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:

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:

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:

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.

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:

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.

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:

, 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.

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

Let $d$ be defined as:

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

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

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.

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

We have: