The complement of a compact plane set has exactly one unbounded connected component

By [L5] the set $K$ is closed and bounded, so by [L6] there are $x_0$ and $s>0$ with $K\subseteq B(x_0,s)$, and then $K\subseteq \{z:\mathrm{\mid }z\mathrm{\mid }\le R\}$ for $R=\mathrm{\mid }{x}_0\mathrm{\mid }+s>0$; if $K=\mathrm{\varnothing }$ any $R>0$ serves. By [L7] the complement $\mathbb{C}\setminus K$ is open.

Fix any $R>0$ with $K\subseteq \{z:\mathrm{\mid }z\mathrm{\mid }\le R\}$ and put $E_R=\{z:\mathrm{\mid }z\mathrm{\mid }>R\}$. Then $E_R\subseteq \mathbb{C}\setminus K$, and $E_R$ is a connected subset of $\mathbb{C}$ by [L1]; it is nonempty, since $2R\in E_R$, and unbounded, since for every real $r>0$ and every $x_0$ the number $R+\mathrm{\mid }{x}_0\mathrm{\mid }+r$ has modulus exceeding $R$ and lies outside $B(x_0,r)$, so no ball of [L6] contains $E_R$.

Let $z_0\in E_R$ and let $U_{\mathrm{\infty }}:=C(z_0)$ be its component in $\mathbb{C}\setminus K$. The set $E_R$ is a connected subset of $\mathbb{C}\setminus K$ containing $z_0$, so $E_R\subseteq U_{\mathrm{\infty }}$ by [L2] and [L3]; hence $U_{\mathrm{\infty }}$ is unbounded by step 1.2 and [L6]. This holds for every admissible $R$, which is the final clause of the statement.

Let $C$ be a component of $\mathbb{C}\setminus K$ with $C\mathrm{\ne }{U}_{\mathrm{\infty }}$. By [L3] the two are disjoint, so $C\cap E_R=\mathrm{\varnothing }$ by step 2.1, that is $C\subseteq \{z:\mathrm{\mid }z\mathrm{\mid }\le R\}\subseteq B(0,R+1)$; hence $C$ is bounded by [L6].

Steps 2.1 and 3.1 give exactly one unbounded component, namely $U_{\mathrm{\infty }}$, with every other component bounded. When $K=\mathrm{\varnothing }$ the complement is $\mathbb{C}$, which is connected by [L8], so by [L3] and [L4] it is its own single component and that component is $U_{\mathrm{\infty }}$.