Metric Space Continuity by Open Ball

Theorem
Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ 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 $\left({d_1, d_2}\right)$-continuous at $a$ in the sense that:
 * $\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in A_1: d_1 \left({x, a}\right) < \delta \implies d_2 \left({f \left({x}\right), f \left({a}\right)}\right) < \epsilon$

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

Then:

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

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

Let $x \in A_1$.

Then:

Also see

 * Metric Space Continuity by Epsilon-Delta