TITLE: For continuous function $ f:\mathbb S^1 \to \mathbb R$ there exists uncountably many distinct points $x,y$ such that $f(x)=f(y)$
QUESTION [9 upvotes]: 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 $f(x)=f(y)$.

REPLY [5 votes]: 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:

This path says inside the region: $\{(x,y):-2\pi<x-y<0\}$, and so if $x,y$ is on the path, then $e^{ix}\neq e^{iy}$. 
$\phi_u(t)\neq \phi_{u'}(t)$ 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 $(x,y)$.