something about convex set

Question

Let $M$ be a convex subset in $\mathbb R^n$ and $\partial M=\emptyset$. Then $M=\emptyset$ or $\mathbb R^n$.

This can be deduced by

Theorem Let $M\subset X$ with $\partial M=\emptyset$ and $X$ is a connected topological space. Then $M=\emptyset$ or $X$.

Let $M$ be a compact convex subset in $\mathbb R^n$. Then there is a homeomorphism $f:(M,\partial M)\rightarrow(\mathbb D^n,\mathbb S^{n-1})$.

There is a proof in Lee's An Introduction on Topological Manifold.

My question arises from two facts above. Now I am considering the noncompact case. That is to say, assume that $M$ is a noncompact convex set in $\mathbb R^n$ with boundary. I guess in this situation, there is a homeomorphim $f:(M,\partial M)\rightarrow(\mathbb H^n,\mathbb R^{n-1})$. So can someone help me prove or disprove my guess? Thank you in advance.

$M=[0,1] \times \mathbb{R}^{n-1}$ is convex but $\partial M$ is not connected. — Seirios, Apr 25 '14 at 04:58
May be of interest: http://math.stackexchange.com/questions/331034/closed-convex-sets-of-mathbbr2 — Seirios, Apr 25 '14 at 04:59
@Seirios Thank you. So is there any classification theory in higher dimension? — gaoxinge, Apr 25 '14 at 05:02

score 1 · Accepted Answer · edited Jun 12 '20 at 10:38

I think your question is answered in Berger's book, Geometry I.

Proposition 11.3.1: Let $X$ be a $d$-dimensional affine space and $A$ a $d$-dimensional convex subset of $X$. Then $\overset{\circ}{A}$ is homeomorphic to $\mathbb{R}^d$.

Proposition 11.3.8: Let $A$ be a convex subset of $X$ such that $\dim A = \dim X =d$ and $\mathrm{Fr}(A) \neq \emptyset$. Then $\mathrm{Fr}(A)$ is homeomorphic to $\mathbb{R}^{d-1}$ or $S^{d-r-1} \times \mathbb{R}^r$ ($0 \leq r \leq d-1$).

something about convex set

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