1
$\begingroup$

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?

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – bodo Jul 13 '12 at 7:29
  • $\begingroup$ @canaaerus I do not understand your comment would you please explain it? $\endgroup$ – Marso Jun 3 '13 at 5:16
  • $\begingroup$ The stronger claim, that every continuous map $f:S^1 \to \mathbb{R}$ has these uncountably many pairs, has been posed and answer here. $\endgroup$ – hardmath Dec 9 '15 at 2:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.