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:

By the definition of the subspace topology induced by $\tau$ on $H$ then:
 * $U \in \tau_H$

Hence:
 * $\tau_{d_H} \subseteq \tau_H$.

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:

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

By Equivalence of Definitions of Set Equality then:
 * $\tau_H = \tau_{d_H}$.