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This is related to a previous question : $A \subset \mathbb{R^n}$, $n \geq 2$, such that $A$ is homeomorphic to $\mathbb{R^n} \setminus A$, and for general $X$

For 2 topological spaces $(X,T_X)$ and $(Y,T_Y)$, I write $X \simeq Y$ if $X$ and $Y$ are homeomorphic. If $A \subset X$, I always endow $A$ with the subspace topology.

Let $n \geq 2$.

I wonder if it is true that :

(1) $\exists A \subset \mathbb{R^n}$ connected such that $A \simeq \mathbb{R}^n \setminus A$.

For $n=1$, I know this is not true for reasons of connectedness.

What is certain is that $A$ can't be an open set (as a consequence of the invariance of domain and the connectedness of $\mathbb{R^n}$), nor a compact set (as $\mathbb{R^n}$ would be the union of two compact sets, so compact), nor countable or co-countable (as $A$ and $\mathbb{R^n} \setminus A$ would have different cardinalities).

My intuition is that it is false. However, if it is true, then I wonder if (still for $n \geq 2$) :

(2) $\exists A \subset \mathbb{R^n}$ simply connected such that $A \simeq \mathbb{R}^n \setminus A$.

(3) $\exists A \subset \mathbb{R^n}$ contractible such that $A \simeq \mathbb{R}^n \setminus A$.

(4) $\exists A \subset \mathbb{R^n}$ star domain such that $A \simeq \mathbb{R}^n \setminus A$.

(5) $\exists A \subset \mathbb{R^n}$ convex such that $A \simeq \mathbb{R}^n \setminus A$.

Thanks.

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    $\begingroup$ Everything of these conditions can hold. Think of the halfplane were half of the boundary is missing. $\endgroup$ – ctst Oct 10 '16 at 11:51
  • $\begingroup$ Do you mean $A = (]0,+\infty[ \times \mathbb{R}) \cup (\{0\} \times ]0, +\infty[)$ ? What is the homeomorphism between $A$ and $\mathbb{R}^2 \setminus A$ ? $\endgroup$ – Vandrin Oct 10 '16 at 11:58
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    $\begingroup$ Close, here you might have a problem with (0,0). Try more with several halfopen intervals at the boundary, s.t. $A^c$ is just $A$ flipped. $\endgroup$ – ctst Oct 10 '16 at 12:12
  • $\begingroup$ Ok. $A = (]0,+\infty[ \times \mathbb{R}) \cup \bigcup_{z\in\mathbb{Z}} (\{ 0 \} \times [2z, 2z+1[)$. This is indeed a star domain, but it is not convex. So how about (5) ? $\endgroup$ – Vandrin Oct 10 '16 at 12:20
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    $\begingroup$ You can dent your boundary in and so get a convex set $A$ (beware that $A^c$ is not convex). Since you can do so continous you don't change $A$ and $A^c$ topologicaly. $\endgroup$ – ctst Oct 10 '16 at 12:35

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