# Metric Space Continuity by Inverse of Mapping between Open Balls

## 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 $A_1$.

$f$ is continuous at $a$ with respect to the metrics $d_1$ and $d_2$ iff:

$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: B_\delta \left({a; d_1}\right) \subseteq f^{-1} \left[{B_\epsilon \left({f \left({a}\right); d_2}\right)}\right]$

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 $B_\delta \left({a; d_1}\right)$.

## Proof

By definition, $f$ is continuous at $a$ with respect to the metrics $d_1$ and $d_2$ iff:

$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: f \left[{B_\delta \left({a; d_1}\right)}\right] \subseteq B_\epsilon \left({f \left({a}\right); d_2}\right)$

For a mapping $f: X \to Y$ we have:

$f \left[{U}\right] \subseteq V \iff U \subseteq f^{-1} \left[{V}\right]$

where $U \subseteq X$ and $V \subseteq Y$.

Hence the result.

$\blacksquare$