Distance Function for Distinct Elements in Metric Space is Strictly Positive

Theorem

Let $A$ be a set.

Let $d: A \times A \to \R$ be a real-valued function on $A$ with the following properties:

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

which can be considered as being an alternative formulation of the metric space axioms.

Then:

$\forall x, y \in A: x \ne y \implies \map d {x, y} > 0$

which is Metric Space Axiom $(\text M 4)$.

Thus $d$ is a distance function, so making $M := \struct {A, d}$ a metric space.

Proof

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

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

\(\ge\)

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

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

\(\ds \leadsto \ \ \)

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

\(\ge\)

\(\ds 0\)

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

\(\ds \leadsto \ \ \)

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

\(\ge\)

\(\ds 0\)

$\blacksquare$