Is it possible that a simply connected domain and a Jordan curve's interior have same boundary but not equal? There is a bounded simply connected domain $\Omega$ whose boundary $\partial\Omega$ is a Jordan curve $\gamma$ (simple closed curve). In Jordan Curve Theorem, we have known that the Jordan curve splits the plain into two simply connected components, the bounded one called "interior". My question is whether $\Omega$ is exactly the interior of the Jordan curve $\gamma$? Is it possible that $\Omega$ and the interior have same boundary but $\Omega$ is not equal to the interior? (Using math words is that whether $\partial\Omega=\gamma$ can imply $\Omega= int(\gamma)$.)
The proposition above seems like obvious. I tried hard to prove but failed. Could anyone help me solve this tough problem? Thanks very much!!
 A: Your concern is a valid subtletly, which I glossed over in my comments. So here's a complete argument. Let $\Omega\subseteq\mathbb{R}^2$ be open, connected and bounded, such that $\partial\Omega$ is a Jordan curve (we do not need to assume that $\Omega$ is simply connected - actually, that follows from these hypotheses). The Jordan Curve Theorem tells us that $\mathbb{R}^2\setminus\partial\Omega=U\cup V$, with $U,V$ connected, open and disjoint, $\partial U=\partial V=\partial\Omega$, $U$ bounded and $V$ unbounded.
Now, $\Omega\subseteq\mathbb{R}^2\setminus\partial\Omega$ (since $\partial\Omega=\overline{\Omega}\setminus\mathrm{Int}(\Omega)=\overline{\Omega}\setminus\Omega$ is disjoint from $\Omega$ by $\Omega$ being open) is connected, so either $\Omega\subseteq U$ or $\Omega\subseteq V$. Let $A\in\{U,V\}$ be the one such that $\Omega\subseteq A$. Note that $A=\Omega\cup A\setminus\Omega$ is a disjoint union and $\Omega\subseteq A$ is open. Now, $\partial A\cap A=\emptyset$ since $A$ is open (same argument as before), but $\partial A=\partial\Omega$. This implies $A\setminus\Omega=A\setminus\overline{\Omega}$, since $\overline{\Omega}=\Omega\cup\partial\Omega$. Thus, $A\setminus\Omega\subseteq A$ is open too. The connectedness of $A$ now forces $\Omega=\emptyset$ or $A\setminus\Omega=\emptyset$. The former is excluded for trivial reasons ($\partial\emptyset=\emptyset$ would not be a Jordan curve), so $A\setminus\Omega=\emptyset$, i.e. $A\subseteq\Omega$. In total, $\Omega=A$. Since $\Omega$ is bounded, it now follows a posteriori that $\Omega=A=U$, i.e. $\Omega$ is the interior of $\partial\Omega$.
If you want a slightly more geometric picture for the above argument, we can utilize path-connectedness of $A$ instead of connectedness of $A$ (these are equivalent for open subsets of the plane). Then, you can argue that if you were to take a path in $A$ from a point in $\Omega$ to a point in $A\setminus\Omega$, this path would have to cross $\partial\Omega=\partial A$, contradicting $\partial A\cap A=\emptyset$. I leave the details as exercise.
