Included Set Topology on Union

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space on a set $S$.

Let $\left\langle{A_i}\right\rangle_{i \in I}$ be a family of subsets of $S$ indexed by the indexing set $I$:
 * $\forall i \in I: A_i \subseteq S$

Let $\forall i \in I: T \left({A_i}\right) = \left({S, \tau_{A_i}}\right)$ be the included set spaces on $S$ by $A_i$.

Let:
 * $\forall i \in I: T \left({A_i}\right) \ge T$

where $T \left({A_i}\right) \ge T$ denotes that $T \left({A_i}\right)$ is finer than $T$.

Then:
 * $T \left({\bigcup A_i}\right) \ge T$

where $T \left({\bigcup A_i}\right)$ is the included set space on $S$ by $\displaystyle \bigcup_{i \in I} A_i$.

Proof
For ease of notation, define:
 * $A := \displaystyle \bigcup_{i \in I} A_i$

and let $\tau_A$ denote the included set topology on $S$ by $A$.

Let $U \in \tau$ be nonempty.

As $T \left({A_i}\right)$ is finer than $T$ it follows by definition that:
 * $\forall i \in I: \tau \subseteq \tau_{A_i}$

Thus:
 * $\forall i \in I: U \in \tau_{A_i}$

Hence for all $i$ there is a subset $Z_i \subseteq S$ of $S$, such that $U = A_i \cup Z_i$; that is:


 * $\displaystyle U = \bigcup_{i \in I} \left({A_i \cup Z_i}\right) = \left({\bigcup_{i \in I} A_i}\right) \cup \left({\bigcup_{i \in I} Z_i}\right)$

where the latter equality follows from associativity and commutativity of set union.

That is:
 * $U = A \cup Z$

where $\displaystyle Z = \left({\bigcup_{i \in I} Z_i}\right)$.

Hence $U \in \tau_A$. by definition of the included set topology.

This comes down to $\tau \subseteq \tau_A$, and hence $T \left({\bigcup A_i}\right) \ge T$, by definition of a finer topology.