Connected-ness of the boundary of convex sets in $\mathbb R^n$ , $n>1$ , under additional assumptions of the convex set being compact or bounded Is the boundary of every compact convex set in $\mathbb R^n$ , ($n>1 $ ) connected ? is it path connected ? What if we assume only that the convex set is bounded , is the boundary connected ( and path-connected ) even then ? 
 A: Let $X$ be a nonempty bounded convex subset of $\mathbb R^n$.
If $X$ has no interior points, then $\partial X=\overline X$ is again convex and hence path connected.
If $X$ does have interior points we may assume wlog. (after translation) that $0$ is an interior point. Consider the map $$\begin{align}f\colon \partial X&\to S^{n-1}\\ x&\mapsto \frac x{|x|}\end{align}$$ 
Since $0\in X^\circ$,  $f$ is defined and continuous on all of $\partial X$.
For each $p\in S^{n-1}$, the set $\{\,t\in[0,\infty)\mid tp\in X\,\}$ is convex and bounded and contains all $t\approx 0$, hence is of the form $[0,a)$ or $[0,a]$ with $a>0$. 
Let $r>0$ with $B(0,r)\subseteq X$. If $0\le t<a'<a$, then $a'p\in X$ implies $B(tp,\frac{a'-t}{a'}r)\subseteq X$ because the point $tp+\frac{a'-t}{a'}v$ with $|v|<r$ is a convex combination of $a'p$ and $v$. And if $a<a''<t$ then $a''\notin X$ implies $B(tp,\frac{t-a''}{a''}r)\cap X=\emptyset$ because now $a''p$ is a convex combination of $tp+ \frac{t-a''}{a''}v$ and $ -v$. Hence for $0\le t<a$ we have $tp\in X^\circ$, for $t>a$ we have $tp\notin \overline X$, whereas each neighbourhood of $ap$ clearly intersects both $X$ and its complement. We conclude that the ray $[0,\infty)\cdot p$ intersects $\partial X$ in exactly one point. This makes $f$ a bijection.
As a continuos bijection between compacts, $f$ is a homeomorphism.
Especially, we conclude that $\partial X$ is path connected.
