How does the "arc tangent metric" $d(x,y) = | \arctan(x) - \arctan(y)| $ work?

Question

I see there are some counterexamples and so forth in metric spaces regarding the metric $$d(x,y) = | \arctan(x) - \arctan(y)| $$

But honestly I have no intuition as to how it works

For example, in the Euclidean metric the intuition here is just length between two points and you have a good visual/physical intuition as to how it works.

Discrete metric fixes that length between two points to be 1 and can be visualized as a unit line segment, or an equilateral triangle.

But what is the intuition behind this metric? $$d(x,y) = | \arctan(x) - \arctan(y)| $$

What is the metric space equipped with $d(x,y)$ referred to? How do people came up with it and is it possible to have some mental picture of how the metric works. Finally, what is so special about this metric?

Let me know if this is just a metric created for sake of counterexample.

Answer 1

Here is a quick graph and how $d(x,y)$ works. Notice that near $0$, $d(x,y)$ evaluates the distance of $x,y$ as if it were the absolute value, whereas for large $|x|$ things are completely different: distances are quite small. In fact the "biggest" distance you'd get is $\pi$ (and is never attained of course).

Answer 2

This metric is, in a way, just the standard metric on $(-\frac{\pi}{2},\frac{\pi}{2})$ in disguise.

To see that, take the interval $(-\frac{\pi}{2},\frac{\pi}{2})$ and imagine that you are "stretching" it, say by mapping every point $a$ to some point $f(a)$. As long as no two points overlap, then we can measure distances between points on the stretched interval, by saying that the distance between $f(a)$ and $f(b)$ is the same as the distance between $a$ and $b$ in the non-stretched interval, which is $|a-b|$.

Now we take the "stretching function" to be $f(a)=\tan a$. In our metric, we define the distance between $\tan(a),\tan(b)$ to be $|a-b|$. Now if we denote $\tan(a)=x,\tan(b)=y$, then this is precisely stating that the distance between $x$ and $y$ is $|\arctan(x)-\arctan(y)|$.

Answer 3

First, you can check that the arctan metric satisfies the triangle inequality, so it does meet the definition of a metric.

Now if you want some intuition, say you measure the distance between two points by the time taken by a ship traveling in Euclidean space along a line connecting $A$ and $B$, with some formula for the ship's path-length velocity. For example, if that velocity is a constant, you would have the Euclidean metric.

But say that on each trip you start off at some velocity $v_0$ (which is the same for every path and then start increasing the velocity more than exponentially in the length traveled thus far. Then you can guess that perhaps by travelling for some finite time you can get arbitrarily far from the original spot.

In the metric induced by this model, there are not points more than (say) $\pi$ apart. Yet the topolgy is still like that of ordinary space, rather than that of a sphere hwich would also put a limit on how far you can travel.

Answer 4

Imagine the vertical line $x = 1$. The metric is just the angle between the rays from the origin to the points on this line at heights $x$ and $y$. For small heights $x$, the ray depends strongly on $x$. For large (positive or negative) heights, the ray doesn't depend much on $x$ because it's mostly vertical.