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Let $X$ be a compact star shaped subset of $\mathbb{R}^n$ with center $x_0\in X$, that is, for every $x\in X$ the line segment $[x_0,x]=\{(1-t)x_0+tx : t\in [0,1]\}$ is contained in $X$.

Suppose $X$ has nonempty interior and that $\partial X$ is homeomorphic to $\mathbb{S}^{n-1}$.

My question is: does $X$ admit a center point in its interior?


I'm trying to generalize the statement "in $\mathbb{R}^n$, a compact convex set with nonempty interior is homeomorphic to the unit ball $\mathbb{D}^n$", which I proved using the fact that we can translate the convex so that the origin belongs to its interior and then the proof does not depend on any other point than the origin (that is a center for the convex).

Therefore, if the answer to my question is "yes", then it's true that "in $\mathbb{R}^n$, a compact star shaped set with nonempty interior and boundary homeomorphic to $\mathbb{S}^{n-1}$ is homeomorphic to $\mathbb{D}^n$.

Thanks in advance!

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I think this has only one center lying on his boundaryexample

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    $\begingroup$ Such a simple counterexample, you got it! However this set is clearly homeomorphic to $\mathbb{D}^2$, its just the case that my proof will not work as I expexted. I'm wondering if what I wrote in quotes in the final of my question is true or not.. Anyway, thanks a bunch! $\endgroup$ Sep 1, 2018 at 15:18
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    $\begingroup$ @rldias In $\mathbb{R}^2$ any compact set bounded by a (topological) circle is homeomorphic to $D^2$ (Jordan–Schoenflies theorem). $\endgroup$
    – Paul Frost
    Sep 1, 2018 at 15:38
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    $\begingroup$ @rldias I believe it is true because being star shaped is a strong requirement. However, if you want to construct a counterexample, you must be aware of the generalized Schoenflies theorem which says that a locally flat embedded sphere $S \subset \mathbb{R}^n$ (i.e. $S \approx S^{n-1}$) bounds a (topological) ball. $\endgroup$
    – Paul Frost
    Sep 1, 2018 at 15:52
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Here is a counterexample. Consider the closed 1st quadrant $Q = \{(x,y) \in \mathbb R^2 \mid x,y \ge 0\}$. Let $I_x = [0,1] \times 0$, and let $I_y = 0 \times [0,1]$.

I'm going to describe a Jordan arc $J$ with endpoints $(0,1)$ and $(1,0)$ such that $J \cup I_x \cup I_y$ is a Jordan curve bounding a compact set $X$, hence $X$ is homeomorphic to $\mathbb D^2$, and $(0,0)$ is a star point for $X$, but no interior point is a star point for $X$.

The arc $J$ is the polar coordinate graph of a function $$r = f(\theta), \quad 0 \le \theta \le \pi/2 $$ with $f(0)=f(\pi/2)=1$. There will be an infinite, strictly increasing sequence $$0 = \theta_0 < \theta_1 < \theta_2 < \cdots $$ with $\lim_{i \to +\infty} \theta_i = \pi/2$, and I'll denote $a_i$ to be the point in the plane represented by polar coordinates $(\theta_i,r(\theta_i))$. The portion of $J$ with $\theta_{i-1} \le \theta \le \theta_i$ will be denoted $J_i$, and it is simply the Euclidean line segment $$J_i = [a_{i-1},a_i] $$ Thus, $J$ is obtained by adding the point $(0,1)$ to the concatenation of the closed line segments $$J_0 J_1 J_2 \cdots $$ As $n \to +\infty$ the segments $J_n$ will decrease in length to zero and approach $(1,0)$.

The idea is to construct the $J_n$'s so that the limit of the Euclidean slope of $J_n$ is equal to $+\infty$, but this limit is acheived in a "zig-zaggy" way that prevents any interior point of $X$ from being a center point.

Here's a few details on how to achieve this result. Consider $n \ge 1$. If $n$ is odd we require that $|a_{n-1}| < |a_n|$, and hence $J_n$ has "positive polar slope", because $r$ is increasing as a function of $\theta$ along the segment $J_n$. Also, if $n$ is even we require that $|a_{n-1}| > |a_n|$, hence $J_n$ has "negative polar slope". Finally, as $n \to +\infty$ we require that the Euclidean slope of $J_n$ approaches $+\infty$; hence for odd $n$ the polar slope is approaching $+\infty$ and for even $n$ the polar slope is approaching $-\infty$.

From these conditions, for any interior point $p \in Q$ one sees that for sufficiently large odd values, the point $p$ lies on the "outside" of the segment $J_n$, hence the line segment from $p$ to any point in the interior of $J_n$ does not lie in $Q$.

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  • $\begingroup$ Well done! Now I'm wondering if the result I was trying to prove is true, since your example is still homeomorphic to $\mathbb{D}^2$.. Thank you! $\endgroup$ Sep 1, 2018 at 15:24

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