Is the complement of the closed unit ball connected? In $\mathbb{R}^2$, how can I show directly that the complement of the closed unit ball is connected?
 A: In case you are looking at $\mathbb{R}^n$ with the standard Euclidean metric, then $\mathbb{R}^n \setminus B(0,1)$ is path connected if and only if $n = 1$, where $B(0, 1)$ denotes the closed unit ball. Now, for $n=1$ you have already the answer. For $n \geq 2$, you can take two points $a, b \in \mathbb{R}^n \setminus B(0,1)$. Thus
$$
a' = \frac{2a}{\lVert a \rVert} \quad \text{and} \quad b' = \frac{2b}{\lVert b \rVert}
$$
ly on the sphere $S^n$ with radius $2$ and center the origin. Clearly, $a$ and $a'$ can be connect via a straight line and the same for $b$ and $b'$. Moreover, $a'$ and $b'$ ly on an $(n-1)$-dimensional sphere $S^{n-1}$ with radius $2$ which lies on the orginial sphere $S^n$. By induction on $n$, there exists now a path between $a'$ and $b'$ in $S^{n-1}$ (where you have to prove that $S^1$ is connected (which can be done explicitely with $\sin$ and $\cos$)).
A: 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.
A: let $B_{1}$ be the closed unit ball in $X$. 
Note that the set $\mathbb{R}^n\setminus B_{1}$ is open since $B_{1}$ is closed. There is no way to partition into two disjoint open sets $U$ and $V$ where $U\cup V=\mathbb{R}^n\setminus B_{1}$. 
in $\mathbb{R}$ this would be equivalent to showing the set $\mathbb{R}\setminus (-1,-1)$ is connected. This is equivalent to $(-\infty,-1-\epsilon)\cup (1+\epsilon, \infty)$. Recall that intervals, both closed and open are connected. For $\mathbb{R}^n$, we can consider the Cartesian product $\mathbb{R}^n\setminus (-1,1)^{n}$. Cartesian products (finite) of connected sets are connected, which proves $\mathbb{R}^n\setminus B_{1}$ is connected. 
