# Metric Space Continuity by Neighborhood

## 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:

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

### Definition by Neighborhood

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

for each neighborhood $N'$ of $f \left({a}\right)$ in $M_2$ there exists a corresponding neighborhood $N$ of $a$ in $M_1$ such that $f \left[{N}\right] \subseteq N'$.

## Proof

### $\epsilon$-Ball Definition implies Definition by Neighborhood

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 $N'$ be a neighborhood of $f \left({a}\right)$ in $M_2$.

Then :

 $\, \displaystyle \exists \epsilon \in \R_{>0}: \,$ $\displaystyle B_\epsilon \left({f \left({a}\right); d_2}\right)$ $\in$ $\displaystyle N'$ Definition of Neighborhood $\displaystyle \implies \ \$ $\, \displaystyle \exists \delta \in \R_{>0}: \,$ $\displaystyle f \left[{B_\delta \left({a; d_1}\right)}\right]$ $\subseteq$ $\displaystyle B_\epsilon \left({f \left({a}\right); d_2}\right)$ as $f$ is $\left({d_1, d_2}\right)$-continuous at $a$

But by Open Ball is Neighborhood of all Points Inside, $B_\delta \left({a; d_1}\right)$ is a neighborhood of $a$ in $M_1$.

Thus letting $N := B_\delta \left({a; d_1}\right)$:

$f \left({N}\right) \subseteq B_\epsilon \left({f \left({a}\right); d_2}\right) \subseteq N'$

and so the condition for $f$ to be $\left({d_1, d_2}\right)$-continuous at $a$ by Neighborhood is fulfilled.

$\Box$

### Definition by Neighborhood implies $\epsilon$-Ball Definition

Suppose that $f$ is $\left({d_1, d_2}\right)$-continuous at $a$ in the sense that:

for each neighborhood $N'$ of $f \left({a}\right)$ in $M_2$ there exists a corresponding neighborhood $N$ of $a$ in $M_1$ such that $f \left[{N}\right] \subseteq N'$.

Let $\epsilon \in \R_{>0}$.

Consider the open $\epsilon$-ball $B_\epsilon \left({f \left({a}\right); d_2}\right)$ of $f \left({a}\right)$.

By Open Ball is Neighborhood of all Points Inside, $B_\epsilon \left({f \left({a}\right); d_2}\right)$ is a neighborhood of $f \left({a}\right)$ in $M_2$.

Let $N' := B_\epsilon \left({f \left({a}\right); d_2}\right)$.

By definition of $\left({d_1, d_2}\right)$-continuity at $a$, there exists a neighborhood $N$ of $a$ in $M_1$ such that:

$f \left[{N}\right] \subseteq N'$

By definition of neighborhood:

$\exists \delta \in \R_{>0}: B_\delta \left({a; d_1}\right) \subseteq N$.

Therefore:

$f \left({B_\delta \left({a; d_1}\right)}\right) \subseteq B_\epsilon \left({f \left({a}\right); d_2}\right)$

and so the condition for $f$ to be $\left({d_1, d_2}\right)$-continuous at $a$ by open $\epsilon$-ball is fulfilled.

$\blacksquare$