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 $\text M 1$ 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.

$\blacksquare$