Projection from Metric Space Product with Taxicab Metric is Continuous
Theorem
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $\AA := A_1 \times A_2$ be the cartesian product of $A_1$ and $A_2$.
Let $\MM = \struct {\AA, d}$ denote the metric space on $\AA$ where $d: \AA \to \R$ is the taxicab metric on $\AA$:
- $\map d {x, y} := \map {d_1} {x_1, y_1} + \map {d_2} {x_2, y_2}$
where $x = \tuple {x_1, x_2}, y = \tuple {y_1, y_2} \in \AA$.
Let $\pr_1: \MM \to M_1$ and $\pr_2: \MM \to M_2$ denote the first projection and second projection respectively on $\MM$.
Then $\pr_1$ and $\pr_2$ are both continuous on $\MM$.
Proof 1
The taxicab metric is an instance of the $p$-product metric:
- $\map {d_p} {x, y} := \paren {\paren {\map {d_1} {x_1, y_1} }^p + \paren {\map {d_2} {x_2, y_2} }^p}^{1/p}$
where $p = 1$.
The result is therefore seen to be an instance of Projection from Metric Space Product with P-Product Metric is Continuous.
$\blacksquare$
Proof 2
We want to show that, for all $a \in \AA$:
- $\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall z \in \AA: \map d {z, a} < \delta \implies \map {d_1} {\map {\pr_1} z, \map {\pr_1} a} < \epsilon$
and:
- $\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall z \in \AA: \map d {z, a} < \delta \implies \map {d_2} {\map {\pr_2} z, \map {\pr_2} a} < \epsilon$
Let $\epsilon \in \R_{>0}$ be arbitrary.
Let $a = \tuple {x_0, y_0} \in \AA$ also be arbitrary.
Let $\delta = \epsilon$.
Let $z = \tuple {x_1, y_1} \in \AA$ such that $\map d {x, a} < \delta$.
We have:
\(\ds \map d {z, a}\) | \(=\) | \(\ds \map {d_1} {x_1, x_0} + \map {d_2} {y_1, y_0}\) | Definition of $d$ | |||||||||||
\(\ds \map {d_1} {\map {\pr_1} z, \map {\pr_1} a}\) | \(=\) | \(\ds \map {d_1} {x_1, x_0}\) | Definition of First Projection | |||||||||||
\(\ds \map {d_2} {\map {\pr_2} z, \map {\pr_2} a}\) | \(=\) | \(\ds \map {d_2} {y_1, y_0}\) | Definition of Second Projection |
Hence:
\(\ds \map d {z, a}\) | \(<\) | \(\ds \delta\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map {d_1} {x_1, x_0} + \map {d_2} {y_1, y_0}\) | \(<\) | \(\ds \delta\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map {d_1} {x_1, x_0}\) | \(<\) | \(\ds \delta\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map {d_1} {\map {\pr_1} z, \map {\pr_1} a}\) | \(<\) | \(\ds \epsilon\) | as $\epsilon = \delta$ |
and:
\(\ds \map d {z, a}\) | \(<\) | \(\ds \delta\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map {d_1} {x_1, x_0} + \map {d_2} {y_1, y_0}\) | \(<\) | \(\ds \delta\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map {d_2} {y_1, y_0}\) | \(<\) | \(\ds \delta\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map {d_2} {\map {\pr_2} z, \map {\pr_2} a}\) | \(<\) | \(\ds \epsilon\) | as $\epsilon = \delta$ |
We have that $a$ and $\epsilon$ are arbitrary.
Hence the result by definition of continuity.
$\blacksquare$
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $2$: Continuity generalized: metric spaces: Exercise $2.6: 15$