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 $T_i$ 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
Let $\tau_\cap$ denote the included set topology on $S$ by $\displaystyle \bigcap_{i=1}^n A_i$.

Thus $T \left({\bigcap A_i}\right) := \left({S, \tau_\cap}\right)$.

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

Thus, by definition of coarser topology:
 * $\forall i \in \left[{1 .. n}\right]: \tau_{A_i} \subseteq \tau$

From Intersection Subset we have that:
 * $\displaystyle \forall i \in \left[{1 .. n}\right]: \bigcap_{i=1}^n A_i \subseteq A_i$

Thus, from Expansion of Included Set Topology:
 * $\forall i \in \left[{1 .. n}\right]: \left({S, \tau_{A_i}}\right) \le \left({S, \tau_\cap}\right)$

Thus, by definition of coarser topology:
 * $\forall i \in \left[{1 .. n}\right]: \tau_{A_i} \subseteq \tau_\cap$

So from Intersection Largest:
 * $\tau_\cap \subseteq \tau$