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 $f$ is $\left({d_1, d_2}\right)$-continuous at $a$ iff:


 * $\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: f \left({ B_\delta \left({a}\right)}\right) \subseteq B_\epsilon \left({f \left({a}\right)}\right)$

where $B_\epsilon \left({a}\right)$ denotes the open $\epsilon$-ball of $a$ in $M_1$.

That is, for every open $\epsilon$-ball of $f \left({a}\right)$ in $M_2$, there exists an open $\delta$-ball of $a$ in $M_1$ whose image is a subset of that open $\epsilon$-ball.

Also used as definition
Many treatments of this subject use this property of continuity to define the concept.

Also see

 * Metric Space Continuity by Epsilon-Delta