Definition:Continuous Mapping (Metric Space)
This page is about continuous mapping in the context of metric spaces. 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 $M_1$.
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.
Also see
- Results about continuous mappings on metric spaces can be found here.