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 $T_1$.

We have by definition of subspace topology that:

$\tau_2 = \set {U \cap S_2: U \in \tau_1}$

Then we have by hypothesis 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 $T_1$.

$\blacksquare$

1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $3$: Continuity generalized: topological spaces: Exercise $3.9: 10$