Is the closure of every bounded convex set , with non-empty interior , in $\mathbb R^n (n>1)$ homeomorphic to a closed ball (by closed ball I mean $B[a,r]:=\{x \in \mathbb R^n : d(x,a)\le r\}$ , where $a \in \mathbb R^n$ and $r>0$ ) ?
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What do you think? – Amitai Yuval Aug 02 '15 at 12:44
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@AmitaiYuval : I think yes ... – Aug 02 '15 at 12:51
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The answer is positive. Note that the closure is a (compact) convex set, and in particularly star-shaped.
(We call a set $U⊆\mathbb{R}^n$ star-shaped if there is a point $x∈U$ such that for all lines $L$ through $x$, $L∩U$ is a connected open line segment).
Now we use the star-shaped structure. My answer is essentially taken from here.
Translating $U$ if necessary we may assume that $x=0$. Now scale each ray from $0$ to the boundary of $U$ appropriately (that is, map it linearly on the unit interval $[0,1]$) , obtaining a homeomorphism from $U$ onto the closed unit ball.
Asaf Shachar
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Simply scaling the rays does not give an homeomorphism in general. For example let $B$ be the open unit ball of $\mathbb{R}^2$, and let $U=B\cup\left{(2x,2y):(x,y)\in B, y<0\right}$. The distance from $0$ to the boundary is not continuous. – Luiz Cordeiro Aug 06 '15 at 01:04
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Of course, the set $U$ in my previous comment is star-shaped but not convex. In the convex case the procedure described here should work without too many problems. – Luiz Cordeiro Aug 06 '15 at 02:16