Proving that $\mathbb{R}^n \setminus K_r(0)$, $r>0$ is connected

Question

$\mathbb{R}^p \setminus K_r(0)$, $r>0$.

How can I show that $\mathbb{R}^p \setminus K_r(0)$ is connected, that is, it cannot be divided into two disjoint nonempty open sets? I tried assuming that it isn't connected and find a contradiction but I can't get any further.

$K_r(0)$ is the closed circle around $0$ with radius $r$.

I know that I can also show path connected instead.

Answer 1

For $n>2$ the $n-1$-sphere $S^{n-1}$ is connected and the map $$R^n\backslash K_r(0)\ \longrightarrow\ S^{n-1}:\ x\ \longmapsto\ \frac{x}{||x||},$$ is continuous and surjective.

Answer 2

It is actually easier to prove directly that your set is path-connected and then to deduce that it is connected.

In order to prove that it is path-connected, take two points $a$ and $b$ from your set. Also, take a vector $v$ with norm $1$ which is orthgonal to the line passing through $a$ and $b$. Finally, considear a polygonal path as follows:

• At first, it goes (in a straight line) from $a$ to $a+2rv$;
• then, it goes (in a straight line) from $a+2rv$ to $b+2rv$;
• finally, it goes (still in a straight line) from $b+2rv$ to $b$.