Show that $d(x,y)= \left|\frac{1}{x} - \frac{1}{y}\right|$ defines a metric on $(0,\infty)$.

Question

Exercise: Show that $d(x,y)= \left|\frac{1}{x} - \frac{1}{y}\right|$ defines a metric on $(0,\infty)$.

What I've tried: I need to check the following properties

• i) $0\leq d(x,y)<\infty,$ for all $x,y\in(0,\infty)$.
• ii) $d(x,y)=0$ if and only if $x=y$
• iii) $d(x,y)=d(y,x),$ for all $x,y\in(0,\infty)$.
• iv) $d(x,z)\leq d(x,y) + d(y,z)$ for all $x,y,z\in(0,\infty)$.

I think I was able to solve the first three properties

i): If $\dfrac{1}{x} - \dfrac{1}{y}$ is positive then $d(x,y)$ is positive, if $\dfrac{1}{x} - \dfrac{1}{y}$ is negative then $d(x,y) = \left|\dfrac{1}{x} - \dfrac{1}{y}\right|$ is positive, so $d(x,y)\geq 0$. Since $x,y\in(0,\infty)$ we have that $\left|\dfrac{1}{x} - \dfrac{1}{y}\right|\in(0,\infty)$.

ii): If $x = y$ then $d(x,y) = 0$, if $x\neq y$ then $\dfrac{1}{x} \neq \dfrac{1}{y}$ and $d(x,y) \neq 0$.

iii): If $\left|\dfrac{1}{x} - \dfrac{1}{y}\right|^2 = \left|\dfrac{1}{y} - \dfrac{1}{x}\right|^2$ then $\left|\dfrac{1}{x} - \dfrac{1}{y}\right| = \left|\dfrac{1}{y} - \dfrac{1}{x}\right|$. I know that $\left|\dfrac{1}{x} - \dfrac{1}{y}\right|^2 = 1/x^2 - 2/xy + 1/y^2 = \left|\dfrac{1}{y} - \dfrac{1}{x}\right|^2,$ hence $\left|\dfrac{1}{x} - \dfrac{1}{y}\right| = \left|\dfrac{1}{y} - \dfrac{1}{x}\right|^2$ and $d(x,y) = d(y,x)$.

I don't know how I should show that the fourth property holds!

Question: How do I show that $\left|\dfrac{1}{x} - \dfrac{1}{y}\right|\leq \left|\dfrac{1}{x} - \dfrac{1}{z}\right| + \left|\dfrac{1}{z} - \dfrac{1}{y}\right|$, or equivalently, that $d(x,y) \leq d(x,z) + d(z,y)$?

Thanks!

Answer 1

You just apply the triangle inequality: for any three real numbers $a$, $b$, and $c$,$$|a-b|\leqslant|a-c|+|c-b|.$$

Answer 2

In general if $(Y,d)$ is a metric space and $f:X\to Y$ is injective then $\overline d:X^2\to\mathbb R$ prescribed by: $$\overline d(a,b)=d(f(a),f(b))$$ is a metric on $X$.

It is not really difficult to prove that $\overline d$ has the characteristic properties of a metric.

Answer 3

If $(X,d_X)$ is a metric space and $f: Y \to X$ is a 1-1 function, then $$d_Y(y_1, y_2) := d_X(f(y_1), f(y_2))$$

is a metric on $Y$.

As $d_X$ has values in $[0,\infty)$ so has $d_Y$.

$d_Y(y_1, y_2) = 0$ iff $d_X(f(y_1), f(y_2)) = 0$ iff $f(y_1) = f(y_2)$ (as $d_X$ is a metric) iff $y_1 = y_2$ (as $f$ is injective).

$d_Y(y_1, y_2) = d_X(f(y_1), f(y_2)) = d_X(f(y_2, f(y_1)) = d_Y(y_2,y_1)$ where in the middle we use the symmetry of $d_X$.

if $y_1, y_2, y_3 \in Y$ then using the triangle inequality of $d_X$ we get

$$d_Y(y_1, y_3) = d_X(f(y_1), f(y_3)) \le d_X(f(y_1), f(y_2)) + d_X(f(y_2), f(y_3)) = d_Y(y_1, y_2) + d_Y(y_2,y_3)$$

Note that $f$ is irrelevant except we need its injectivity once. Here we apply it to $f(x) = \frac{1}{x}$ and $d_X(x, x') = |x-x'|$, the usual metric on the reals.

Also note that this makes $f$ an isometry by definition between $(Y,d_Y)$ and $(X,d_X)$.

Answer 4

The following equality holds in an absolute value:$$\left|\frac{1}{x}-\frac{1}{y}\right|=\left|\frac{1}{x}-\frac{1}{z}+\frac{1}{z}-\frac{1}{y}\right|.$$

Now just break it apart using the triangle inequality.