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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.

I wonder if it is true that :

$\exists A \subset \mathbb{R}$ such that $A \simeq \mathbb{R} \setminus A$.

It doesn't hold for connected sets (i.e intervals, either because $\mathbb{R} \setminus A$ would not be connected or one of $A$ or $\mathbb{R} \setminus A$ would not be connected after removing a point whereas the other one may still be), nor for compact sets (as $\mathbb{R}$ would be the union of two compact sets, so compact), nor for countable or co-countable sets (as $A$ and $\mathbb{R} \setminus A$ would have different cardinalities).

If this is true, I wonder if moreover :

$\exists A \subset \mathbb{R}$ open such that $A \simeq \mathbb{R} \setminus A$.

Thanks.

Jeyrome Sapin
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2 Answers2

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For your first question, the answer is ye: consider $$\bigcup_{z\in\mathbb{Z}} [2z, 2z+1).$$

For your second question, the answer is no. HINT: Show that if $A$ is open and $\mathbb{R}\setminus A\cong A$, then $(a)$ every connected component of $A$ is homeomorphic to $(0, 1)$, but $(b)$ no connected component of $\mathbb{R}\setminus A$ is.

Noah Schweber
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Noah answered the first question.

For the second question, the answer is no.

Because if $A$ were homeomorphic to $\mathbb{R} \backslash A$ with homeomorphism being $f$, it would follow that $A \to \mathbb{R} \backslash A \hookrightarrow \mathbb{R}$ is an open map (invariance of domain), hence $\mathbb{R} \backslash A$ would be open in $\mathbb{R}$. Hence, it should be $\mathbb{R}$ (since it is also closed, being the complement of an open set, and clearly cannot be the empty set). It follows that $A$ is empty, an absurd.

Aloizio Macedo
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