# Metric Subspace Induces Subspace Topology

## Theorem

Let $M = \struct {A,d}$ be a metric space.

Let $H \subseteq A$.

Let $\tau$ be the topology induced by the metric $d$.

Let $\tau_H$ be the subspace topology induced by $\tau$ on $H$.

Let $d_H$ be the subspace metric induced by $d$ on $H$.

Let $\tau_{d_H}$ be the topology induced by the metric $d_H$.

Then:

$\tau_{d_H} = \tau_H$

## Proof

Let $\mathcal B$ be the set of open $\epsilon$-balls in $M$.

Let $\mathcal B_H$ be the set of open $\epsilon$-balls in $\struct {H, d_H}$.

Let $U \in \tau_{d_H}$.

By the Definition of the topology induced by the metric $d_H$ then:

$\exists \mathcal A_H \subseteq \mathcal B_H:U = \bigcup \mathcal A_H$

Let $\mathcal A = \set {B': B' \in \mathcal B, B' \cap H \in \mathcal A_H}$

Let $V = \bigcup \set {B' : B' \in \mathcal A}$

By the definition of the topology induced by the metric $d$ then:

$V \in \tau$

By the definition of the subspace metric then:

$\forall B \in \mathcal B_H: \exists B' \in \mathcal B: B = B' \cap H$

Hence:

 $\displaystyle U$ $=$ $\displaystyle \bigcup \set {B' \cap H : B' \in \mathcal A}$ $\displaystyle$ $=$ $\displaystyle \paren {\bigcup \set {B' : B' \in \mathcal A} } \cap H$ Intersection Distributes over Union $\displaystyle$ $=$ $\displaystyle V \cap H$

By the definition of the subspace topology induced by $\tau$ on $H$ then:

$U \in \tau_H$

Hence:

$\tau_{d_H} \subseteq \tau_H$.

$\Box$

Let $U \in \tau_H$.

By the definition of the subspace topology induced by $\tau$ on $H$ then:

$\exists V \in \tau : U = V \cap H$

By the definition of the topology induced by the metric $d$ then:

$\exists \mathcal A \subseteq \mathcal B: V = \bigcup \set {B' : B' \in \mathcal A}$

Hence:

 $\displaystyle U$ $=$ $\displaystyle \paren {\bigcup \set {B' : B' \in \mathcal A} } \cap H$ $\displaystyle$ $=$ $\displaystyle \bigcup \set {B' \cap H : B' \in \mathcal A}$ Intersection Distributes over Union

By the definition of the subspace metric then:

$\forall B' \in \mathcal B: B' \cap H \in \mathcal B_H$

By the Definition of the topology induced by the metric $d_H$ then:

$U \in \tau_{d_H}$

Hence:

$\tau_H \subseteq \tau_{d_H}$.

$\tau_H = \tau_{d_H}$.

$\blacksquare$