Equivalence of Definitions of Topology Induced by Metric

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Theorem

The following definitions of the concept of Topology Induced by Metric are equivalent:

Definition 1

The topology on the metric space $M = \struct {A, d}$ induced by (the metric) $d$ is defined as the set $\tau$ of all open sets of $M$.

Definition 2

The topology on the metric space $M = \struct {A, d}$ induced by (the metric) $d$ is defined as the topology $\tau$ generated by the basis consisting of the set of all open $\epsilon$-balls in $M$.


Proof

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


$(1)$ implies $(2)$

Let $T = \struct {A, \tau_d}$ be the topological space of which $\tau_d$ is the topology induced on $M$ by $d$ by definition 1.

Then by definition:

$\tau_d$ is the set of all open sets of $M$.


Let $U \in \tau_d$.

From definition:

$\forall y \in U: \exists \epsilon_y \in \R_{>0}: \map {B_{\epsilon_y} } y \subseteq U$

where $\map {B_{\epsilon_y}} y$ is the open $\epsilon$-ball of $y$.

We find these $\epsilon$ for every $y \in U$.

Consider the set $\ds \bigcup_{y \mathop \in U} \map {B_{\epsilon_y} } y$.

From $y \in \map {B_{\epsilon_y} } y \subseteq U$ for each $y$:

$\ds \bigcup_{y \mathop \in U} \map {B_{\epsilon_y} } y = U$


This shows that any $U \in \tau_d$ is a union of open $\epsilon$-balls in $M$.

So open $\epsilon$-balls in $M$ form a basis for $\tau_d$.

Thus $\tau_d$ is the topology induced on $M$ by $d$ by definition 2.

$\Box$


$(2)$ implies $(1)$

Let $T = \struct {A, \tau_d}$ be the topological space of which $\tau_d$ is the topology induced on $M$ by $d$ by definition 2.

Then by definition:

$\tau$ is the topology generated by the basis consisting of the set of all open $\epsilon$-balls in $M$.


Let $U \in \tau_d$.

Since open $\epsilon$-balls in $M$ form a basis for $\tau_d$, $U$ is a union of open $\epsilon$-balls in $M$.

Hence for each $x \in U$:

$\exists y \in U: \exists \epsilon \in \R_0: x \in \map {B_\epsilon} y$

By Open Ball is Subset of Open Ball:

$x \in \map {B_{\epsilon - \map d {x, y} } } x \subseteq \map {B_\epsilon} y \subseteq U$

Therefore $U$ is an open set of $M$.


Thus $\tau_d$ is the topology induced on $M$ by $d$ by definition 1.

$\blacksquare$