3
$\begingroup$

I was attempting to do an exercise on my topology book. It states the following:

Let $f:\mathbb{R} \mapsto \mathbb{R}$ be a homeomorphism such that $\exists p \in \mathbb{N}: f^p =$ identity. Prove that $f$ has a fixed point.

I think I have a solution but I never really use the fact that $f$ is a homeomorphism. That is:

Let $g(x)=f(x)-x$. Then $g$ is a continuous function. Now, let's say that $f(x)\geq x \ \forall x \in \mathbb{R}$. Then $f^p(x) \geq f(x) \geq x = f^p(x) \forall x$, which implies $f(x)=x \forall x$.

In the same way, if $f(x)\leq x \forall x$ we have $f^p(x) \leq f(x) \leq x = f^p(x) \implies f(x) = x$.

So, if $f(x)$ is not the identity, then $\exists x_1, x_2 \in \mathbb{R}: f(x_1) < x_1, f(x_2) > x_2$, and because $g$ is continuous and $g(x_1)<0, g(x_2)>0$ then there's a point $\bar{x}: g(\bar{x})=0 \implies f(\bar{x})=\bar{x}$, so it is a fixed point.

I've not actually used the fact that $f$ is a homeomorphism, I only used the fact that it is continuous and has finite order, so is my proof wrong?

$\endgroup$
  • 8
    $\begingroup$ I think your proof is fine. Note that being continuous and having finite order already implies that $f$ is a homeomorphism, since $f^{p-1}$ is a left- and right-inverse. $\endgroup$ – PhoemueX Jul 27 at 16:24
  • $\begingroup$ That's a fair point indeed $\endgroup$ – DottorMaelstrom Jul 27 at 16:29
1
$\begingroup$

Your proof is correct, but in my opinion it could be formulated more transparently. In fact, you consider three cases.

Case 1: $\forall x \in \mathbb{R} : f(x) \ge x$.

Case 2. $\forall x \in \mathbb{R} : f(x) \le x$.

Case 3: Neither 1 nor 2 is satisfied. Then $\exists x_1, x_2 \in \mathbb{R} : f(x_1) < x_1, f(x_2) > x_2$. Now introduce $g$.

$\endgroup$

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.