$U$ $\setminus$ $K$ is an open connected set

Question

Let $U$ be an open connected subset of $\Bbb{C}$ and $K$ $\subset$ $U$ a closed ball around $p \in U$. Show that $U\setminus K$is an open connected set.

I probably have to prove it by contradiction but I don‘t know how.

Answer 1

As $K$ is closed, $U\setminus K$ is open. It remains to show that $U\setminus K$ is connected.

Wlog. $K=\overline{B_1(0)}$ is the closed unit disk and wlog. $U\subseteq B_2(0)$. The function $K\to (0,1]$, $x\mapsto d(x,\Bbb C\setminus U)$ is continuous on the compact $\partial K$, hence assumes its minimum $m>0$. Then for all $\rho$ with $1<\rho<1+m$, it follows that $\rho S^1\subset U$. Note that for open sets in $\Bbb C$, connected implies path-connected. Let $a,b\in U\setminus K$. Then there exists a path $\gamma$ in $U$ from $a$ to $b$. If that path avoids $K$, we are done. And if the path passes through $K$, pick $\rho$ with $1<\rho<\min\{|a|,|b|\}$. Then $\gamma$ will have a first and a last intersection with $\rho S^1$. Replace the part between these to points with an arc of $\rho S^1$ to obtain a path from $a$ to $b$ in $U\setminus K$.