Taxicab Metric is Metric/Proof 1

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The taxicab metric is a metric.


From the definition, the taxicab metric is as follows:

Let $M_{1'} = \left({A_{1'}, d_{1'}}\right), M_{2'} = \left({A_{2'}, d_{2'}}\right), \ldots, M_{n'} = \left({A_{n'}, d_{n'}}\right)$ be a finite number of metric spaces.

Let $\mathcal A$ be the Cartesian product $\displaystyle \prod_{i \mathop = 1}^n A_{i'}$.

The taxicab metric on $\displaystyle \mathcal A$ is:

$\displaystyle d_1 \left({x, y}\right) = \sum_{i \mathop = 1}^n d_{i'} \left({x_{i'}, y_{i'}}\right)$

for $x = \left({x_1, x_2, \ldots, x_n}\right), y = \left({y_1, y_2, \ldots, y_n}\right) \in \mathcal A$.

Proof of $M1$

\(\displaystyle d_1 \left({x, x}\right)\) \(=\) \(\displaystyle \sum_{i \mathop = 1}^n d_{i'} \left({x_{i'}, x_{i'} }\right)\) Definition of $d_1$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop = 1}^n 0\) as $d_{i'}$ fulfils axiom $M1$
\(\displaystyle \) \(=\) \(\displaystyle 0\)

So axiom $M1$ holds for $d_1$.


Proof of $M2$

\(\displaystyle d_1 \left({x, y}\right) + d_1 \left({y, z}\right)\) \(=\) \(\displaystyle \sum_{i \mathop = 1}^n d_{i'} \left({x_{i'}, y_{i'} }\right) + \sum_{i \mathop = 1}^n d_{i'} \left({y_{i'}, z_{i'} }\right)\) Definition of $d_1$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop = 1}^n \left({d_{i'} \left({x_{i'}, y_{i'} }\right) + d_{i'} \left({y_{i'}, z_{i'} }\right)}\right)\)
\(\displaystyle \) \(\ge\) \(\displaystyle \sum_{i \mathop = 1}^n d_{i'} \left({x_{i'}, z_{i'} }\right)\) as $d_{i'}$ fulfils axiom $M2$
\(\displaystyle \) \(=\) \(\displaystyle d \left({x, z}\right)\) Definition of $d_1$

So axiom $M2$ holds for $d_1$.


Proof of $M3$

\(\displaystyle d_1 \left({x, y}\right)\) \(=\) \(\displaystyle \sum_{i \mathop = 1}^n d_{i'} \left({x_{i'}, y_{i'} }\right)\) Definition of $d_1$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop = 1}^n d_{i'} \left({y_{i'}, x_{i'} }\right)\) as $d_{i'}$ fulfils axiom $M3$
\(\displaystyle \) \(=\) \(\displaystyle d \left({y, x}\right)\) Definition of $d_1$

So axiom $M3$ holds for $d_1$.


Proof of $M4$

\(\displaystyle x\) \(\ne\) \(\displaystyle y\)
\(\displaystyle \implies \ \ \) \(\, \displaystyle \exists i \in \left\{ {1, 2, \ldots, n}\right\}: \, \) \(\displaystyle x_i\) \(\ne\) \(\displaystyle y_i\)
\(\displaystyle \implies \ \ \) \(\displaystyle d_{i'} \left({x_{i'}, y_{i'} }\right)\) \(>\) \(\displaystyle 0\) as $d_{i'}$ fulfils axiom $M4$
\(\displaystyle \implies \ \ \) \(\displaystyle \sum_{i \mathop = 1}^n d_{i'} \left({x_{i'}, y_{i'} }\right)\) \(>\) \(\displaystyle 0\)
\(\displaystyle \implies \ \ \) \(\displaystyle d_1 \left({x, y}\right)\) \(>\) \(\displaystyle 0\) Definition of $d_1$

So axiom $M4$ holds for $d_1$.