Constructing a Set with Connected Interior

Question

Suppose that $K\subset\mathbb C$ is a compact set with non-empty interior and suppose that $a\in\operatorname{int} K$. I want to construct a set $M$ with the following properties:

1. $M\subseteq K$;
2. $M$ is compact;
3. $a\in\operatorname{int} M$;
4. $\operatorname{int} M$ is connected;
5. $\partial M\subseteq \partial K$.

An obvious candidate for $M$ is the connected component of $a$ in $K$, which is closed (and hence compact), but there are easy examples for cases in which the interior of the connected component is not connected. Should I perhaps consider the connected component of $a$ in the subspace $\operatorname{int} K$? Any hints will be much appreciated.

EDIT: $x$ in the comments corresponds to $a$ here and in the answer.

Answer 1

Let $C_a$ be the connected component of $a$ in the relative topology generated by $\operatorname{int} K$; this set is relatively closed, meaning that $C_a=C\cap\operatorname{int} K$ for some absolutely closed set $C\subseteq \mathbb C$. Let $M\equiv \operatorname{cl} C_a$. It is clear that $M\subseteq K$ and $M$ is compact. Moreover, $C_a$ is open in the absolute topology. (By absolute topology, I mean the Euclidean topology on $\mathbb C$, to differentiate it from relative topologies.) To see this, let $y\in C_a\subseteq\operatorname{int} K$. Then, since $\operatorname{int} K$ is open and not empty, there exists an open circle $B_y\subseteq\operatorname{int} K$ such that $y\in B_y$. Since $B_y$ is obviously connected in the relative topology generated by $\operatorname{int} K$ and $y\in C_a\cap B_y$, it follows that $C_a\cup B_y$ is connected in the relative topology as well. By definition of the connected component, it follows that $C_a\cup B_y\subseteq C_a$, which implies that $y\in B_y\subseteq C_a$. Hence, $C_a$ is open in the absolute topology. It follows easily also that $C_a$ is connected in the absolute topology. Next, I will show that $C_a=\operatorname{int} M$. That $C_a\subseteq \operatorname{int} M$ is clear. Moreover, $C_a=C\cap\operatorname{int} K$, from which it follows that \begin{align*}{M=\operatorname{cl} C_a= \operatorname{cl}\left(C\cap \operatorname{int} K\right)\subseteq C\cap\operatorname{cl}(\operatorname{int} K)\subseteq C\cap K.\quad(*)}\end{align*} Therefore, \begin{align}{\operatorname{int} M\subseteq\operatorname{int}(C\cap K)=\operatorname{int} C\cap\operatorname{int} K\subseteq C\cap\operatorname{int} K=C_a.}\end{align} In particular, $\operatorname{int} M=C_a$ is connected in the absolute topology and $a\in C_a=\operatorname{int} M$. Finally, I will show that $\partial M\subseteq \partial K$. If $y\in \partial M$, then $y\in M\subseteq K$ but $y\notin \operatorname{int} M=C_a=C\cap\operatorname{int} K$. The formula (*) implies that $y\in C$, so it follows that $y\notin \operatorname{int} K$. Hence, $y\in K\setminus \operatorname{int} K=\partial K$.