3
$\begingroup$

A homeomorphism f is said to be orientation reversing if for any $x<y<z$ we have $f(z)<f(y)<f(x)$. Show that every orientation reversing homeomorphism of the real line has a fixed point.

This is a question on my assignment sheet (not for credit) that I've been thinking about for days but not made any progress on. I feel like this will be easy to answer once I find the trick, any hints would be great!

$\endgroup$
  • $\begingroup$ IVT${}{}{}{}{}$? $\endgroup$ – Lord Shark the Unknown Oct 13 '18 at 18:21
  • $\begingroup$ I was thinking that but I'm not quite sure how to apply it $\endgroup$ – SpaghettiMonsterMan Oct 13 '18 at 18:23
5
$\begingroup$

If $f(0)=0$, we are done.

If $f(0)>0$, then $f(f(0))<f(0)$, and if $f(0)<0$, then $f(f(0))>f(0)$. In both cases, we have poitns $x_1,x_2$ with $f(x_1)<x_1$ and $f(x_2)>x_2$. Then the Intermediate Value Theorem tells us that the continuous function $x\mapsto f(x)-x$ has a zero between $x_1$ and $x_2$, i.e., $f$ has a fixed point.

$\endgroup$
  • $\begingroup$ So we don't need homeomorphism, just continuous. $\endgroup$ – lhf Oct 13 '18 at 18:46
  • $\begingroup$ @ihf I'm not sure what your comment means, but the hypothesis that it $f$ is an orientation reversing homeomorphism has been used in this proof, i.e. in the case $f(0)>0$ we may conclude $f(f(0))<f(0)$. This proof would break down if you only assumed $f$ was continuous. $\endgroup$ – Lee Mosher Oct 13 '18 at 20:32
  • $\begingroup$ @LeeMosher, I meant, does $f$ need to be a bijection? With continuous inverse? Sure, orientation reversing is essential. $\endgroup$ – lhf Oct 14 '18 at 10:56
  • 1
    $\begingroup$ What should be the meaning of "orientation reversing" for a map which is not a homeomorphism? $\endgroup$ – Paul Frost Oct 14 '18 at 12:13
  • 1
    $\begingroup$ Okay. The only assumption you need is that $f$ is strictly decreasing. In this case $f(\mathbb{R})$ is an open interval (bounded or unbounded), and $f$ is an orientation reversing homeomorphism from $\mathbb{R}$ to $f(\mathbb{R})$. $\endgroup$ – Paul Frost Oct 15 '18 at 14:58

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.