Taxicab Metric is Metric

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 $\displaystyle \prod_{i=1}^n M_{i'}$.

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

The taxicab metric on $\displaystyle \mathcal A$ is $\displaystyle d_1 \left({x, y}\right) = \sum_{i=1}^n d_{i'} \left({x_{i'}, y_{i'}}\right)$.

It is easy to see that conditions M0, M1 and M2 of the conditions for being a metric are satisfied. So all we need to do is check M3.

That is:
 * $\displaystyle \sum_{i=1}^n d_{i'} \left({x_{i'}, z_{i'}}\right) \le \sum_{i=1}^n d_{i'} \left({x_{i'}, y_{i'}}\right) + \sum_{i=1}^n d_{i'} \left({y_{i'}, z_{i'}}\right)$

Proof by induction:

For all $n \in \N^*$, let $P \left({n}\right)$ be the proposition that the taxicab metric is a metric on $\displaystyle \prod_{i=1}^n M_{i'}$.


 * $P(1)$ is true, as this just says $d \left({x, z}\right) \le d \left({x, y}\right) + d \left({y, z}\right)$.

This is the metric space axiom M3 itself.

Basis for the Induction

 * $P(2)$ is the following case:

This is our basis for the induction.

Induction Hypothesis

 * Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:


 * $\displaystyle \sum_{i=1}^k d_{i'} \left({x_i, z_i}\right) \le \sum_{i=1}^k d_{i'} \left({x_i, y_i}\right) + \sum_{i=1}^k d_{i'} \left({y_i, z_i}\right)$.

Then we need to show:


 * $\displaystyle \sum_{i=1}^{k+1} d_{i'} \left({x_i, z_i}\right) \le \sum_{i=1}^{k+1} d_{i'} \left({x_i, y_i}\right) + \sum_{i=1}^{k+1} d_{i'} \left({y_i, z_i}\right)$.

Induction Step
This is our induction step:

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore $d_1 \left({x, y}\right)$ is a metric on $\displaystyle \prod_{i=1}^n \left({A_{i'}, d_{i'}}\right)$ for all $n$.

Therefore $\displaystyle \left({\prod_{i=1}^n \left({A_{i'}, d_{i'}}\right), d_1}\right)$ is a metric space.