The boundary of connected component of complement of compact set if $K$ is a compact set of $\mathbb{C}$, the connected components of $\mathbb{C}\backslash K$ are connected open sets of $\mathbb{C}$, that is, a countable quantity of domains of $\mathbb{C}$. Every component $C$ of $\mathbb{C}\backslash K$ satisfies the condition that $\partial C\subset K$ where $\partial C$ denote the boundary of $C$ .
How to proof this result .I need the proofs of such result.I will be thankful if someone can suggest me some reference book for the proof of such kind of results.
 A: Since $K$ is compact, it is closed. Therefore $\bar K:=\Bbb C\setminus K$ is open.

Proposition 1. Connected component of open sets are open.

Proof.
Let $C\subseteq\bar K$ a connected component of $\bar K$. Choose some $x\in C$. Then $x\in \bar K$ and we can choose some $\epsilon>0$ so that  $U_\epsilon(x)\subseteq \bar K$. Since $U_\epsilon(x)$ is connected, it is completely contained in the same connected component as $x$, hence $x\in U_\epsilon(x)\subseteq C$. Thus $C$ is open. $\square$

Proposition 2. $U_i,i\in I$ are pair-wise disjoint open sets in $\Bbb C$, then $I$ is at most countable.

Proof.
$\Bbb C$ is second countable, i.e. there are countably many open sets $V_j,j\in J$ so that any open set in $\Bbb C$ can be written as the union of some $V_j$. For exmaple, choose the countably many open sets $U_\epsilon(x)$ with $\epsilon\in\Bbb Q^+$ and $x\in\Bbb Q[i]=\{a+ib\mid a,b\in\Bbb Q\}$.
Now, this means that every $U_i$ is the union of some of these $V_j$. Also, no two different disjoint $U_i$ share the same $V_j$. This means that each of the $U_i$ represents a group of $V_j$, and all the groups are disjoint. But since there are only countably many $V_j$, we cannot partition them into uncountably many disjoint groups. Hence there are at most countably many $U_i$. $\square$
This proves the first part, namely, that $\bar K$ is the union of countably many connected open sets. The open sets in question are just the connected components of $\bar K$. That these are open is shown in Proposition 1. That there are at most countably many is shown in Proposition 2.

Proposition 3. If $A,B\subset \Bbb C$ are two disjoint open sets, then $\partial A\cap B=\varnothing$.

Proof.
If $x\in B$, then there is some $U_\epsilon(x)\subseteq B$. But when also $x\in\partial A$, then every such $U_\epsilon(x)$ must intersect $A$. This contradicts that $A$ and $B$ are disjoint. $\square$
This shows the second part. An open set does not contain any of its boundary points (by definition). Hence $\partial C\cap C=\varnothing$. Now, if $D\subseteq\bar K$ is another one of these open connected components of $\bar K$, then Proposition 3 shows that $\partial C\cap D=\varnothing$. Hence the complete boundary must be in $K$.
