# Taxicab Metric is Metric/Proof 1

## Theorem

The taxicab metric is a metric.

## Proof

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 $d_1$.

$\Box$

### 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 $d_1$.

$\Box$

### 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 $d_1$.

$\Box$

### 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 $d_1$.

$\blacksquare$