Equivalence of Definitions of Limit of Mapping between Metric Spaces

Theorem
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.

Let $c$ be a limit point of $M_1$.

Let $f: A_1 \to A_2$ be a mapping from $A_1$ to $A_2$ defined everywhere on $A_1$ except possibly at $c$.

Let $L \in M_2$.

$\epsilon$-$\delta$ Condition implies $\epsilon$-Ball Condition
Suppose that $f$ satisfies the $\epsilon$-$\delta$ condition:
 * $\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: 0 < \map {d_1} {x, c} < \delta \implies \map {d_2} {\map f x, L} < \epsilon$

Let $y \in f \sqbrk {\map {B_\delta} {c; d_1} \setminus \set c}$.

By definition of open $\epsilon$-ball, this means:
 * $\exists x \in A_1: 0 < \map {d_1} {x, c} < \delta: y = \map f x$

By hypothesis, it follows that $\map {d_2} {y, L} < \epsilon$

That is:
 * $y \in \map {B_\epsilon} {L; d_2}$

By definition of subset:
 * $f \sqbrk {\map {B_\delta} {c; d_1} \setminus \set c} \subseteq \map{B_\epsilon} {L; d_2}$

Thus it follows that $f$ satisfies the $\epsilon$-ball condition.

$\epsilon$-Ball Condition implies $\epsilon$-$\delta$ Condition
Suppose that $f$ satisfies the $\epsilon$-ball condition:
 * $\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: f \sqbrk {\map {B_\delta} {c; d_1} \setminus \set c} \subseteq \map {B_\epsilon} {L; d_2}$

Let $0 < \map {d_1} {x, c} < \delta$.

By definition of open $\epsilon$-ball, this means:
 * $x \in \map {B_\delta} {c; d_1} \setminus \set c$

By hypothesis, it follows that:
 * $\map f x \in \map {B_\epsilon} {L; d_2}$

Thus by definition of open $\epsilon$-ball:
 * $\map {d_2} {\map f x, L} < \epsilon$

That is, $f$ satisfies the $\epsilon$-$\delta$ condition.