Metric Subspace Induces Subspace Topology

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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 $\BB$ be the set of open $\epsilon$-balls in $M$.

Let $\BB_H$ 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 $d_H$:

$\exists \AA_H \subseteq \BB_H: U = \bigcup \AA_H$

Let $\AA = \set {B': B' \in \BB, B' \cap H \in \AA_H}$.

Let $V = \bigcup \set {B' : B' \in \AA}$.

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

$V \in \tau$


By the definition of the subspace metric:

$\forall B \in \BB_H: \exists B' \in \BB: B = B' \cap H$

Hence:

\(\ds U\) \(=\) \(\ds \bigcup \set {B' \cap H : B' \in \AA}\)
\(\ds \) \(=\) \(\ds \paren {\bigcup \set {B' : B' \in \AA} } \cap H\) Intersection Distributes over Union
\(\ds \) \(=\) \(\ds V \cap H\)

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

$U \in \tau_H$

Hence:

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

$\Box$


Let $U \in \tau_H$.

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

$\exists V \in \tau : U = V \cap H$

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

$\exists \AA \subseteq \BB: V = \bigcup \set {B' : B' \in \AA}$

Hence:

\(\ds U\) \(=\) \(\ds \paren {\bigcup \set {B' : B' \in \AA} } \cap H\)
\(\ds \) \(=\) \(\ds \bigcup \set {B' \cap H : B' \in \AA}\) Intersection Distributes over Union

By the definition of the subspace metric:

$\forall B' \in \BB: B' \cap H \in \BB_H$

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

$U \in \tau_{d_H}$

Hence:

$\tau_H \subseteq \tau_{d_H}$


By Equivalence of Definitions of Set Equality:

$\tau_H = \tau_{d_H}$

$\blacksquare$


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