Proving Rn minus the unit disk is connected

Question

I'm trying to show that $\mathbb{R}^n - \{x | norm(x) \leq 1 \}$ is connected. I was thinking of trying to show that it's path connected but I got stuck.

Answer 1

Consider any two arbitrary points in $A,B\in\mathbb{R}^n - \{x\ |\ norm(x) \leq 1 \}$. Then $|A| \equiv \mbox{norm }(A) >1$ and $|B| \equiv \mbox{norm }(B) > 1$.

Choose any positive $\epsilon < \min(|A|-1,|B|-1)$ and let $A^\star = \frac{1+\epsilon}{\lvert A \rvert}A$ and $B^\star = \frac{1+\epsilon}{\lvert B \rvert}B$.

Now consider the path $P$ consisting of: the straing line $AA^\star$; followed by the great circle along the $n$-sphere of radius $1+\epsilon$ centered at the origin, passing through $A^\star$ and $B^\star$; followed by a straight line from $B^\star$ to $B$. $P$ is continuous, goes from $A$ to $B$, and is always in $\mathbb{R}^n - \{x\ |\ norm(x) \leq 1 \}$.

Since $A$ and $B$ were arbitray points in $\mathbb{R}^n - \{x\ |\ norm(x) \leq 1 \}$, this shows $\mathbb{R}^n - \{x\ |\ norm(x) \leq 1 \}$ is path connected.

Answer 2

Let $B$ be the unit ball and choose distinct $x,y \notin B$.

It is clear that there is some $R$ such that the sphere of radius $R$ does not intersect $B$.

The path $t \mapsto t x$, $t \in [1,{R \over \|x\|}]$ connects $x$ to the $R$ sphere and lies in $B^c$. Similarly for $y$. It is clear that the $R$ sphere is path connected. Hence there is a path from $x \to y$. Hence $B^c$ is path connected.