# 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? Jan 16 '14 at 1:44

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.

$S^1$ 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. Jan 16 '14 at 2:53

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 $(x,y)$.