Let be $K$ a non-empty compact, convex subset of $\mathbb{R}^n$, where $n>1$. Show that $S:=\mathbb{R}^n\setminus K$ is connected.
As $K$ is bounded there exists a real number $c\geq 0$ such that $\Vert x\Vert\leq \frac{c}{2}$ for all $x\in K$.
Of course, we can approach this problem by choosing two arbitrary points $a,b\not \in K$ and $a\neq b$ and a third point $c:=(c,c,c\dots,c)\notin K$. Then, we construct a continuous path along the $n$-many axes. Consequently, this proves path-connectedness which implies our statement. However, to prove that each of the $n$-many paths doens't contain any points of $K$ seems a bit tedious as it requires to look at several different cases.
So I was wondering if there is an easier and swifter way to prove the statement?
Suppose $r>0$ is such that $K \subset B(0,r)$. Note that $\partial B(0,r)$ is path connected and does not intersect $K$. Choose $x,y \in S$. Pick $d \neq 0$, then at most one of the rays $\{ x+td \}_{t \ge 0}$, $\{ x-td \}_{t \ge 0}$ can intersect $S$. In particular, there is a line connecting $x$ to a point in $\partial B(0,r)$. Similarly for $y$. Hence $x,y$ are connected by a path.