Metric on the line (1-Dimensional space)

Question

Is the Euclidean metric the only possible metric in real line? Recall that the Euclidean metric is given by $d(x,y) = |y-x|$, and if there is another metric, are these two equal? what about the metric on the one dimensional sphere?

Answer 1

Given continuous $f(x) > 0,$ take $$ d(x,y) = \left| \int_x^y f(t) dt \right| $$ If $x < y < z$ you always get $d(x,y) + d(y,z) = d(x,z).$ if you take something along the lines of $f(t) = 1 / (1 + t^2)$ the length of the entire line becomes finite.

Meanwhile, when $$ \int_{- \infty}^\infty f(t) dt < \infty, $$ we also have provided a metric on the circle... let the distance between $x < y$ going "the other way around" be $$ \int_{- \infty}^x f(t) dt + \int_y^\infty f(t) dt. $$ This can now be pulled back to the standard unit circle by stereographic projection out of The North Pole $(0,1).$ Either case, you get a genuine metric space by taking the smaller value between two points, through the north pole or not.

Answer 2

Let $(M,d)$ be a metric space with cardinality greater than or equal to that of the continuum and let $f\colon \mathbb{R}\to M$ be an injective function. Define $d'\colon\mathbb{R}\times\mathbb{R}\to\mathbb{R}$ by $$d'(x,y)=d(f(x),f(y)).$$

$d'$ is a metric.