Density of points on a circle

Question

Given a circle with radius 1. Take an arbitrary starting point. Then go around the circle an infinite number of times, always drawing a point when you are one unit further. So the next point is always an arc length 1 step further. Easy to prove that all points will be different! But my question is: is each point randomly close approached? I believe this is true, but I can’t prove it!

Answer 1

As mentioned in the comments, this has to do with $\pi$ being irrational. Let's assume that, and go on to prove that you'll end up arbitrarily close to any given point on the circle.

Represent the circle as the interval $[0,2\pi)$ -- basically, a point is represented by how far counterclockwise you have to go from your starting point to get there. In your process, you're starting at $0$ and adding $1$ to your location every step, and when the location becomes greater than $2\pi$, you subtract $2\pi$ so that it lies between $0$ and $2\pi$.

Assume that there's some $0\leq x<2\pi$ and some $\epsilon>0$ for which no point in the interval $(x-\epsilon,x+\epsilon)$ is ever reached. Now, consider the first $$n=\left\lceil \frac{\pi}{\epsilon}\right\rceil$$ steps, and say the locations reached (including the starting point $0$) are $$0=x_0<x_1<x_2<\cdots<x_n<2\pi$$ (note that $x_i$ is not the location after $i$ steps, but the $i$th smallest location in $(0,2\pi)$ that you reach). Since they're all in $[0,2\pi)$, there can't be a distance of greater than $2\epsilon$ between each pair -- otherwise, $$2\pi>x_n=(x_n-x_{n-1})+(x_{n-1}-x_{n-2})+\cdots+(x_2-x_1)+(x_1-x_0)\geq 2n\epsilon\geq 2\pi.$$ So, there must be some integers $a<b<n$ for which the $a$th step and the $b$th step are very close to each other -- within $2\epsilon$. So, the $(b-a)$th step is within $2\epsilon$ of $0$, since going $b-a$ units forward from the $a$th location gets you to a point within $2\epsilon$ of the $a$th location. Let this location be $\delta$ units away from $0$, so it is either $\delta$ or $2\pi-\delta$.

Finally, consider the points reached after $k(b-a)$ steps for various integers $k$. As you add $1$ to $k$, this point moves $\delta$ units around the circle, either clockwise (if the $b-a$th location is $\delta$ units clockwise from $0$) or counterclockwise (if it's counterclockwise from $0$). Because $\delta<2\epsilon$, you can't ever "jump over" the interval $(x-\epsilon,x+\epsilon)$ in this way -- you'll hit some point inside it at step $k(b-a)$ for some integer $k$. So, there is some step that drops you in $(x-\epsilon,x+\epsilon)$, as desired.

Note: This proof is quite similar to a proof of Dirichlet's approximation theorem, which states that irrational numbers are rather well-approximated by rational numbers. Rational approximation, specifically how well certain classes of numbers can be approximated by rationals, is a very deep area of mathematics!