# $\Bbb{C}\setminus B(o,r)$ is connected

There is a result that is often use in Complexe Analysis the fact that given an open disc $D(0,r)$ for $r>0$, $\Bbb{C}\setminus D(0,r)$ is connected.

It has been some time since this "trivial" result holds me in check. Never mind I was thinking to find two connected sets $A,B$ such that their closures $\overline{A},\overline{B}$ have points in commun and their unions is $\Bbb{C}\setminus D(0,r)$. I was thinking something like $A=\{(x,y)\in\Bbb{R^2}:\vert x\vert^2+y^2>R, y>0\}$ and similarly $B=\{(x,y)\in\Bbb{R^2}:\vert x\vert^2+y^2>R, y<0\}.$ But not sure how can I prove that $A$ and $B$ are connected.

I tried also to prove it's path connected, but not sure how can I formalize it. It remains to formalize the case that the straight line passing through two points in $\Bbb{C}\setminus D(0,r)$ passe through $D(0,r).$

I admit it it's obviously true but I cannot formalize it.

Iif $|x|\ge r$ and $|y|\ge r$ and wlog $|x|\le |y|$, then $|x|\cdot S^1\cup |y|\cdot S^1\cup [|x|,|y|]$ is path connected in an obvious fashien, is a subset of $\Bbb C\setminus B(0,r)$ and contains $x$ and $y$.
To prove that $\Bbb{C}\setminus D(0,r)$ is path connected, you have You have to discuss the two cases:
• In fact I juste need to prove that I can find a point $z$ that is not in $D(0,r)$ such that the straight line $x->z$ and $z->y$ never passes in $D(0,r)$ – Alex Apr 27 '17 at 11:55
Given two points $(r_1, \theta_1)$ and $(r_2, \theta_2)$ in polar coordinates (with $r_1, r_2 > r)$, there exists a straight line (radial) path from $(r_1, \theta_1)$ to $(r_2, \theta_1)$ and an arc from $(r_2, \theta_1)$ to $(r_2, \theta_2)$. Composing these two maps gives a path from one point to the other. Since a path-connected space is always connected, done.