Expansion of Included Set Topology

Theorem
Let $S$ be a set.

Let $A_1 \subseteq S$ and $A_2 \subseteq S$.

Let $T_1 = \left({S, \tau_{A_1}}\right)$ and $T_2 = \left({S, \tau_{A_2}}\right)$ be included set spaces on $S$.

Then:
 * $(1): \quad T_1 \ge T_2 \iff A_1 \subseteq A_2$
 * $(2): \quad T_1 > T_2 \iff A_1 \subsetneq A_2$

where:
 * $T_1 \ge T_2$ denotes that $T_1$ is finer than $T_2$
 * $T_1 > T_2$ denotes that $T_1$ is strictly finer than $T_2$.

Proof
Let $T_1 = \left({S, \tau_{A_1}}\right)$ and $T_2 = \left({S, \tau_{A_2}}\right)$ be included set spaces on $S$.

Necessary Condition
Let $T_1 \ge T_2$.

We have that $A_2 \in \tau_{A_2}$ by definition of included set topology.

But by definition of finer topology, $A_2 \in \tau_{A_1}$.

So by definition of included set topology, $A_1 \subseteq A_2$.

Now suppose that $T_1 > T_2$.

That is, $T_1 \ge T_2$ and $T_1 \ne T_2$.

It follows that $\exists X \in \tau_{A_1}: X \notin \tau_{A_2}$.

Then $A_1 \subseteq X \subsetneq A_2$ from which it follows that $A_1 \subsetneq A_2$.

Sufficient Condition
Let $A_1 \subseteq A_2$.

Then:

So $T_1 \ge T_2$ by definition of finer topology.

Now suppose that $A_1 \subsetneq A_2$.

Then $A_2 \not \subseteq A_1$ and so $A_1 \notin \tau_{A_2}$.

Thus $\tau_{A_1} \not \subseteq \tau_{A_2}$ and so $T_1 > T_2$.

Both necessity and sufficiency have been proved, hence the result.