# Metric Induced by Norm is Metric

## Theorem

Let $V$ be a normed vector space, and denote $\norm{\,\cdot\,}$ for its norm.

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

Then $d$ is a metric.

## Proof

### Proof of $M1$ and $M4$

Let $x, y \in V$.

Then $d \left({x, y}\right) = \left\Vert{x - y}\right\Vert \ge 0$, and furthermore:

 $\displaystyle d \left({x, y}\right)$ $=$ $\displaystyle 0$ $\displaystyle \iff \ \$ $\displaystyle \left\Vert{x - y}\right\Vert$ $=$ $\displaystyle 0$ $\displaystyle \iff \ \$ $\displaystyle x - y$ $=$ $\displaystyle \mathbf{0}_V$ Norm Axioms: Axiom $N1$ $\displaystyle \iff \ \$ $\displaystyle x$ $=$ $\displaystyle y$

$\Box$

### Proof of $M2$

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

Then:

 $\displaystyle d \left({x, z}\right)$ $=$ $\displaystyle \left\Vert{x - z}\right\Vert$ $\displaystyle$ $=$ $\displaystyle \left\Vert{x - y + y - z}\right\Vert$ $\displaystyle$ $\le$ $\displaystyle \left\Vert{x - y}\right\Vert + \left\Vert{y - z}\right\Vert$ Norm Axioms: Axiom $N3$ $\displaystyle$ $=$ $\displaystyle d \left({x, y}\right) + d \left({y, z}\right)$

$\Box$

### Proof of $M3$

Let $x, y \in V$.

Then:

 $\displaystyle d \left({x, y}\right)$ $=$ $\displaystyle \left\Vert{x - y}\right\Vert$ $\displaystyle$ $=$ $\displaystyle \left\Vert{-1 \left({y - x}\right)}\right\Vert$ $\displaystyle$ $=$ $\displaystyle \left\vert{-1}\right\vert \times \left\Vert{y - x}\right\Vert$ Norm Axioms: Axiom $N2$ $\displaystyle$ $=$ $\displaystyle \left\Vert{y - x}\right\Vert$ $\displaystyle$ $=$ $\displaystyle d \left({y, x}\right)$

$\Box$

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

$\blacksquare$