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Let $\bf S^1$ denote the unit circle in the plane $\bf R^2$. True/False ?

For every continuous function $f:\bf S^1 \to \bf R$, there exist uncountably many pairs of distinct points $x$ and $y$ in $\bf S^1$ such that $f(x)=f(y)$

Borsuk-Ulam or by taking the function $g(x)=f(x)-f(-x)$, IVT implies that there exist $x$ such that $f(x)=f(-x)$. But I'm unable to show the existence of uncountably many pairs. I think the fact $RP^1 \cong \bf S^1$ may be helpful. Any ideas?

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marked as duplicate by Macavity, user91500, mrp, Arnaldo, Daniel W. Farlow May 29 '17 at 12:46

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Further search failed to turn up a reasonable duplicate candidate, so I retracted my close vote. $\endgroup$ – hardmath Dec 9 '15 at 2:13
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    $\begingroup$ Dear @hardmath,existence is of course trivial,one can just take the identity map.Regards, $\endgroup$ – Arpit Kansal Dec 9 '15 at 2:14
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If $f$ is constant then this is obviously true.

Otherwise, choose points $a,b \in \Bbb S^1$ such that $f(a)\ne f(b)$ and "split the circle" at $a$, so that $f$ becomes a continuous map $[a,a+2\pi] \to \Bbb R$ with $f(a)=f(a+2\pi)\ne f(b)$ for $b\in(a,a+2 \pi)$. Then for any value $M$ strictly between $f(a)$ and $f(b)$, the intermediate value theorem gives us the existence of $x \in (a,b)$ and $y \in (b,a+ 2\pi)$ such that $f(x)=f(y)=M$. Since $f(a)\ne f(b)$, there are uncountably many such values $M$ and thus such pairs, so the proposition is true.

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Here is "almost" same solution as of @Anthony Carapetis written in slightly different way:

Assume $f$ is not constant.Since $\bf {S}^1$ is connected and compact so let $f(\bf S^1)=[a,b]. $Suppose pre image of some $y \in (a,b) $ consists only one point,say $x$.Then $f( \bf {S}^1 / {x})$ is connected.Contradiction!

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    $\begingroup$ I like the appeal to compactness at the outset, and the way you finish with an appeal to connectedness argument is compelling. However I think I would add an observation that $f(S^1/\{x\}) = [a,b]/\{y\}$ makes the image disconnected, so contradiction. $\endgroup$ – hardmath Dec 9 '15 at 23:33
  • $\begingroup$ Dear @hardmath Thank you,I am pleased that you really enjoyed reading my solution.Best Regards, $\endgroup$ – Arpit Kansal Dec 10 '15 at 2:19
  • $\begingroup$ @ArpitKansal where r u now? I know u were cmi stdnt... $\endgroup$ – sani May 29 '17 at 3:57
  • $\begingroup$ @sani, Hi Sani,currently i am at IMSc. $\endgroup$ – Arpit Kansal May 29 '17 at 7:09
  • $\begingroup$ good...Best of luck@ArpitKansal $\endgroup$ – sani May 29 '17 at 10:39

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