# Continuous Mapping (Metric Space)/Examples/Composition of Arbitrary Mappings

## Examples of Continuous Mappings in the Context of Metric Spaces

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.

## Proof

By Composition of Mappings is Associative and General Associativity Theorem, $m \circ j \circ h \circ g$ is well-defined and unambiguous.

Let $d : \R^2 \times \R^2 \to \R$ be defined as:

$\map d {\tuple {a, b}, \tuple {c, d} } = \max \set {\size {a - c}, \size {b - d} }$

and $d': \paren {\R^2 \times \R^2} \times \paren {\R^2 \times \R^2} \to \R$ be defined as:

 $\ds \map {d'} {\tuple {\tuple {a, b}, \tuple {c, d} }, \tuple {\tuple {k, l}, \tuple {m, n} } }$ $=$ $\ds \max \set {d {\tuple {a, b}, \tuple {k, l} }, d {\tuple {c, d}, \tuple {m, n} } }$ $\ds$ $=$ $\ds \max \set {\size {a - k}, \size {b - l}, \size {c - m}, \size {d - n} }$

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

Let $d$ be constrained by some $\delta \in \R_{>0}$ such that $\delta < \epsilon$:

$\map d {\tuple {x, y}, \tuple {a, b} } < \delta$

and so:

$\size {x - a} <\delta$ and $\size {y - b} < \delta$

Then:

 $\ds \map {d'} {\map g {x, y}, \map g {a, b} }$ $=$ $\ds \max \set {\map d {\tuple {x, y}, \tuple {a, b} }, \map d {\tuple {x, y}, \tuple {a, b} } }$ $\ds$ $=$ $\ds \map d {\tuple {x, y}, \tuple {a, b} }$ $\ds$ $<$ $\ds \delta$ $\ds$ $<$ $\ds \epsilon$

Hence:

$\map d {x, a} < \delta \implies \map {d'} {\map g x, \map g a} < \epsilon$

where $x, a \in \R^2$.

Hence $g$ is continuous.

$\Box$

We have that:

$h: \R^2 \times \R^2 \to \R \times \R: \map h {\tuple {a, b}, \tuple {c, d} } = \tuple {a + b, c - d}$

Let $d$ be defined as elements of $\R^2 \times \R^2$ as above.

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

Let $d$ be constrained by some $\delta \in \R_{>0}$ such that $\delta < \dfrac \epsilon 2$:

$\map d {\tuple {\tuple {a, b}, \tuple {c, d} }, \tuple {\tuple {k, l}, \tuple {m, n} } } = \max \set {\size {a - k}, \size {b - l}, \size {c - m}, \size {d - n} } < \delta$

Then:

 $\ds$  $\ds \map {d'} {\map h {\tuple {a, b}, \tuple {c, d} }, \map h {\tuple {k, l}, \tuple {m, n} } }$ $\ds$ $=$ $\ds \map {d'} {\tuple {a + b, c - d}, \tuple {k + l, m - n} }$ $\ds$ $=$ $\ds \max \set {\size {\size {a - k} + \size {b - l} }, \size {\size {c - m} + \size {d - n} } }$ $\ds$ $<$ $\ds 2 \delta$ $\ds$ $<$ $\ds \epsilon$

Hence:

$\map d {x, a} < \delta \implies \map {d'} {\map h x, \map h a} < \epsilon$

where $x, a \in \R^2 \times \R^2$.

Hence $h$ is continuous.

$\Box$

We have that:

$k: \R \times \R \to \R \times \R: \map k {u, v} = \tuple {u^2, v^2}$

Let $d$ be defined as:

 $\ds \map d {\tuple {u, v}, \tuple {a, b} }$ $=$ $\ds \max \set {\size {u - a}, \size {v - b} }$ $\ds \leadsto \ \$ $\ds \map {d'} {\map k {u, v}, \map k {a, b} }$ $=$ $\ds \max \set {\size {u^2 - a^2}, \size {v^2 - b^2} }$

Without loss of generality, let it be assumed that $\size {u - a} \ge \size {v - b}$.

Let $u > a$.

Then:

$\size {u - a} = u - a$

Setting $\map d {\tuple {u, v}, \tuple {a, b} } < \delta$ gives:

$u - a < \delta$

and so:

$u < a + \delta$

Hence:

$\size {u^2 - a^2} < \size {\delta^2 + 2 a \delta}$

Setting $\epsilon = \size {\delta^2 + 2 a \delta}$, it is noted that this can be made arbitrarily small for any given finite $a$.

Also:

$\size {v - b} < \delta$

and so:

$\size {v^2 - b^2} < \size {\delta^2 + 2 b \delta}$

As all the elements are even, $a > u$ gives similar results.

By the same argument, $\size {v - b} \ge \size {u - a}$ gives the same result.

So in all cases:

$\map d {\tuple {u, v}, \tuple {a, b} } < \delta \implies \map {d'} {\map k {u, v}, \map l {a, b} } < \epsilon$

for whatever $\epsilon$ is chosen.

Hence $k$ is continuous.

$\Box$

We have that:

$m: \R \times \R \to \R: \map k {x, y} = \dfrac {x - y} 4$

Let $d$ be defined as:

 $\ds \map d {\tuple {x, y}, \tuple {a, b} }$ $=$ $\ds \max \set {\size {x - a}, \size {y - b} }$ $\ds \leadsto \ \$ $\ds \map {d'} {\map m {x, y}, \map m {a, b} }$ $=$ $\ds \size {\dfrac {x - y} 4 - \dfrac {a - b} 4}$ $\ds$ $=$ $\ds \dfrac 1 4 \size {x + b - y - a}$ $\ds$ $<$ $\ds \size {x + b - y - a}$ $\ds$ $\le$ $\ds \size {\size {x - a} + \size {y - b} }$

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

Let $d$ be constrained by some $\delta \in \R_{>0}$ such that $\delta < \dfrac \epsilon 2$:

 $\ds \map d {\tuple {x, y}, \tuple {a, b} }$ $<$ $\ds \delta$ $\ds \leadsto \ \$ $\ds \size {x - a}$ $<$ $\ds \delta \land \size {y - b} < \delta$ $\ds \leadsto \ \$ $\ds \map {d'} {\map m {x, y}, \map m {a, b} }$ $<$ $\ds 2 \delta$ $\ds$ $<$ $\ds \epsilon$

Hence:

$\map d {\tuple {x, y}, \tuple {a, b} } < \delta \implies \map {d'} {\map m {x, y}, \map m {a, b} } < \epsilon$

Hence $m$ is continuous.

$\Box$

Consider $\tuple {x, y} \in \R^2$.

We have:

 $\ds \map {\paren {m \circ k \circ h \circ g} } {x, y}$ $=$ $\ds \map {\paren {m \circ k \circ h} } {\map g {x, y} }$ Definition of Composition of Mappings $\ds$ $=$ $\ds \map {\paren {m \circ k \circ h} } {\tuple {x, y}, \tuple {x, y} }$ Definition of $g$ $\ds$ $=$ $\ds \map {\paren {m \circ k} } {\map h {\tuple {x, y}, \tuple {x, y} } }$ Definition of Composition of Mappings $\ds$ $=$ $\ds \map {\paren {m \circ k} } {x + y, x - y}$ Definition of $h$ $\ds$ $=$ $\ds \map m {\map k {x + y, x - y} }$ Definition of Composition of Mappings $\ds$ $=$ $\ds \map m {\paren {x + y}^2, \paren {x - y}^2}$ Definition of $k$ $\ds$ $=$ $\ds \map m {x^2 + 2 x y + y^2, x^2 - 2 x y + y^2}$ expanding $\ds$ $=$ $\ds \dfrac {\paren {x^2 + 2 x y + y^2} - \paren {x^2 - 2 x y + y^2} } 4$ Definition of $m$ $\ds$ $=$ $\ds x y$ simplification

$\blacksquare$