0

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

What is certain is that $A$ can't be 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).

It can be shown using the invariance of domain and the connectedness of $\mathbb{R}^n$ that the following is false :

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

However, (2) doesn't of course solve (1).

In a preceding question ($A \subset \mathbb{R}$ such that $A$ is homeomorphic to $\mathbb{R} \setminus A$), I asked for the case $n=1$. The answer was that (1) is true and (2) is false.

Likewise, what are some spaces $X$ (or maybe a characterization of all of them ?) in which :

(3) $\exists A \subset X$ such that $A \simeq X \setminus A$. ?

For example, $X = \mathbb{R}$ is fine, $X$ discrete and of finite and even cardinality is fine, and $X$ of finite and odd cardinality is never fine.

or,

(4) $\exists A \subset X$ open such that $A \simeq X \setminus A$.

For example, $X$ discrete and of finite and even cardinality is fine, and $X$ of finite and odd cardinality is never fine, $X = \mathbb{R}^n$, $n \geq 0$ is never fine.

Thanks.

Community
  • 1
Jeyrome Sapin
  • 853

1 Answers1

2

For (1) the answer is yes, by a trivial modification of the argument I gave in my answer to your previous question: instead of half-open intervals, do half-open slices. E.g. for $n=2$, look at $$A=\bigcup_{z\in\mathbb{Z}}[\mathbb{R}\times [2z, 2z+1)],$$ and more generally for arbitrary $n$ set $$A=\bigcup_{z\in\mathbb{Z}}[\mathbb{R}^{n-1}\times [2z, 2z+1)]$$ (ignoring the very small abuse of notation).

Noah Schweber
  • 245,398