1

Question: Let $B^n \subset \mathbb{R^n}$ be open ball in the Euclidean metric. Prove that the complement of $B^n$ in $\mathbb{R^n}$ has exactly one unbounded component (components of a set are class partitions defining largest connected sets...).

This is an exercise of the book 'C. Adam - Topology'. Obviously, it does not hold for $n=1$, since $\mathbb{R}-(-a,a)$ is disconnected and is made of TWO unbounded components. But for the case of $n=2,3$ it is easy to prove that the conjecture holds (each 'side' of the open balls is homeomorphic to $\mathbb{R^n}$ and has non-empty intersection with its neighbour-side...)

How to elevate the conjecture for the case of $n\ge 4$, since it is not possible by intuition?

Thank you.

EDIT - This is not duplicate, since my question is about complement of an open ball not a bounded set in general. I read here before I wrote my question; the answer doesn't prove $\mathbb{R^n}−B^n$ is connected, which I need to prove.

J. W. Tanner
  • 60,406
  • I fail to see why your question is not a duplicate. – "... since my question is about complement of an open ball not a bounded set in general" Your question, as I understand it, is about the complement of a bounded set $B$. – Martin R May 10 '15 at 09:43
  • @Dosidis&Martin R: I edited all. –  May 10 '15 at 09:47

1 Answers1

0

Hint: Given two points in the complement of the ball, you can explicitly write down a path connecting them. This shows the complement is path connected, which shows it is connected. (And it's clearly unbounded.)

Potato
  • 40,171