Metric Space Continuity by Open Ball

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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 $A_2$.

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 $f$ is $\tuple {d_1, d_2}$-continuous at $a$ in the sense that:

$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in A_1: \map {d_1} {x, a} < \delta \implies \map {d_2} {\map f x, \map f a} < \epsilon$

where $\R_{>0}$ denotes the set of all strictly positive real numbers.


Then:

\(\displaystyle \text{Let} \ \ \) \(\displaystyle \epsilon\) \(\in\) \(\displaystyle \R_{>0}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \exists \delta \in \R_{>0} : \forall x \in A_1 : \map {d_1} {x, a} < \delta\) \(\implies\) \(\displaystyle \map {d_2} {\map f x, \map f a} < \epsilon\) by hypothesis
\(\displaystyle \text{Let} \ \ \) \(\displaystyle y\) \(\in\) \(\displaystyle f \sqbrk {\map {B_\delta} {a; d_1} }\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \exists x \in \map {B_\delta} {a; d_1}: y\) \(=\) \(\displaystyle \map f x\) Definition 1 of Image of Subset under Mapping
\(\displaystyle \leadsto \ \ \) \(\displaystyle \map {d_1} {x, a}\) \(<\) \(\displaystyle \delta\) Definition of Open Ball
\(\displaystyle \leadsto \ \ \) \(\displaystyle \map {d_2} {\map f x, \map f a}\) \(<\) \(\displaystyle \epsilon\) by hypothesis
\(\displaystyle \leadsto \ \ \) \(\displaystyle \map {d_2} {y, \map f a}\) \(<\) \(\displaystyle \epsilon\) as $y = \map f x$
\(\displaystyle \leadsto \ \ \) \(\displaystyle y\) \(\in\) \(\displaystyle \map {B_\epsilon} {\map f a; d_2}\) Definition of Open Ball
\(\displaystyle \leadsto \ \ \) \(\displaystyle f \sqbrk {\map {B_\delta} {a; d_1} }\) \(\subseteq\) \(\displaystyle \map {B_\epsilon} {\map f a; d_2}\) by hypothesis

$\Box$


$\epsilon$-Ball Definition implies $\epsilon$-$\delta$ Definition

Suppose that $f$ is $\tuple {d_1, d_2}$-continuous at $a$ in the sense that:

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

where $\map {B_\epsilon} {\map f a; d_2}$ denotes the open $\epsilon$-ball of $\map f a$ with respect to the metric $d_2$, and similarly for $\map {B_\delta} {a; d_1}$.


Then:

\(\displaystyle \text{Let} \ \ \) \(\displaystyle \epsilon\) \(\in\) \(\displaystyle \R_{>0}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \exists \delta \in \R_{>0} : \map f {\map {B_\delta} {a; d_1} }\) \(\subseteq\) \(\displaystyle \map {B_\epsilon} {\map f a; d_2}\) by hypothesis
\(\displaystyle \text{Let} \ \ \) \(\displaystyle x \in A_1\) \(\text{ such that }\) \(\displaystyle \map {d_1} {x, a} < \delta\) (*)
\(\displaystyle \leadsto \ \ \) \(\displaystyle x\) \(\in\) \(\displaystyle \map {B_\delta} {a; d_1}\) Definition of Open $\epsilon$-Ball
\(\displaystyle \leadsto \ \ \) \(\displaystyle \map f x\) \(\in\) \(\displaystyle \map {B_\epsilon} {\map f a; d_2}\) by hypothesis
\(\displaystyle \leadsto \ \ \) \(\displaystyle \map {d_2} {\map f x, \map f a}\) \(<\) \(\displaystyle \epsilon\) Definition of Open $\epsilon$-Ball

$\blacksquare$

(*) Only consider this case, since otherwise $\map {d_1} {x, a} < \delta \implies \map {d_2} {\map f x, \map f a} < \epsilon$ is vacuously true.


Also see


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