Taxicab Metric on Real Vector Space is Metric

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Theorem

The taxicab metric on the real vector space $\R^n$ is a metric.


Proof 1

This is an instance of the taxicab metric on the cartesian product of $A_{1'}, A_{2'}, \ldots, A_{n'}$.

This is proved in Taxicab Metric is Metric.

$\blacksquare$


Proof 2

The taxicab metric on $\R^n$ is:

$\displaystyle d_1 \left({x, y}\right) = \sum_{i \mathop = 1}^n \left\vert{x_i - y_i}\right\vert$

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


Proof of $M1$

\(\displaystyle d_1 \left({x, x}\right)\) \(=\) \(\displaystyle \sum_{i \mathop = 1}^n \left\vert{x_i - x_i}\right\vert\) Definition of $d_1$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop = 1}^n 0\)
\(\displaystyle \) \(=\) \(\displaystyle 0\)

So axiom $M1$ holds for $d_1$.

$\Box$


Proof of $M2$

\(\displaystyle d_1 \left({x, y}\right) + d_1 \left({y, z}\right)\) \(=\) \(\displaystyle \sum_{i \mathop = 1}^n \left\vert{x_i - y_i}\right\vert + \sum_{i \mathop = 1}^n \left\vert{y_i - z_i}\right\vert\) Definition of $d_1$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop = 1}^n \left({\left\vert{x_i - y_i}\right\vert + \left\vert{y_i - z_i}\right\vert}\right)\)
\(\displaystyle \) \(\ge\) \(\displaystyle \sum_{i \mathop = 1}^n \left\vert{x_i - z_i}\right\vert\) Triangle Inequality for Real Numbers
\(\displaystyle \) \(=\) \(\displaystyle d_1 \left({x, z}\right)\) Definition of $d_1$

So axiom $M2$ holds for $d_1$.

$\Box$


Proof of $M3$

\(\displaystyle d_1 \left({x, y}\right)\) \(=\) \(\displaystyle \sum_{i \mathop = 1}^n \left\vert{x_i - y_i}\right\vert\) Definition of $d_1$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop = 1}^n \left\vert{y_i - x_i}\right\vert\)
\(\displaystyle \) \(=\) \(\displaystyle d \left({y, x}\right)\) Definition of $d_1$

So axiom $M3$ holds for $d_1$.

$\Box$


Proof of $M4$

\(\displaystyle x\) \(\ne\) \(\displaystyle y\)
\(\displaystyle \implies \ \ \) \(\, \displaystyle \exists i \in \left\{ {1, 2, \ldots, n}\right\}: \, \) \(\displaystyle x_i\) \(\ne\) \(\displaystyle y_i\)
\(\displaystyle \implies \ \ \) \(\displaystyle \left\vert{x_i - y_i}\right\vert\) \(>\) \(\displaystyle 0\) Definition of Absolute Value
\(\displaystyle \implies \ \ \) \(\displaystyle \sum_{i \mathop = 1}^n \left\vert{x_i - y_i}\right\vert\) \(>\) \(\displaystyle 0\)
\(\displaystyle \implies \ \ \) \(\displaystyle d_1 \left({x, y}\right)\) \(>\) \(\displaystyle 0\) Definition of $d_1$

So axiom $M4$ holds for $d_1$.

$\blacksquare$


Sources