# Definition:Continuous Mapping (Metric Space)

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 $\struct {A_1, d_1}$ to $\struct {A_2, d_2}$ if and only if it is continuous at every point $x \in A_1$.

## Metric Subspace

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 $Y \subseteq A_1$.

By definition, $\struct {Y, d_Y}$ is a metric subspace of $A_1$.

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

Then $f$ is $\tuple {d_Y, d_2}$-continuous at $a$ if and only if:

$\forall \epsilon > 0: \exists \delta > 0: \map {d_Y} {x, a} < \delta \implies \map {d_2} {\map f x, \map f a} < \epsilon$

Similarly, $f$ is $\tuple {d_Y, d_2}$-continuous if and only if:

$\forall a \in Y: f$ is $\tuple {d_Y, d_2}$-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.

## Examples

### Composition of Arbitrary Mappings

Let the following mappings be defined:

 $\ds g: \R^2 \to \R^2 \times \R^2: \,$ $\ds \map g {x, y}$ $=$ $\ds \tuple {\tuple {x, y}, \tuple {x, y} }$ $\ds h: \R^2 \times \R^2 \to \R \times \R: \,$ $\ds \map h {\tuple {a, b}, \tuple {c, d} }$ $=$ $\ds \tuple {a + b, c - d}$ $\ds k: \R \times \R \to \R \times \R: \,$ $\ds \map k {u, v}$ $=$ $\ds \tuple {u^2, v^2}$ $\ds m: \R \times \R \to \R: \,$ $\ds \map k {x, y}$ $=$ $\ds \dfrac {x - y} 4$

where $\R$ and $\R^2$ denote the real number line and real number plane respectively, under the usual (Euclidean) metric.

Then:

each of $g, h, k, m$ are continuous
$x y = \map {\paren {m \circ k \circ h \circ g} } {x, y}$

where $\circ$ denotes composition of mappings.

### Identity Function with Discontinuity

Let $f: \R \to \R$ be the real function defined as:

$\forall x \in \R: \map f x = \begin {cases} x & : x \ne 0 \\ 1 & : x = 0 \end {cases}$

Then $f$ is not continuous.