If the boundary of a convex set in $\mathbb R^n$ ( where $n>1$) is connected , is it necessarily also path-connected ?
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The case of a convex and bounded set is straightforward. In general, you could use the Minkowski seminorm $p$ given by the set. Perhaps factor off the subspace where $p$ is null and get a norm. Then it's more or less a ball. – orangeskid May 25 '15 at 09:05
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@orangeskid : could you please say it in detail ? – May 25 '15 at 14:36
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Assume $X\subseteq \mathbb R^n$ is a nonempty convex set. If $X$ has no interior points then $\partial X=\overline X$ is convex and hence path-connected. Hence assume wlog. that $0\in X^\circ$. For $a\in\partial X$ let $P(a)$ be the set of points $p\in \partial X$ that can be reached from $p$ by a path in $\partial X$.
Claim. $P(a)$ is a clopen subset of $\partial X$.
Proof. For $r>0$ let $X_r=X\cap B(0,r)$. Then $X_r$ is a nonempty bounded convex set, hence there exists a homeomorphism $f\colon \partial X_r\to S^{n-1}$ . For $r>|a|$, the image of the open set $B(a,r-|a|)\cap \partial X_r=B(a,r-|a|)\cap \partial X$ under $f$ the contains an open $\let\epsilon\varepsilon \epsilon$-ball around $f(a)$, i.e., $B(f(a),\epsilon)\cap S^{n-1}$. As this is path-connected, we conclude that $P(a)$ contains a $\partial X$-open neighbourhood of $a$. Since for $a,b\in\partial X$ we have either $P(a)=P(b)$ or $P(a)\cap P(b)=\emptyset$, we conclude that $P(a)$ is open and closed. $_\square$
Corollary. If $\partial X$ is connected, then it is also path-connected.
Hagen von Eitzen
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