Openness Relation on Topological Spaces is Transitive

Theorem
Let $T_1 = \struct {S_1, \tau_1}$ be a topological space

Let $S_2 \subseteq S_1$ be a subset of $S_1$.

Let $S_3 \subseteq S_2$ be a subset of $S_2$.

Let $T_2 = \struct {S_2, \tau_2}$ be the topological subspace of $T_1$ such that $\tau_2$ is the subspace topology induced by $\tau_1$.

Let $T_3 = \struct {S_3, \tau_3}$ be the topological subspace of $T_2$ such that $\tau_3$ is the subspace topology induced by $\tau_2$.

Let:
 * $S_2$ be an open set of $T_1$
 * $S_3$ be an open set of $T_2$.

Then:
 * $S_3$ is an open set of $T_1$.

Proof
We have by definition of subspace topology that:
 * $\tau_2 = \set {U \cap S_2: U \in \tau_1}$

Then we have that:
 * $S_3 \in \tau_2$

and so:
 * $S_3 \in \set {U \cap S_2: U \in \tau_1}$

That is, $S_3$ is the intersection of $U$ and $S_2$, both of which are open sets of $T_1$.

Hence by, $S_3$ is an open set of $T_1$.