Arc Length on Circle forms Metric

Theorem
Let $A \subseteq \R^2$ be the set defined as:
 * $A = \set {\tuple {x_1, x_2}: x_1^2 + y_2^2 = 1}$

Thus from Equation of Unit Circle, $A$ is the unit circle embedded in the Cartesian plane.

Let $d: A^2 \to \R$ be the mapping defined as:
 * $\forall \tuple {x, y} \in A^2: \map d {x, y} = \begin {cases} 0 & : x = y \\ \pi & : x = -y \\ l & : \text {otherwise} \end {cases}$

where:
 * $x$ and $y$ are of the form $\tuple {x_1, x_2}$
 * $l$ denotes the length of the minor arc of $A$ between $x$ and $y$.

Then $d$ is a metric for $A$.

Proof
It is to be demonstrated that $d$ satisfies all the metric space axioms.

Proof of

 * $\map d {x, x} = 0$ by definition.

So holds for $d$.

Proof of
Let $x$, $y$ and $z$ be points on $A$.

Suppose $z$ be such that either:
 * it lies on the minor arc from $x$ and $y$
 * $x$ and $y$ lie on a diameter of $A$.

Then $x z$ and $z y$ are both minor arcs of $A$.

Hence:
 * $\map d {x, z} + \map d {z, y} = \map d {x, y}$

Suppose otherwise, that $z$ lies on the major arc of $A$.

Then by inspection:
 * $\map d {x, z} + \map d {z, y} > \map d {x, y}$

So holds for $d$.

Proof of
By inspection:
 * $\map d {x, y} = \map d {y, x}$

for all $x, y \in A$.

So holds for $d$.

Proof of
By inspection:
 * $\map d {x, y} \ge 0$

for all $x, y \in A$.

So holds for $d$.

Thus $d$ satisfies all the metric space axioms and so is a metric.