Distance on Real Numbers is Metric

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Theorem

Let $x, y \in \R$ be real numbers.

Let $\map d {x, y}$ be the distance between $x$ and $y$:

$\map d {x, y} = \size {x - y}$


Then $\map d {x, y}$ is a metric on $\R$.

Thus it follows that $\tuple {\R, d}$ is a metric space.


Proof

We check the metric space axioms in turn.


Metric Space Axiom $(\text M 1)$

The statement of this axiom is:

Metric Space Axiom $(\text M 1)$: $\forall x \in X: \size {x - x} = 0$


This follows from the definition of absolute value.

$\Box$


Metric Space Axiom $(\text M 2)$: Triangle Inequality

The statement of this axiom is:

Metric Space Axiom $(\text M 2)$: Triangle Inequality: $\forall x, y, z \in X: \size {x - y} + \size {y - z} \ge \size {x - z}$


We have:

$\paren {x - y} + \paren {y - z} = \paren {x - z}$

The result follows from the Triangle Inequality for Real Numbers.

$\Box$


Metric Space Axiom $(\text M 3)$

The statement of this axiom is:

Metric Space Axiom $(\text M 3)$: $\forall x, y \in X: \size {x - y} = \size {y - x}$


As $x - y = -\paren {y - x}$, it follows from the definition of absolute value that $\size {x - y} = \size {y - x}$.

$\Box$


Metric Space Axiom $(\text M 4)$

The statement of this axiom is:

Metric Space Axiom $(\text M 4)$: $\forall x, y \in X: x \ne y \implies \size {x - y} > 0$


This follows from the definition of absolute value.

$\Box$


Having verified all the axioms, we conclude $d$ is a metric.

$\blacksquare$


Sources

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