TITLE: On any continuous map $f:S^1 \to \mathbb R$
QUESTION [5 upvotes]: Let $f:S^1 \to \mathbb R$ be any continuous map , where $S^1$ is the unit circle in the plane . Let $A:=\{(x,y) \in S^1 \times S^1 : x \ne y , f(x)=f(y)\}$ ; then how to prove $A$ is uncountable , or to atleast prove $A$ is infinite ?

REPLY [3 votes]: Regarding $S^1$ as a subset of the complex plane: If $f(z)=f(-z)$ for all $z$ you're done. Suppose not. WLOG $f(-1)\ne f(1)$. Define $\phi,\psi:[0,\pi]\to\Bbb R$ by $\phi(t)=f(e^{it})$, $\psi(t)=f(e^{-it})$. If $r$ is any real in the interval with endpoints $f(1)$ and $f(-1)$ there exist $s$ and $t$ with $\phi(t)=\psi(s)=r$.

REPLY [2 votes]: The image is compact, hence has a minimum and a maximum, i.e., is an interval of the form $[a,b]$. Assume $f$ is nonconstant (otherwise the statement is trivially true). Let $x_0$ and $y_0$ be elements that map to the minimum/maximum, respectively, and consider either arc along $S^1$ joining these two points (both arcs will be nondegenerate, hence uncountable); let's say $S^1=L_1\cup L_2$ and take $L_1$. Then along $L_2$, the image still must belong to $[a,b]$, and each point in $L_2$ is distinct from $L_1$, hence $A$ is uncountable.
In particular, we can look at $L_2\times L_1\subseteq S^1\times S^1$ to see that $A$ is uncountable.