Expansion of Included Set Topology

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Theorem

Let $S$ be a set.

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

Let $T_1 = \struct {S, \tau_{A_1} }$ and $T_2 = \struct {S, \tau_{A_2} }$ 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 $T_1 = \struct {S, \tau_{A_1} }$ and $T_2 = \struct {S, \tau_{A_2} }$ 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 $A_1 \subsetneq A_2$.

$\Box$


Sufficient Condition

Let $A_1 \subseteq A_2$.


Then:

\(\ds U\) \(\in\) \(\ds \tau_{A_2}\)
\(\ds \leadsto \ \ \) \(\ds A_2\) \(\subseteq\) \(\ds U\) Definition of Included Set Topology
\(\ds \leadsto \ \ \) \(\ds A_1\) \(\subseteq\) \(\ds U\) Subset Relation is Transitive
\(\ds \leadsto \ \ \) \(\ds U\) \(\in\) \(\ds \tau_{A_1}\) Definition of Included Set Topology

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


Now suppose that $A_1 \subsetneq A_2$.

Then $A_2 \nsubseteq A_1$ and so $A_1 \notin \tau_{A_2}$.

Thus $\tau_{A_1} \nsubseteq \tau_{A_2}$ and so $T_1 > T_2$.

$\Box$


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

$\blacksquare$


Sources

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