$f:S^1 \to \mathbb R$ 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: There is likely a better way, but you could choose some distinct $x$ and $y$ such that $f(x)$ and $f(y)$ are maximized and minimized respectively. These exist by compactness of $S_1$. Then, one can split the circle into the "intervals" $A=[x,y]$ and $B=[y,x]$ - that is, the closures of the two connected components of the circle minus $\{x,y\}$. These are homeomorphic to intervals, so by the intermediate value theorem, one finds that the image of $A$ equals the image of $B$ under $f$. In particular, this means that for any $z\in A\setminus\{x,y\}$, there exists a $z'\in B$ such that $f(z)=f(z')$. It is clear that $A\setminus \{x,y\}$ is infinite and disjoint from $B$, so there are infinitely many solutions with $z$ and $z'$ distinct.
A: 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 $J$.
A: 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.
