Equivalence of Definitions of Limit of Mapping between Metric Spaces

Theorem
Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ 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:
 * $\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: 0 < d_1 \left({x, c}\right) < \delta \implies d_2 \left({f \left({x}\right), L}\right) < \epsilon$

Let $y \in f \left({B_\delta \left({c; d_1}\right) \setminus \left\{{c}\right\}}\right)$.

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

By hypothesis, it follows that $d_2 \left({y, L}\right) < \epsilon$

That is, that $y \in B_\epsilon \left({L; d_2}\right)$.

By definition of subset:
 * $f \left({B_\delta \left({c; d_1}\right) \setminus \left\{{c}\right\}}\right) \subseteq B_\epsilon \left({L; d_2}\right)$

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:
 * $\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: f \left({B_\delta \left({c; d_1}\right) \setminus \left\{{c}\right\}}\right) \subseteq B_\epsilon \left({L; d_2}\right)$.

Let $0 < d_1 \left({x, c}\right) < \delta$.

Then by definition of open $\epsilon$-ball, this means:
 * $x \in B_\delta \left({c; d_1}\right) \setminus \left\{{c}\right\}$

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

Thus by definition of open $\epsilon$-ball, this means:
 * $d_2 \left({f \left({x}\right), L}\right) < \epsilon$

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