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Let $(V, \|\cdot\|)$ be a finite dimensional Banach space, with $n:=\dim(V) \geq 2$, and consider a non-empty, closed and convex subset $A \subset V$. Suppose that there is no $(n-1)$-dimensional subspace of $V$ that contains $A$. Show that $A$ is homeomorphic to the unit ball in $V$.

I have seen similar results for compact convex subsets of the Eucilidan space $\mathbb{R}^n$, but I don't see how to prove this statement.

Arctic Char
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userr777
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1 Answers1

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This is not true. Take $V=\mathbb R^2$ and $A=[0,\infty )\times [0,\infty )$. Then $A$ is closed and convex, but it's not homeomorphic to $\mathcal B:=\{(x,y)\mid x^2+y^2\leq 1\}$.

Surb
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  • I see now, thanks! Do you know maybe what additional condition is necessary in order to have such an homeomorphism? – userr777 Aug 31 '20 at 08:38
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    @userr777 In finite dimension, the unit ball is always convex, symmetric, nonempty and bounded. See my answer (and its remark) to this question https://math.stackexchange.com/questions/1495726/norm-induced-by-convex-open-symmetric-bounded-set-in-bbb-rn – Surb Aug 31 '20 at 08:44