Definition:Continuous Mapping (Metric Space)

From ProofWiki
Jump to navigation Jump to search

This page is about Continuous Mapping in the context of Metric Space. For other uses, see Continuous Mapping.

Definition

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 $A_1$.


Continuous at a Point

$f$ 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.


Continuous on a Space

$f$ is continuous from $\left({A_1, d_1}\right)$ to $\left({A_2, d_2}\right)$ if and only if it is continuous at every point $x \in A_1$.


Metric Subspace

Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be metric spaces.

Let $f: A_1 \to A_2$ be a mapping from $A_1$ to $A_2$.


Let $Y \subseteq A_1$.

By definition, $\left({Y, d_Y}\right)$ is a metric subspace of $A_1$.


Let $a \in Y$ be a point in $Y$.

Then $f$ is $\left({d_Y, d_2}\right)$-continuous at $a$ if and only if:

$\forall \epsilon > 0: \exists \delta > 0: d_Y \left({x, a}\right) < \delta \implies d_2 \left({f \left({x}\right), f \left({a}\right)}\right) < \epsilon$


Similarly, $f$ is $\left({d_Y, d_2}\right)$-continuous if and only if:

$\forall a \in Y: f$ is $\left({d_Y, d_2}\right)$-continuous at $a$


Also known as

A mapping which is continuous from $\struct {A_1, d_1}$ to $\struct {A_2, d_2}$ can also be referred to as $\tuple {d_1, d_2}$-continuous.