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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.

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  • $\begingroup$ Yes, I think your final suggestion is the best idea. Look at the closure of the connected component of $x$ in $\text{int} K$. It obviously satisfies one through four, and you only need to check the last bit. $\endgroup$ – user98602 Apr 2 '14 at 0:45
  • $\begingroup$ @Mike The headache is that the connected component of $x$ in $\operatorname{int} K$ may not be closed (it's closed in the relative topology generated by $\operatorname{int} K$, but not necessarily in the absolute topology on $\mathbb C$). If I consider its closure, I can see 1 to 3, but 4 is not as obvious to me. $\endgroup$ – triple_sec Apr 2 '14 at 0:50
  • $\begingroup$ You're right! It's not as obvious as I thought. I still think it's correct, though. $\endgroup$ – user98602 Apr 2 '14 at 0:57
  • $\begingroup$ @Mike I got it. I do think that the (absolute) closure of the connected component of $x$ in $\operatorname{int}K$ does work. Thank you for your help! $\endgroup$ – triple_sec Apr 2 '14 at 3:00
  • $\begingroup$ You should write up your solution as an answer to your own question and accept it. For one, it takes this question off the unanswered list, and for two, you might get upvotes. $\endgroup$ – user98602 Apr 2 '14 at 3:51
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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, $y\in K\setminus \operatorname{int} K=\partial K$.

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