2
$\begingroup$

Let $A$ be an arc in the plane (so there is a homeomorphism $h:[0,1]\to A\subseteq \mathbb R ^2$).

Assume that $h(0)$ is the origin $o$ in the plane.

Say that a point $p\in A$ is a turning point if there is a neighborhood $U:=h[(a,b)]$ of $p$ such that $$\{p\}=U\cap B(o,\epsilon),$$ and either $U\subseteq \overline{B(o,\epsilon)}$ or $U\subseteq \mathbb R ^2\setminus B(o,\epsilon)$, where $\epsilon=d(o,p)$.

enter image description here

Let $T$ be the set of turning points in $A$.

Question: Is $\{\epsilon>0:T\cap \partial B(o,\epsilon)\neq\varnothing\}$ necessarily a countable set?

Easier Question: Does there exists $\epsilon>0$ such that $A\cap \partial B(o,\epsilon)\neq \varnothing$ and $T\cap \partial B(o,\epsilon)=\varnothing$?

$\endgroup$
  • $\begingroup$ Do you mean for $h(0)$ to be the origin? That seems like it contradicts your picture. $\endgroup$ – Rocket Man Jun 24 '17 at 3:29
  • $\begingroup$ $h(0)$ is the center of the circle in the picture, which you can think of as the origin in the plane (technically there should be a line leaving it) $\endgroup$ – Forever Mozart Jun 24 '17 at 3:30
  • $\begingroup$ I don't suppose that $h$ is known to be continuously differentiable? $\endgroup$ – Chris Culter Jun 24 '17 at 5:08
  • $\begingroup$ ...Never mind, I don't need any differentiability. $\endgroup$ – Chris Culter Jun 24 '17 at 5:16
1
$\begingroup$

In fact, $T$ itself is countable! Define $f:[0,1]\to\mathbb R$ by $f(t)=|h(t)|$. Then we can identify $T$ with the strict local extrema of $f$. But there are only countably many of those, since each minimum or maximum can be uniquely bounded by a pair of rational numbers. (For details on this step, see: Does there exist a continuous function from [0,1] to R that has uncountably many local maxima?)

$\endgroup$
  • $\begingroup$ oh, how clever! $\endgroup$ – Forever Mozart Jun 24 '17 at 17:29

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.