Two subsets with homeomorphic parts of boundaries Let $X$ be a topological space and let $U$ be an open subset. Denote by $\simeq$ "being homeomorphic". Assume that $A, B\subseteq X$ are such that
$$U\subseteq A\subseteq\overline{U}$$
$$U\subseteq B\subseteq\overline{U}$$
$$A\backslash U \simeq B\backslash U$$
So $A\backslash U$ is not really a boundary of $A$ but a piece of boundary that belongs to $A$.
Question: Does it follow that $A\simeq B$?
I'm pretty sure that there will be some pathological $X$ that doesn't satisfy it. So here are some variants I am interested in as well:


*

*Both $A\backslash U$ and $B\backslash U$ are closed in $X$

*$X$ is Hausdorff or metrizable or even finite dimensional real space

*$U$ is connected

*$\overline{U}$ is compact

 A: It does not follow. A simple example is
$$U = \{ z \in \mathbb{C} : 0 < \lvert z\rvert < 1\},\quad A = U \cup \{0\}, \quad B = U \cup \{1\}.$$
An example where $\partial U$ is a smooth manifold: Let $U = \{ z : \lvert z\rvert < 1\}$ and
\begin{align}
A &= U \cup \bigl\{e^{i\varphi} : \varphi \in [-1,1] \cup \{\pi/2\} \cup [\pi - 1, \pi + 1] \cup \{3\pi/2\}\bigr\}, \\
B &= U \cup \bigl\{e^{i\varphi} : \varphi \in [-1,1] \cup \{3/2, 2\} \cup [\pi - 1, \pi + 1]\bigr\}.
\end{align}
A putative homeomorphism between $A$ and $B$ would induce a homeomorphism of $U$ and thus it is impossible that it switches the order of the parts on $\partial U$, but in $B$ the two singleton components are neighbours, while in $A$ they aren't.
In both examples, $X = \mathbb{C}$ is an extremely nice topological space (a Riemann surface), $U$ is connected with compact closure, and $A\setminus U$ resp. $B\setminus U$ are also compact.
A: The answer is no. Let $X=\mathbb{R}^2,$ $U=V\cup W$ $V=(0,1)\times (0,1) ,W=(-1,0)\times (-1,0), K=\left\{\left( 0,\frac{1}{2} \right)\right\}, L=\{ (0,t): 0<t<1 \wedge t\in \mathbb{Q}\}$. Then take $A=V\cup L\cup W, B=V\cup K\cup W.$
