# 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

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

$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in A_1: \map {d_1} {x, a} < \delta \implies \map {d_2} {\map f x, \map f a} < \epsilon$

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

### $\epsilon$-Ball Definition

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

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

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

## Proof

### $\epsilon$-$\delta$ 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:

 $\displaystyle \text{Let} \ \$ $\displaystyle \epsilon$ $\in$ $\displaystyle \R_{>0}$ $\displaystyle \leadsto \ \$ $\displaystyle \exists \delta \in \R_{>0} : \forall x \in A_1 : \map {d_1} {x, a} < \delta$ $\implies$ $\displaystyle \map {d_2} {\map f x, \map f a} < \epsilon$ by hypothesis $\displaystyle \text{Let} \ \$ $\displaystyle y$ $\in$ $\displaystyle f \sqbrk {\map {B_\delta} {a; d_1} }$ $\displaystyle \leadsto \ \$ $\displaystyle \exists x \in \map {B_\delta} {a; d_1}: y$ $=$ $\displaystyle \map f x$ Definition 1 of Image of Subset under Mapping $\displaystyle \leadsto \ \$ $\displaystyle \map {d_1} {x, a}$ $<$ $\displaystyle \delta$ Definition of Open Ball $\displaystyle \leadsto \ \$ $\displaystyle \map {d_2} {\map f x, \map f a}$ $<$ $\displaystyle \epsilon$ by hypothesis $\displaystyle \leadsto \ \$ $\displaystyle \map {d_2} {y, \map f a}$ $<$ $\displaystyle \epsilon$ as $y = \map f x$ $\displaystyle \leadsto \ \$ $\displaystyle y$ $\in$ $\displaystyle \map {B_\epsilon} {\map f a; d_2}$ Definition of Open Ball $\displaystyle \leadsto \ \$ $\displaystyle f \sqbrk {\map {B_\delta} {a; d_1} }$ $\subseteq$ $\displaystyle \map {B_\epsilon} {\map f a; d_2}$ by hypothesis

$\Box$

### $\epsilon$-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:

 $\displaystyle \text{Let} \ \$ $\displaystyle \epsilon$ $\in$ $\displaystyle \R_{>0}$ $\displaystyle \leadsto \ \$ $\displaystyle \exists \delta \in \R_{>0} : \map f {\map {B_\delta} {a; d_1} }$ $\subseteq$ $\displaystyle \map {B_\epsilon} {\map f a; d_2}$ by hypothesis $\displaystyle \text{Let} \ \$ $\displaystyle x \in A_1$ $\text{ such that }$ $\displaystyle \map {d_1} {x, a} < \delta$ (*) $\displaystyle \leadsto \ \$ $\displaystyle x$ $\in$ $\displaystyle \map {B_\delta} {a; d_1}$ Definition of Open $\epsilon$-Ball $\displaystyle \leadsto \ \$ $\displaystyle \map f x$ $\in$ $\displaystyle \map {B_\epsilon} {\map f a; d_2}$ by hypothesis $\displaystyle \leadsto \ \$ $\displaystyle \map {d_2} {\map f x, \map f a}$ $<$ $\displaystyle \epsilon$ Definition of Open $\epsilon$-Ball

$\blacksquare$

(*) 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.