Metric Space Continuity by Open Ball

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

Let $f: A_1 \to A_2$ be a mapping from $A_1$ to $A_2$.

Let $a \in A_1$ be a point in $A_1$.

Then the following definitions of continuity of $f$ at $a$ with respect to $d_1$ and $d_2$ are equivalent:

$\epsilon$-$\delta$ Definition implies $\epsilon$-Ball Definition
Suppose that $f$ is $\tuple {d_1, d_2}$-continuous at $a$ in the sense that:
 * $\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in A_1: \map {d_1} {x, a} < \delta \implies \map {d_2} {\map f x, \map f a} < \epsilon$

where $\R_{>0}$ denotes the set of all strictly positive real numbers.

Then:

$\epsilon$-Ball Definition implies $\epsilon$-$\delta$ Definition
Suppose that $f$ is $\tuple {d_1, d_2}$-continuous at $a$ in the sense that:
 * $\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \map f {\map {B_\delta} {a; d_1} } \subseteq \map {B_\epsilon} {\map f a; d_2}$

where $\map {B_\epsilon} {\map f a; d_2}$ denotes the open $\epsilon$-ball of $\map f a$ with respect to the metric $d_2$, and similarly for $\map {B_\delta} {a; d_1}$.

Then:

(*) Only consider this case, since otherwise $\map {d_1} {x, a} < \delta \implies \map {d_2} {\map f x, \map f a} < \epsilon$ is vacuously true.

Also see

 * Metric Space Continuity by Epsilon-Delta