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

Theorem
Let $\left({\R, \tau_d}\right)$ be the real number line considered as a topological space under the usual (Euclidean) topology.

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

y - x & : y \ge x \\ 2 \left({x - y}\right) & : y < x \end{cases}$

Thus $\left({\R, \tau_d}\right)$ is a quasimetric space.

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


 * M1: $d \left({x, x}\right) = 0$
 * M2: $d \left({x, y}\right) + d \left({y, z}\right) \ge d \left({x, z}\right)$
 * M4: $x \ne y \implies d \left({x, y}\right) > 0$


 * M1:
 * $d \left({x, x}\right) = x - x = 0$

So M1 is shown to hold.

Suppose $x \le y \le z$.
 * M2:

Then:

Equality holds, hence so does the inequality.

Suppose $x \ne y$.
 * M4:

There are two possibilities:


 * $(1): \quad y \ge x$: In this case:
 * $d \left({x, y}\right) = y - x > 0$
 * $(2): \quad y < x$: In this case:
 * $d \left({x, y}\right) = 2 \left({x - y}\right) > 0$

So M4 is shown to hold.

All criteria M1, M2 and M4 are shown to hold, and so $\left({\R, \tau_d}\right)$ is shown to be a quasimetric space.