Real Number Line with Off-Center Distance Function is Quasimetric Space

Theorem
Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.

Then $\tau_d$ can be given by a quasimetric defined as:
 * $\map d {x, y} = \begin {cases}

y - x & : y \ge x \\ 2 \paren {x - y} & : y < x \end {cases}$

Thus $\struct {\R, \tau_d}$ is a quasimetric space.

Proof
To show that $\map d {x, y}$ is a quasimetric, we need to show that $d: \R \times \R \to \R$ satisfies the following conditions for all $x, y, z \in \R$:


 * $(\text M 1): \quad \map d {x, x} = 0$
 * $(\text M 2): \quad \map d {x, y} + \map d {y, z} \ge \map d {x, z}$
 * $(\text M 4): \quad x \ne y \implies \map d {x, y} > 0$

Proof of $\text M 1$

 * $\map d {x, x} = x - x = 0$

So $(\text M 1)$ is shown to hold.

Proof of $\text M 2$
Suppose $x \le y \le z$.

Then:

Suppose $x > y > z$.

Then:

Equality holds, hence so does the inequality.

Proof of $\text M 4$
Suppose $x \ne y$.

There are two possibilities:


 * $(1): \quad y \ge x$, in which case:
 * $\map d {x, y} = y - x > 0$


 * $(2): \quad y < x$, in which case:
 * $\map d {x, y} = 2 \paren {x - y} > 0$

So $(\text M 4)$ is shown to hold.

All criteria $(\text M 1)$, $(\text M 2)$ and $(\text M 4)$ are shown to hold, and so $\struct {\R, \tau_d}$ is shown to be a quasimetric space.