# Complement of a bounded set $B$ in $\mathbb{R}^{n}$ has exactly one unbounded component.

I'm working on a problem that I think has a very intuitive result, but I'm having a hard time coming up with a rigorous proof. The problem reads

If $B$ is a bounded subset of $\mathbb{R}^{n}$ where $n\geq2$, then the complement of $B$ in $\mathbb{R}^{n}$ has exactly one unbounded component.

I naively intuit that because $B$ is bounded, the complement is of course unbounded and the complement of $B$ must be connected. I guess I think of this as making a hole in $\mathbb{R}^{n}$ which of course leaves one connected set and therefore exactly one unbounded component.

• There's no reason that the complement of $B$ has to be connected - consider the complement of the sphere for example. – Zev Chonoles Apr 18 '12 at 3:25
• Oh right. I knew that, I don't know why I said that. Should I edit? @ZevChonoles – David K. Apr 18 '12 at 3:26
• Yup. And don't worry, happens to everyone sometimes :) – Zev Chonoles Apr 18 '12 at 3:37

## 1 Answer

If $B$ is bounded then there is a ball $S$ of some radius $r$ and $B\subseteq S$. Therefore, $\neg S\subseteq \neg B$ and since $\neg S$ is connected (for $n\geq 2$), $\neg S$ lies within one component of $\neg B$. All other components of $\neg B$ (if exist) must be inside $S$ and bounded.