# For continuous function $f:\mathbb S^1 \to \mathbb R$ there exists uncountably many distinct points $x,y$ such that $f(x)=f(y)$

Let $\mathbb S^1$ denote the unit circle in $\mathbb R^2$.

Then prove that for every continuous function $f:\mathbb S^1 \to \mathbb R$, there exist uncountably many pairs of distinct points $x, y$ in $S^1$, such that $f(x)=f(y)$.

• Where did you get this question from? What are your thoughts? – tomasz Jan 16 '14 at 1:44

$S^1$ is compact and connected; the continuous image of a compact and connected set is also compact and connected, and hence a bounded interval $[a, b]$ in $\mathbb R$. Let $(\cos \theta, \sin \theta)$ be a point of $S^1$ such that $f(\theta) = a$. Define $$g: [0, 2 \pi ] \to S^1: t \mapsto (\cos (\theta + t), \sin(\theta + t))\\ h = f \circ g.$$

Then $h(0)) = h(2\pi) = a$, and the image of $h$ is $[a, b]$.

For some point $0 < u < 2 \pi$, we must have $h(u) = b$.

Now consider the intervals $[0, u]$ and $[u, 2\pi]$. $h$ maps each of these intervals continuously to the whole interval $[a, b]$. For any $c$ between $a$ and $b$ there are points $x_L$ and $x_R$ in the two intervals, respectively, with $h(x_L) = h(x_R) = c$, by the intermediate value theorem. Each value $c$ provides and instance of a point-pair that maps to the same value in $[a, b]$ under $f$.

• By the way, @clark's comment below makes essentially the same argument, but in short form. I wanted to make it really explicit and use only basic theorems like IVT, but his use of "continuous image of connected set is connected" is far more pithy. – John Hughes Jan 16 '14 at 2:53

Assume $f$ is non constant. Take a point $y_0$ between the minimum and maximum of $f$.

Assume the $f^{-1} (y_0)= \{x_0 \}$ i.e. it consists of only one point of $\mathbb{S}^1$.

Then $f(\mathbb{S}^1/{x_0} )$ must be connected which leads to a contradictions.

So the preimage of $y_0$ has at least two points.

A useful way to look at this is topologically. If you take the space $S^1\times S^1$ and remove the diagonal, the space is still path-connected. Define $h(x,y)=f(x)-f(y)$. Then any path between $(x_0,y_0)$ and $(y_0,x_0)$ inside this set will contain a zero of $h$. (Why?)

Then, show there is an uncountable collection of curves between $(x_0,y_0)$ and $(y_0,x_0)$ that are 100% distinct - no pair of paths intersect (except at the end points.)

Verbose details

If $f(e^{ix_0})\neq f(e^{iy_0})$ for $0<x_0\leq y_0< 2\pi$, then for each $u\in(0,1)$ define the define a path $\phi_u$ which goes linearly from $(x_0,y_0)$ to $(0,t)$ and then from there to $(y_0-2\pi,x_0)$.

You need to show that:

1. This path says inside the region: $\{(x,y):-2\pi<x-y<0\}$, and so if $x,y$ is on the path, then $e^{ix}\neq e^{iy}$.

2. $\phi_u(t)\neq \phi_{u'}(t)$ unless $t=0,1$ or $u=u'$.

Then defining $h(x,y)=f(e^{ix})-f(e^{iy})$ you get that $h(\phi_u(0))=-h(\phi_u(1))$ and thus, there must be a $\phi_u(t)$ for some $t\in(0,1)$ with $h(\phi_u(t))=0$. Since $(x,y)=\phi_u(t)$ has the property that $e^{ix}\neq e^{iy}$, this means that $f(e^{ix}=e^{iy}$ for some $(x,y)$ inside this curve.

But there are uncountably many $u$, and since the curves are disjoint except on the endpoints, there are thus uncountably many such pairs $(x,y)$.