Metric Space Continuity by Neighborhood

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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:

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

Definition by Neighborhood

$f$ is continuous at (the point) $a$ (with respect to the metrics $d_1$ and $d_2$) if and only if:

for each neighborhood $N'$ of $\map f a$ in $M_2$ there exists a corresponding neighborhood $N$ of $a$ in $M_1$ such that $f \sqbrk N \subseteq N'$.


Proof

$\epsilon$-Ball Definition implies Definition by Neighborhood

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 $N'$ be a neighborhood of $\map f a$ in $M_2$.

Then:

\(\ds \exists \epsilon \in \R_{>0}: \, \) \(\ds \map {B_\epsilon} {\map f a; d_2}\) \(\subseteq\) \(\ds N'\) Definition of Neighborhood (Metric Space)
\(\ds \leadsto \ \ \) \(\ds \exists \delta \in \R_{>0}: \, \) \(\ds f \sqbrk {\map {B_\delta} {a; d_1} }\) \(\subseteq\) \(\ds \map {B_\epsilon} {\map f a; d_2}\) as $f$ is $\tuple {d_1, d_2}$-continuous at $a$

But by Open Ball is Neighborhood of all Points Inside, $\map {B_\delta} {a; d_1}$ is a neighborhood of $a$ in $M_1$.

Thus letting $N := \map {B_\delta} {a; d_1}$:

$f \sqbrk N \subseteq \map {B_\epsilon} {\map f a; d_2} \subseteq N'$

and so the condition for $f$ to be $\tuple {d_1, d_2}$-continuous at $a$ by Neighborhood is fulfilled.

$\Box$


Definition by Neighborhood implies $\epsilon$-Ball Definition

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

for each neighborhood $N'$ of $\map f a$ in $M_2$ there exists a corresponding neighborhood $N$ of $a$ in $M_1$ such that $f \sqbrk N \subseteq N'$.

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

Consider the open $\epsilon$-ball $\map {B_\epsilon} {\map f a; d_2}$ of $\map f a$.

By Open Ball is Neighborhood of all Points Inside, $\map {B_\epsilon} {\map f a; d_2}$ is a neighborhood of $\map f a$ in $M_2$.

Let $N' := \map {B_\epsilon} {\map f a; d_2}$.

By definition of $\tuple {d_1, d_2}$-continuity at $a$, there exists a neighborhood $N$ of $a$ in $M_1$ such that:

$f \sqbrk N \subseteq N'$

By definition of neighborhood:

$\exists \delta \in \R_{>0}: \map {B_\delta} {a; d_1} \subseteq N$.

Therefore:

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

and so the condition for $f$ to be $\tuple {d_1, d_2}$-continuous at $a$ by open $\epsilon$-ball is fulfilled.

$\blacksquare$


Also see


Sources

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