$f:S^1 \to \mathbb R$ 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 ? 

If $f$ is constant, then the answer is clearly yes.
Suppose that $f$ is not constant, so there are $p,q\in S^1$ with $f(p) < f(q)$.
There are two paths joining $p$ and $q$ in $S^1$; call them $I$ and $J$.  Applying the Intermediate Value Theorem to $f|_I$ and $f|_J$, we can see that every value in the open interval $(f(p),f(q))$ is achieved at least twice: once in $I$ and once in $J$.
