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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?

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up vote 5 down vote accepted

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.

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Actually there are no other cases where this can happen, as any such uncountable set would have to be dense somewhere but is also closed. – canaaerus Jul 13 '12 at 7:29
@canaaerus I do not understand your comment would you please explain it? – Un Chien Andalou Jun 3 '13 at 5:16
The stronger claim, that every continuous map $f:S^1 \to \mathbb{R}$ has these uncountably many pairs, has been posed and answer here. – hardmath Dec 9 '15 at 2:26

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