$f:S^1 \to \mathbb R$ be continuous , is the set $\{(x,y) \in S^1 \times S^1 : x \ne y , f(x)=f(y)\}$ infinite ? Let $f:S^1 \to \mathbb R$ be a continuous function , I know that $\exists y \in S^1 : f(y)=f(-y)$ where $y \ne -y $ (since $||y||=1$) , so that the set $A:=\{(x,y) \in S^1 \times S^1 : x \ne y , f(x)=f(y)\}$ is non-empty ; my question is , is the set $A$ infinite ? 

A more combinatorial solution:
Since the set of "bad pairs" $B=\{(x, y)\in S^1\times S^1: x\not=y, f(x)=f(y)\}$ is finite and nonempty, we can find some $a, b\in B$ and a path $P\subset S^1$ joining $a$ to $b$ such that


*

*$f(a)=f(b)$, but

*for any $c, d\in P\setminus\{a, b\}$, we have $c\not=d\implies f(c)\not=f(d)$.
Fixing such $a, b, P$, look at the quotient space $X=P/(a\sim b)$. This is homeomorphic to $S^1$, and the map $\hat{f}: X\rightarrow\mathbb{R}$ induced by $f$ is continuous and injective. This gives a contradiction.
