# Metric Induced by Norm is Metric

## 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.

$\blacksquare$