# 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 $\text M 1$ and $\text M 4$

Let $x, y \in V$.

Then:

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

and furthermore:

 $\displaystyle \map d {x, y}$ $=$ $\displaystyle 0$ $\displaystyle \leadstoandfrom \ \$ $\displaystyle \norm {x - y}$ $=$ $\displaystyle 0$ $\displaystyle \leadstoandfrom \ \$ $\displaystyle x - y$ $=$ $\displaystyle \mathbf 0_V$ Norm Axioms: Axiom $\text N 1$ $\displaystyle \leadstoandfrom \ \$ $\displaystyle x$ $=$ $\displaystyle y$

$\Box$

### Proof of $\text M 2$

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

Then:

 $\displaystyle \map d {x, z}$ $=$ $\displaystyle \norm {x - z}$ $\displaystyle$ $=$ $\displaystyle \norm {x - y + y - z}$ $\displaystyle$ $\le$ $\displaystyle \norm {x - y} + \norm {y - z}$ Norm Axioms: Axiom $\text N 3$ $\displaystyle$ $=$ $\displaystyle \map d {x, y} + \map d {y, z}$

$\Box$

### Proof of $\text M 3$

Let $x, y \in V$.

Then:

 $\displaystyle \map d {x, y}$ $=$ $\displaystyle \norm {x - y}$ $\displaystyle$ $=$ $\displaystyle \norm {-1 \paren {y - x} }$ $\displaystyle$ $=$ $\displaystyle \norm {-1} \times \norm {y - x}$ Norm Axioms: Axiom $\text N 2$ $\displaystyle$ $=$ $\displaystyle \norm {y - x}$ $\displaystyle$ $=$ $\displaystyle \map d {y, x}$

$\Box$

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

$\blacksquare$