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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?

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  • $\begingroup$ I wonder if something is missing from the statement. What if $n=1$, and $K=\{0,1\}$. Then $K$ is non-empty and compact, but $\mathbb R\setminus K=(-\infty,0)\cup(0,1)\cup(1,\infty)$ is not connected. Perhaps $K$ must also be convex? $\endgroup$
    – Plutoro
    Commented Mar 6 at 17:41
  • $\begingroup$ @Plutoro, yes you are right, it should be $n>1$. $\endgroup$
    – Philipp
    Commented Mar 6 at 17:47
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    $\begingroup$ Let $n=2$ and $K$ be a circle... contradiction. In fact if OP's statement were true, every winding number for every closed curve would have to be 0. $\endgroup$
    – user8675309
    Commented Mar 6 at 17:52
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    $\begingroup$ @user8675309, no there is nothing else missing. I guess the problem is simply not well stated. As Plutoro suggested, let's assume that $K$ is convex! $\endgroup$
    – Philipp
    Commented Mar 6 at 17:58
  • $\begingroup$ Wlog $0\in K$ (otherwise $K=\emptyset$ or we can just translate). Pick $x\in K^c$, then $\lambda x\in K^c$ for all $\lambda\geq 1$. Let $R=2 diam(K)$. To connect $x,y\in K^c$ first walk along a straight line until $(R/\vert x\vert)x$, rotate to $(R/\vert y\vert)y$ and walk down to $y$. $\endgroup$
    – Severin Schraven
    Commented Mar 6 at 18:08

1 Answer 1

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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.

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    $\begingroup$ I think one of the rays should be $\{ x-td\}_{t\geq 0}$. $\endgroup$
    – Severin Schraven
    Commented Mar 6 at 21:43
  • $\begingroup$ @SeverinSchraven thanks for catching that! $\endgroup$
    – copper.hat
    Commented Mar 6 at 21:53
  • $\begingroup$ How do you show that the rays /the lines do not contain elements of $K$? $\endgroup$
    – Philipp
    Commented Mar 7 at 0:59
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    $\begingroup$ Assume that $x+t_0 d, x-t_1 d \in K$. Then the entire line segment between those two elements is in $K$ (by convexity) and hence $x\in K$ (as it lives in said line segment) which is a contradiction. $\endgroup$
    – Severin Schraven
    Commented Mar 7 at 1:53

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