Included Set Topology on Finite Intersection

Theorem
Let $T = \struct {S, \tau}$ be a topological space on a set $S$.

Let $A_1, A_2, \ldots, A_n$ be a finite set of subsets of $S$:
 * $\forall i \in \closedint 1 n: A_i \subseteq S$

Let $\forall i \in \closedint 1 n: \map T {A_i} = \struct {S, \tau_{A_i} }$ be the included set spaces on $S$ by $A_i$.

Let:
 * $\forall i \in \closedint 1 n: \map T {A_i} \le T$

where $\map T {A_i} \le T$ denotes that $\map T {A_i}$ is coarser than $T$.

Then:
 * $\ds \map T {\bigcap A_i} \le T$

where $\ds \map T {\bigcap A_i}$ is the included set space on $S$ by $\ds \bigcap_{i \mathop = 1}^n A_i$.

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

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:


 * $U = \ds \paren {\bigcap_{i \mathop = 1}^n A_i} \cup Z = \bigcap_{i \mathop = 1}^n \paren {A_i \cup Z}$

where the latter equality follows from Union Distributes over Intersection.

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

Further, by assumption, $\map T {A_i} \le T$, that is $\tau_{A_i} \subseteq \tau$.

It follows that $A_i \cup Z \in \tau$ for all $i \in \closedint 1 n$.

Hence $U = \ds \bigcap_{i \mathop = 1}^n \paren {A_i \cup Z}$ is in $\tau$, because $\tau$ is a topology.

This comes down to:
 * $\tau_A \subseteq \tau$

Hence by definition of coarser topology:
 * $\map T {\bigcap A_i} \le T$