, \left\{{0, 2}\right\}, S}\right\}$

be topologies on $S$.

Then:
 * $\tau := \tau_1 \cup \tau_2 = \left\{{\varnothing, \left\{{0}\right\}, \left\{{1}\right\}, \left\{{2}\right\}, \left\{{0, 1}\right\}, \left\{{0, 2}\right\}, S}\right\}$

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

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

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 $\left|{S}\right| \le 2$.