# Metric Space Continuity by Inverse of Mapping between Open Balls

## 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 $A_1$.

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

$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \map {B_\delta} {a; d_1} \subseteq f^{-1} \sqbrk {\map {B_\epsilon} {\map f a; d_2} }$

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 $\map {B_\delta} {a; d_1}$.

## Proof

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

$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: f \sqbrk {\map {B_\delta} {a; d_1} } \subseteq \map {B_\epsilon} {\map f a; d_2}$

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

$f \sqbrk U \subseteq V \iff U \subseteq f^{-1} \sqbrk V$

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

Hence the result.

$\blacksquare$