TITLE: Is the complement of the closed unit ball connected?
QUESTION [5 upvotes]: In $\mathbb{R}^2$, how can I show directly that the complement of the closed unit ball is connected?

REPLY [3 votes]: Here's a way to prove it. 
Note that the cartesian product of two connected spaces is connected.
So $D:=(1,\infty)\times\mathbb{R}$ is connected
Consider the map $f:D\to \mathbb{R}^2$ by 
$(x,y)\to (xcos(y),xsin(y))$
$f$ is clearly continuous and $f(D)$ is the compliment of the unit ball. Since connectedness is preserved under continuous maps, connectedness of $f(D)$ follows.