Taxicab Metric is Metric
Theorem
The taxicab metric is a metric.
Proof 1
From the definition, the taxicab metric is as follows:
Let $M_{1'} = \struct {A_{1'}, d_{1'} }, M_{2'} = \struct {A_{2'}, d_{2'} }, \ldots, M_{n'} = \struct {A_{n'}, d_{n'} }$ be a finite number of metric spaces.
Let $\AA$ be the Cartesian product $\ds \prod_{i \mathop = 1}^n A_{i'}$.
The taxicab metric on $\AA$ is:
- $\ds \map {d_1} {x, y} = \sum_{i \mathop = 1}^n \map {d_{i'} } {x_{i'}, y_{i'} }$
for $x = \tuple {x_1, x_2, \ldots, x_n}, y = \tuple {y_1, y_2, \ldots, y_n} \in \AA$.
Proof of Metric Space Axiom $(\text M 1)$
| \(\ds \map {d_1} {x, x}\) | \(=\) | \(\ds \sum_{i \mathop = 1}^n \map {d_{i'} } {x_{i'}, x_{i'} }\) | Definition of $d_1$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 1}^n 0\) | as $d_{i'}$ fulfills Metric Space Axiom $(\text M 1)$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds 0\) |
So Metric Space Axiom $(\text M 1)$ holds for $d_1$.
$\Box$
Proof of Metric Space Axiom $(\text M 2)$: Triangle Inequality
| \(\ds \map {d_1} {x, y} + \map {d_1} {y, z}\) | \(=\) | \(\ds \sum_{i \mathop = 1}^n \map {d_{i'} } {x_{i'}, y_{i'} } + \sum_{i \mathop = 1}^n \map {d_{i'} } {y_{i'}, z_{i'} }\) | Definition of $d_1$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 1}^n \paren {\map {d_{i'} } {x_{i'}, y_{i'} } + \map {d_{i'} } {y_{i'}, z_{i'} } }\) | ||||||||||||
| \(\ds \) | \(\ge\) | \(\ds \sum_{i \mathop = 1}^n \map {d_{i'} } {x_{i'}, z_{i'} }\) | as $d_{i'}$ fulfills Metric Space Axiom $(\text M 2)$: Triangle Inequality | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map {d_1} {x, z}\) | Definition of $d_1$ |
So Metric Space Axiom $(\text M 2)$: Triangle Inequality holds for $d_1$.
$\Box$
Proof of Metric Space Axiom $(\text M 3)$
| \(\ds \map {d_1} {x, y}\) | \(=\) | \(\ds \sum_{i \mathop = 1}^n \map {d_{i'} } {x_{i'}, y_{i'} }\) | Definition of $d_1$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 1}^n \map {d_{i'} } {y_{i'}, x_{i'} }\) | as $d_{i'}$ fulfills Metric Space Axiom $(\text M 3)$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map {d_1} {y, x}\) | Definition of $d_1$ |
So Metric Space Axiom $(\text M 3)$ holds for $d_1$.
$\Box$
Proof of Metric Space Axiom $(\text M 4)$
| \(\ds x\) | \(\ne\) | \(\ds y\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \exists i \in \set {1, 2, \ldots, n}: \, \) | \(\ds x_i\) | \(\ne\) | \(\ds y_i\) | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \map {d_{i'} } {x_{i'}, y_{i'} }\) | \(>\) | \(\ds 0\) | as $d_{i'}$ fulfills Metric Space Axiom $(\text M 4)$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \sum_{i \mathop = 1}^n \map {d_{i'} } {x_{i'}, y_{i'} }\) | \(>\) | \(\ds 0\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \map {d_1} {x, y}\) | \(>\) | \(\ds 0\) | Definition of $d_1$ |
So Metric Space Axiom $(\text M 4)$ holds for $d_1$.
$\blacksquare$
Proof 2
Follows directly from P-Product Metric is Metric, where in this case $p = 1$.
$\blacksquare$
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $2$: Continuity generalized: metric spaces: $2.2$: Examples: Example $2.2.7$