Does there exist continuous map from $S^1$ to $\mathbb{R}$ such that $f(x)=f(y)$ for uncountably many $x,y\in S^1$?
By the Borsuk-Ulam theorem, I know there is no injective map from $S^1\rightarrow \mathbb{R}^1$.
If we consider some map like $g(x)=f(x)-f(-x)$ for $x\in S^1$, I can say by the intermediate value theorem that there is at least one point where $g$ vanishes; is that vanishing set uncountable?
Best Answer
Any constant map $f:\mathbb{S}^1\to\mathbb{R}$ is continuous and would satisfy $f(x)=f(y)$ for every $x,y\in\mathbb{S}^1$, of which there are certainly uncountably many.
In fact, a continuous map $f$ that is only constant on a non-empty open set of $\mathbb{S}^1$ would have $f(x)=f(y)$ for uncountably many $x,y\in\mathbb{S}^1$, because any non-empty open set of $\mathbb{S}^1$ is uncountable.