Included Set Topology on Finite Intersection

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

Let $A_1, A_2, \ldots, A_n$ be a finite set of subsets of $S$:
 * $\forall i \in \left[{1 .. n}\right]: A_i \subseteq S$

Let $\forall i \in \left[{1 .. n}\right]: T \left({A_i}\right) = \left({S, \tau_{A_i}}\right)$ be the included set spaces on $S$ by $A_i$.

Let:
 * $\forall i \in \left[{1 .. n}\right]: T \left({A_i}\right) \le T$

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

Then:
 * $T \left({\bigcap A_i}\right) \le T$

where $T \left({\bigcap A_i}\right)$ is the included set space on $S$ by $\displaystyle \bigcap_{i=1}^n A_i$.

Proof
For ease of notation, define:
 * $A := \displaystyle \bigcap_{i=1}^n A_i$

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

Let $U \in \tau_A$ be nonempty.

Then by definition, $A \subseteq U$.

Hence there is a subset $Z \subseteq S$ of $S$, such that $U = A \cup Z$; that is:


 * $U = \displaystyle \left({\bigcap_{i=1}^n A_i}\right) \cup Z = \bigcap_{i=1}^n \left({A_i \cup Z}\right)$

where the latter equality follows from Union Distributes over Intersection.

For every $i \in \left[{1 .. n}\right]$, one has $A_i \cup Z \in \tau_{A_i}$ by definition of included set topology.

Further, by assumption, $T \left({A_i}\right) \le T$, i.e. $\tau_{A_i} \subseteq \tau$.

It follows that $A_i \cup Z \in \tau$ for all $i \in \left[{1 .. n}\right]$.

Hence $U = \displaystyle \bigcap_{i=1}^n \left({A_i \cup Z}\right)$ is in $\tau$, because $\tau$ is a topology.

This comes down to $\tau_A \subseteq \tau$, and hence $T \left({\bigcap A_i}\right) \le T$, by definition of coarser.