Let $D$ be a bounded domain (open connected) in $ \mathbb C$ and assume that complement of $D$ is connected.Then show that $\partial D$ is connected

Solution 1:

Let us continue your approach, assuming by contradiction that $K_n$ is not connected for some $n$. Cover $K_{4n}$ with finitely many open balls of radius $1/(4n)$ and take a connected component $C$ of the union of these balls. Note that this union $B$ is not connected, since a descomposition of $K_n$ yields a decomposition of $K_{4n}$ and of $B$.

Let $E_1$ be the connected component of $\partial C$, which is a closed Jordan curve. Then $E_1$ lies entirely in $D$ or entirely in $\bar D^c$. By the Jordan curve theorem, we have a disjoint union $\Bbb{C}=E_1\cup A_1\cup A_2$, where $A_1$ and $A_2$ are open. Each of the open sets $A_1,A_2$ contains points of $\partial D$, so each contains points of $D$ and of $D^c$. Hence, if $E_1\subset D$, then $D^c$ is not connected, and if $E_1\subset \bar D^c$, then $D$ is not connected.

This contradiction shows that $\partial D$ is connected.

If you want to avoid using the Jordan curve theorem, you have to prove a weak version of it, for curves that are part of the boundary of a union of balls of the same radius.

$\bf{Theorem}$ (weak Jordan curve theorem for boundaries of union of balls): Assume you have a covering of the plane by $\varepsilon$ balls, such that two of them intersect if and only if their centers can be connected by a straight line that intersects no other ball of the covering. Take $B$ to be the union of a finite number of this balls. Then there is a connected component of $\partial B$, such that we have a disjoint union $\Bbb{R}^2= A_1\cup A_2\cup \partial B$, where $A_1$ and $A_2$ are open, and $B\subset A_1$.