# Equivalence of Definitions of Continuity on Metric Spaces

## 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 $A_2$.

The following definitions of the concept of continuity in the context of metric spaces are equivalent:

### Definition 1

$f$ is continuous from $\struct {A_1, d_1}$ to $\struct {A_2, d_2}$ if and only if it is continuous at every point $x \in A_1$.

### Definition 2

$f$ is continuous from $\struct {A_1, d_1}$ to $\struct {A_2, d_2}$ if and only if:

for every $U \subseteq A_2$ which is open in $M_2$, $f^{-1} \sqbrk U$ is open in $M_1$.

## Proof

### Definition by Points implies Definition by Open Sets

Suppose that $f$ is continuous at every point $x \in A_1$.

Let $U \subseteq M_2$ be open in $M_2$.

Let $x \in f^{-1} \sqbrk U$.

Since $U$ is open in $M_2$:

$\exists \epsilon \in \R_{>0}: \map {B_\epsilon} {\map f x; d_2} \subseteq U$

where $\map {B_\epsilon} {\map f x; d_2}$ denotes the open $\epsilon$-ball of $\map f x$ in $M_2$.

By the definition of continuity at a point:

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

So:

$f \sqbrk {\map {B_\delta} {x; d_1} } \subseteq U$

and so:

$\map {B_\delta} {x; d_1} \subseteq f^{-1} \sqbrk U$

Thus $f^{-1} \sqbrk U$ is open in $M_1$.

$\Box$

### Definition by Open Sets implies Definition by Points

Suppose $f$ is defined to be continuous in the sense that:

for every $U \subseteq A_2$ which is open in $M_2$, $f^{-1} \sqbrk U$ is open in $M_1$.

Let $x \in A_1$.

$\exists \epsilon \in \R_{>0}: \map {B_\epsilon} {\map f x; d_2}$ is open in $M_2$

So by hypothesis, $f^{-1} \sqbrk {\map {B_\epsilon} {\map f x; d_2} }$ is open in $M_1$.

As $\map f x \in \map {B_\epsilon} {\map f x; d_2}$, it follows that:

$x \in f^{-1} \sqbrk {\map {B_\epsilon} {\map f x; d_2} }$

So by hypothesis:

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

Then:

$f \sqbrk {\map {B_\delta} {x; d_1} } \subseteq \map {B_\epsilon} {\map f x; d_2}$

Thus by the $\epsilon$-Ball definition, $f$ is continuous at $x$.

$\blacksquare$