Euclidean Metric on Real Number Line is Metric/Proof 2

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Theorem

The Euclidean metric on the real number line $\R$ is a metric.


Proof

Consider the real number line under the Euclidean metric:

$M = \struct {\R, d}$

where $d$ is the distance function given by:

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


Proof of Metric Space Axiom $(\text M 1)$

\(\ds \map d {x, x}\) \(=\) \(\ds \size {x - x}\) Definition of $d$
\(\ds \) \(=\) \(\ds \size 0\)
\(\ds \) \(=\) \(\ds 0\)

So Metric Space Axiom $(\text M 1)$ holds for $d$.

$\Box$


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

\(\ds \map d {x, y} + \map d {y, z}\) \(=\) \(\ds \size {x - y} + \size {y - z}\) Definition of $d$
\(\ds \) \(\ge\) \(\ds \size {\paren {x - y} + \paren {y - z} }\) Triangle Inequality for Real Numbers
\(\ds \) \(=\) \(\ds \size {x - z}\)
\(\ds \) \(=\) \(\ds \map d {x, z}\) Definition of $d$

So Metric Space Axiom $(\text M 2)$: Triangle Inequality holds for $d$.

$\Box$


Proof of Metric Space Axiom $(\text M 3)$

\(\ds \map d {x, y}\) \(=\) \(\ds \size {x - y}\) Definition of $d$
\(\ds \) \(=\) \(\ds \size {y - x}\)
\(\ds \) \(=\) \(\ds \map d {y, x}\) Definition of $d$

So Metric Space Axiom $(\text M 3)$ holds for $d$.

$\Box$


Proof of Metric Space Axiom $(\text M 4)$

\(\ds x\) \(\ne\) \(\ds y\)
\(\ds \leadsto \ \ \) \(\ds x - y\) \(\ne\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \size {x - y}\) \(>\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \map d {x, y}\) \(>\) \(\ds 0\) Definition of $d$

So Metric Space Axiom $(\text M 4)$ holds for $d$.

$\blacksquare$

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