Induced Topology on Subspace of Subspace Coincides with Induced Topology on Subspace

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 on $T_2$ induced by $\tau_1$.

Let $T_3 = \struct {S_3, \tau_3}$ be the topological subspace of $T_1$ such that $\tau_3$ is the subspace topology on $T_3$ induced by $\tau_1$.

Then $\tau_3$ is the same set of subsets of $S_1$ as is the subspace topology on $T_3$ induced by $\tau_2$.

Proof
Let $\tau_P$ denote the subspace topology on $T_3$ induced by $\tau_1$.

Let $\tau_Q$ denote the subspace topology on $T_3$ induced by $\tau_2$.

The object of this exercise is to demonstrate that $\tau_P = \tau_Q$.

This will be done by showing that an arbitrary set $V$ is in $\tau_P$ $V$ is in $\tau_Q$.

We have that:
 * $\tau_P = \set {U \cap S_3: U \in \tau_1}$
 * $\tau_Q = \set {U \cap S_3: U \in \tau_2}$

Then:

Hence the result.