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

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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.

$\Box$


Proof of $\text M 2$

Suppose $x \le y \le z$.

Then:

\(\ds \map d {x, y} + \map d {y, z}\) \(=\) \(\ds \paren {y - x} + \paren {z - y}\)
\(\ds \) \(=\) \(\ds \paren {z - x}\)
\(\ds \) \(=\) \(\ds \map d {x, z}\)

Suppose $x > y > z$.

Then:

\(\ds \map d {x, y} + \map d {y, z}\) \(=\) \(\ds 2 \times \paren {x - y} + 2 \times \paren {y - z}\)
\(\ds \) \(=\) \(\ds 2 \times \paren {x - z}\)
\(\ds \) \(=\) \(\ds \map d {x, z}\)

Equality holds, hence so does the inequality.



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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.

$\Box$


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.

$\blacksquare$


Sources

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