# 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 $x \in A_1$.

Then by Open Ball of Point Inside Open Ball:

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

$\blacksquare$

## Sources

- 1967: George McCarty:
*Topology: An Introduction with Application to Topological Groups*... (previous) ... (next): $\text{III}$: More About Continuity - 1975: Bert Mendelson:
*Introduction to Topology*(3rd ed.) ... (previous) ... (next): Chapter $2$: Metric Spaces: $\S 6$: Open Sets and Closed Sets: Theorem $6.3$ - 1975: W.A. Sutherland:
*Introduction to Metric and Topological Spaces*... (previous) ... (next): $2$: Continuity generalized: metric spaces: $2.3$: Open sets in metric spaces: Proposition $2.3.13$

- 1953: Walter Rudin:
*Principles of Mathematical Analysis*... (previous) ... (next): $4.8$