Distance in Pseudometric is Non-Negative

Theorem

Let $X$ be a set on which a pseudometric $d: X \times X \to \R$ has been imposed.

Then:

$\forall x, y \in X: \map d {x, y} \ge 0$

Proof

By definition of pseudometric, we have that:

\((\text M 1)\)	$:$			\(\ds \forall x \in A:\)	\(\ds \map d {x, x} = 0 \)
\((\text M 2)\)	$:$			\(\ds \forall x, y, z \in A:\)	\(\ds \map d {x, y} + \map d {y, z} \ge \map d {x, z} \)
\((\text M 3)\)	$:$			\(\ds \forall x, y \in A:\)	\(\ds \map d {x, y} = \map d {y, x} \)

Hence:

\(\ds \forall x, y \in X: \, \)

\(\ds \map d {x, y} + \map d {y, x}\)

\(\ge\)

\(\ds \map d {x, x}\)

Metric Space Axiom $(\text M 2)$: Triangle Inequality

\(\ds \leadsto \ \ \)

\(\ds 2 \map d {x, y}\)

\(\ge\)

\(\ds 0\)

Metric Space Axiom $(\text M 1)$ and Metric Space Axiom $(\text M 3)$

\(\ds \leadsto \ \ \)

\(\ds \map d {x, y}\)

\(\ge\)

\(\ds 0\)

$\blacksquare$