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

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.

• It may help to consider the "$f$ metric" given by $f(x,y)=|f(x)-f(y)|$ for any injective function $f$ instead. It says that the distance between two points is the ordinary distance between their images under $f$ -- in a certain (very real) sense the metric space we get is the same space as the range of $f$ with the usual distance, except that we have given its points new names! The arctan metric is then just this construction with $f=\arctan$ -- no particular properties of arctan are relevant here apart from the fact that it is injective! – hmakholm left over Monica Mar 17 '16 at 21:23
• If I have to say one thing that is special about the arctan metric then it is that it is equivalent to the Euclidean metric (in that they always agree on whether a given subset of the real line is open), and they even agree on which sets are open balls (although they do not agree about where the centre of such a ball is), but at the same time the real line is bounded under the arctan metric (no point is more than $\pi$ away from another). – Arthur Mar 17 '16 at 21:27

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 compeltely different: distances are quite small. In fact the "biggest" distance you'd get is $\pi$ (and is never attained of course).

• I don't know how in the world you just happened to have that diagram ready but it is epic. – Olórin Mar 17 '16 at 21:49
• @MSEisadatingsite from request, to demand! Thanks ;-)! Hope it helps a bit! – Nikolaos Skout Mar 17 '16 at 21:54
• By the way, this metric has some interesting properties: for example, $\mathbb{R}$ has diameter $\pi$ in that metric (so it is bounded), $d$ is equivalent to the absolute-value metric and also $(\mathbb{R},d)$ is not complete (while $(\mathbb{R},|\cdot|$ is complete). To summarize: this metric gives great, and more important, natural counterexamples!! – Nikolaos Skout Mar 17 '16 at 21:59

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 $|\arctan(x)-\arctan(y)|$.

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.

• Take a line parallel to x-axis as x= 1,starting from it. If the ends are distant $X,Y$ from x-axis, $d (X,Y)= arctan Y - arctan X. – Narasimham Mar 17 '16 at 21:36 Imagine the vertical line$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.