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 $A_1$.

The following definitions of the concept 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:

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

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 $\map {B_\delta} {a; d_1}$.

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.

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

Let $\delta$ be such that:

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

as is known to exist by hypothesis.

Then:

 $\ds y$ $\in$ $\ds f \sqbrk {\map {B_\delta} {a; d_1} }$ $\ds \leadsto \ \$ $\ds \exists x \in \map {B_\delta} {a; d_1}: \,$ $\ds y$ $=$ $\ds \map f x$ Definition 1 of Image of Subset under Mapping $\ds \leadsto \ \$ $\ds \map {d_1} {x, a}$ $<$ $\ds \delta$ Definition of Open Ball $\ds \leadsto \ \$ $\ds \map {d_2} {\map f x, \map f a}$ $<$ $\ds \epsilon$ by hypothesis $\ds \leadsto \ \$ $\ds \map {d_2} {y, \map f a}$ $<$ $\ds \epsilon$ as $y = \map f x$ $\ds \leadsto \ \$ $\ds y$ $\in$ $\ds \map {B_\epsilon} {\map f a; d_2}$ Definition of Open Ball $\ds \leadsto \ \$ $\ds f \sqbrk {\map {B_\delta} {a; d_1} }$ $\subseteq$ $\ds \map {B_\epsilon} {\map f a; d_2}$ Definition of Subset

As $\epsilon$ is arbitrary, it follows that:

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

$\Box$

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

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

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

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 $\map {B_\delta} {a; d_1}$.

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

Let $\delta$ be such that:

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

as is known to exist by hypothesis.

Then:

 $\ds \map {d_1} {x, a}$ $<$ $\ds \delta$ $\ds \leadsto \ \$ $\ds x$ $\in$ $\ds \map {B_\delta} {a; d_1}$ Definition of Open $\epsilon$-Ball $\ds \leadsto \ \$ $\ds \map f x$ $\in$ $\ds \map {B_\epsilon} {\map f a; d_2}$ Definition 1 of Image of Subset under Mapping $\ds \leadsto \ \$ $\ds \map {d_2} {\map f x, \map f a}$ $<$ $\ds \epsilon$ Definition of Open $\epsilon$-Ball

As $\epsilon$ is arbitrary, it follows that:

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

$\blacksquare$