Metric Induced by Norm is Metric

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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.

$\blacksquare$