Intuition for the Universal Chord Theorem

Question

So the Universal Chord theorem is a statement and proof that;

The numbers of the form $r = \displaystyle \frac{1}{n} \ \ n \ge 1$ are the only numbers such that for any continuous function $\displaystyle f:[0,1] \to \mathbb{R}$ such that $\displaystyle f(0) = f(1)$, there is some point $\displaystyle c \in [0,1]$ such that $\displaystyle f(c) = f(c+r)$.

The proof is straightforward to understand. I don't have difficulty with any step in it or anything. As well, I understand the counterexamples for any non rational, and why they fail. But even still it seems absurd that this is how it is. I've been drawing a bunch of graphs here on a page and I just can't see any sort of insight as to why rationals should work but not non rationals. I heard once that there was intuition to be found with topology, but I can't find it anywhere online. Could someone please enlighten me as to why this theorem makes sense intuitively? (without just giving a proof)

Answer 1

I saw it proved topologically by imagining wrapping the graph around a cylinder of circumference 1/n (which you can't do evenly for any other number). Then on the cylinder, the graph starts at a base point, loops around $n$ times, then returns to the be point. It's not hard to see that the curve must intersect itself somewhere (otherwise, it wouldn't be looped); thus, we get two points of distance $k/n$ apart. If they weren't $1/n$ apart, then the graph loops around the cylinder at least twice between these two points, so there is another point of intersection giving a pair of strictly closer points.

So the intuitive reason that only these numbers work is that they are the only possible circumferences of cylinders that the graph could wrap evenly around.

For the other direction, try wrapping the unit interval around a cylinder of some other length; the endpoints won't match up. Then construct a monotonic path (I.e. lways wrapping the same way) of length one that dives down, then circles around diagonally several times and ends at the second point (try it and see!). This unwraps to a path with the desired properties.

Answer 2

Universal Chord Theorem (UCT) is a special case of certain topological property of plane continuums (a bounded, closed, connected sets) represented in this specific case by a graph of some continuous function $f$. In his paper from 1936 H.Hopf generalized Levy's result of UCT by proving the following property of plane continuums:

Let $K$ be a plane continuum and $S(K)$ be the set of horizontal chordal lengths for $K$. Then any given set $M$ is $S(K)$ for some $K$ if and only if complemented set $\mathbb R^+\setminus M$ is open, non-empty, and closed under addition.

In his paper from 1971 J.T.Rosenbaum considered connection of UCT with this fundamental property by showing that if for some $K$ we consider any chord of length $c\in S(K)$, then all chords with lengths $c/2$, $c/3$, ..., $c/n$, ... will exist in $K$ too, i.e. all these lengths will also belong to $S(K)$. This is because if $M^*$ = $\mathbb R^+\setminus S(K)$ and $\exists n: c/n \in M^*$, then $c/n$ + ... + $c/n$ ($n$ times) = $c$ and $c \in M^*$ too, which is not true because $c \in S(K)$. That, the paper deduced the first half of UCT from the Hopf's result.

For the second half, Rosenbaum conducted special construction of $M^*$ defining its $S(K)$ and showing that only these integerly-divided values will present in $S(K)$.

Getting back to original UCT wording, we are given $K=f$, which condition $f(0)=f(1)$ denotes a horizontal chord of length $c=1$. Then $c/n=1/n$ and $S(K)=\{1/n|n \in \mathbb N\}$.

Summarising, the intuition behind values $1/n$ is to represent all harmonics of the originally given chord of length 1. Nothing at all about some "better kind of rational numbers" as it might seem by first glance.

Answer 3

Intuitively, the problem begs for us to consider the function $g$ where $g(x) = f(x+r)-f(x)$ and see whether it is ever 0. If it is never 0, then $g$ is always of the same sign on $[0,1-r]$, and it is natural to hop from $f(0)$ to $f(r)$ to $f(2r)$ and so on, which must be a strictly monotonic sequence, which is impossible if $1$ is an integer multiple of $r$. On the other hand, if $1$ is not an integer multiple of $r$, then we can actually let $f$ be such that $g$ is a constant $c$, which makes the hop points a regularly spaced collinear sequence, by drawing a straight line segment from the last hop point to $(1,f(1))$ and filling in what follows and finally the rest of the graph with parallel line segments.