Metric Induced by Norm is Metric

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Theorem

Let $V$ be a normed vector space.

Let $\norm{\,\cdot\,}$ denote its norm.

Let $d$ be the metric induced by $\norm {\,\cdot\,}$.


Then $d$ is a metric.


Proof

Proof of Metric Space Axiom $(\text M 1)$ and Metric Space Axiom $(\text M 4)$

Let $x, y \in V$.

Then:

$\map d {x, y} = \norm {x - y} \ge 0$

and furthermore:

\(\ds \map d {x, y}\) \(=\) \(\ds 0\)
\(\ds \leadstoandfrom \ \ \) \(\ds \norm {x - y}\) \(=\) \(\ds 0\)
\(\ds \leadstoandfrom \ \ \) \(\ds x - y\) \(=\) \(\ds \mathbf 0_V\) Norm Axiom $\text N 1$: Positive Definiteness
\(\ds \leadstoandfrom \ \ \) \(\ds x\) \(=\) \(\ds y\)

$\Box$


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

Let $x, y, z \in V$.

Then:

\(\ds \map d {x, z}\) \(=\) \(\ds \norm {x - z}\)
\(\ds \) \(=\) \(\ds \norm {x - y + y - z}\)
\(\ds \) \(\le\) \(\ds \norm {x - y} + \norm {y - z}\) Norm Axiom $\text N 3$: Triangle Inequality
\(\ds \) \(=\) \(\ds \map d {x, y} + \map d {y, z}\)

$\Box$


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

Let $x, y \in V$.

Then:

\(\ds \map d {x, y}\) \(=\) \(\ds \norm {x - y}\)
\(\ds \) \(=\) \(\ds \norm {-1 \paren {y - x} }\)
\(\ds \) \(=\) \(\ds \norm {-1} \times \norm {y - x}\) Norm Axiom $\text N 2$: Positive Homogeneity
\(\ds \) \(=\) \(\ds \norm {y - x}\)
\(\ds \) \(=\) \(\ds \map d {y, x}\)

$\Box$


As $d$ satisfies the metric space axioms, it is a metric.

$\blacksquare$


Sources

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