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): Chapter $\text{III}$: Metric Spaces: 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 $2.3.13$
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- 1953: Walter Rudin: Principles of Mathematical Analysis ... (previous) ... (next): $4.8$