Is the closure of an open bounded convex set already a ball?

Question

Does the "closure of an open bounded convex set in ${R}^n$ symmetric wrt. the origin" has to be already homeomorphic to a ball?

(My motivation is this: one version of Borsuks theorem says that if $f:B\to R^n$ is continuous and $f(x)\neq \lambda f(-x)$ for $x\in\partial B$ and $\lambda>0$, then $f(x)=0$ has a solution in $B$. If $\partial B$ is a sphere, one can easilly derive that the degree of $f/|f|: \partial B\to S^{n-1}$ is odd -- but I'm not sure if $\partial B$ can be anything else then the sphere..)

Answer 1

If you have such a set $D$ then there exists a norm $p$ on $R^n$ so that $D=\{x: p(x)\le 1\}$, i.e., $D$ is the closed unit ball with respect to $p$, you simply define $p(x)$ to be the value of $t$ such that $t^{-1}x$ belongs to the boundary of $D$ if $x\ne 0$ and $p(0)=0$ (see this wikipedia article).

Recall that all norms on $R^n$ are equivalent to each other (see the same article). Now, consider the homeomorphism $f$ of $R^n$ given by $$ f(x)= \frac{p(x)}{|x|} x, x\ne 0;\quad f(0)=0, $$ where $|x|$ is the Euclidean norm. Equivalence of norms implies that $f$ is indeed a homeomorphism at the origin. Then $f$ sends $D$ homeomorphically to the unit round ball.