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

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

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, that $$\sum_{i=1}^n d_{i'} \left({x_{i'}, z_{i'}}\right) \le \sum_{i=1}^n d_{i'} \left({x_{i'}, y_{i'}}\right) + \sum_{i=1}^n d_{i'} \left({y_{i'}, z_{i'}}\right)$$.

Proof by induction:

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


 * $$P(1)$$ 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:

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

Then we need to show:

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

Induction Step
This is our induction step:

$$ $$ $$ $$ $$

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

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

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