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

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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 $\tau_P$ if and only if $V$ 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:

\(\ds V\) \(\in\) \(\ds \tau_P\)
\(\ds \exists U_1 \in \tau_1: \, \) \(\ds V\) \(=\) \(\ds U_1 \cap S_3\)
\(\ds \) \(=\) \(\ds U_1 \cap \paren {S_2 \cap S_3}\) Intersection with Subset is Subset
\(\ds \) \(=\) \(\ds \paren {U_1 \cap S_2} \cap S_3\) Intersection is Associative
\(\ds \leadstoandfrom \ \ \) \(\ds \exists U_2 \in \tau_2: \, \) \(\ds V\) \(=\) \(\ds U_2 \cap S_3\) Definition of Subspace Topology: $U_1 \cap S_2 \in \tau_2$
\(\ds \leadstoandfrom \ \ \) \(\ds V\) \(\in\) \(\ds \tau_Q\) Definition of $\tau_Q$

Hence the result.

$\blacksquare$


Sources

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