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A star-shaped (relative to $0$) domain $U$ is a bounded open subset of $\mathbb{R}^n$ such that for each $x \in U$, the line segment from $0$ to $x$ lies entirely in $U$.

Is there a star-shaped domain whose closure $\bar{U}$ is not homeomorphic to $B^n$(the unit $n$-ball)?

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  • $\begingroup$ your definition of a star shaped subset is highly nonstandard $\endgroup$
    – Alexander Grothendieck
    Nov 24, 2013 at 16:30
  • $\begingroup$ @ooo I apologize for my casualness. maybe I should give it another name.. $\endgroup$
    – Ezra
    Nov 24, 2013 at 16:34
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    $\begingroup$ Note the closure of a star-shaped domain is also star-shaped. $\endgroup$
    – Dan Rust
    Nov 24, 2013 at 17:03
  • $\begingroup$ @DanielRust can you elaborate on this? $\endgroup$
    – Ezra
    Nov 24, 2013 at 17:05
  • $\begingroup$ The more I think about it, the more i'm not sure the above statement is very useful for this problem. I shall leave it in any case as it is true. $\endgroup$
    – Dan Rust
    Nov 24, 2013 at 17:22

1 Answer 1

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It would seem the answer is yes. Let $U$ be the region bounded by a topologist's sine curve wrapped once around the origin in the plane. I.e. take $$U := \{re^{it}: r<2+\sin(\frac{-\pi^2}{t})\} \\ 0<t \leq 2\pi$$

and note that the topological boundary is not locally path connected.

http://tinyurl.com/org9vkl

Edit: I realized a previous definition of the curve did not result in an open set (points of $U$ on the positive real axis were not in the interior). This was fixed by requiring the curve to have minimal norm ($=1$) at $t= 2\pi$. This results in more points in the boundary than are in the image of the curve (the closure also contains an interval on the positive real axis), but the boundary still fails to be locally path connected.

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