Equivalence of Definitions of Topology Induced by Metric

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 $\displaystyle \bigcup_{y \mathop \in U} \map {B_{\epsilon_y} } y$.

From $y \in \map {B_{\epsilon_y} } y \subseteq U$ for each $y$:
 * $\displaystyle \bigcup_{y \mathop \in U} \map {B_{\epsilon_y} } y = U$.

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.

$(2)$ 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.