# 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

$f$ 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 [[Definition:Continuous Mapping (Metric Space)/Point/Definition 1 |$f$ 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:

 $\displaystyle y$ $\in$ $\displaystyle f \left({B_\delta \left({a; d_1}\right)}\right)$ $\displaystyle \implies \ \$ $\displaystyle \exists x \in A_1: y = f \left({x}\right): x$ $\in$ $\displaystyle B_\delta \left({a; d_1}\right)$ Definition of Open $\epsilon$-Ball $\displaystyle \implies \ \$ $\displaystyle d_1 \left({x, a}\right)$ $<$ $\displaystyle \delta$ Definition of Open $\epsilon$-Ball $\displaystyle \implies \ \$ $\displaystyle d_2 \left({f \left({x}\right), f \left({a}\right)}\right)$ $<$ $\displaystyle \epsilon$ by hypothesis $\displaystyle \implies \ \$ $\displaystyle d_2 \left({y, f \left({a}\right)}\right)$ $<$ $\displaystyle \epsilon$ as $y = f \left({x}\right)$ $\displaystyle \implies \ \$ $\displaystyle y$ $\in$ $\displaystyle B_\epsilon \left({f \left({a}\right); d_2}\right)$ by hypothesis $\displaystyle \implies \ \$ $\displaystyle f \left({B_\delta \left({a; d_1}\right)}\right)$ $\subseteq$ $\displaystyle B_\epsilon \left({f \left({a}\right); d_2}\right)$ by hypothesis

$\Box$

### $\epsilon$-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:

 $\displaystyle d_1 \left({x, a}\right)$ $<$ $\displaystyle \delta$ $\displaystyle \implies \ \$ $\displaystyle x$ $\in$ $\displaystyle B_\delta \left({a; d_1}\right)$ Definition of Open $\epsilon$-Ball $\displaystyle \implies \ \$ $\displaystyle f \left({x}\right)$ $\in$ $\displaystyle B_\epsilon \left({f \left({a}\right); d_2}\right)$ by hypothesis $\displaystyle \implies \ \$ $\displaystyle d_2 \left({f \left({x}\right), f \left({a}\right)}\right) < \epsilon$ $<$ $\displaystyle \epsilon$ Definition of Open $\epsilon$-Ball

$\blacksquare$