-1

A nonempty open set of $\mathbb{R}^n$ is not homeomorphism with a nonempty open set of $\mathbb{R}^n, n\geq 3$

To prove this problem , I suppose there exists $f:\Omega \to U$ homeomorphism, with $\Omega, U$ open set of $\mathbb{R}^n,\mathbb{R}$.

Because $\Omega$ is open of $\mathbb{R}^n$ so exists $p\in\mathbb{R}^n, r>0$ such that $B_r(p)\subset \Omega$. Let $r'=\frac{r}{2}$, then $\overline{B_{\frac{r}{2}(p)}}\subset B_r(p)$.

The next of prove, I want to show that $S^2_{\frac{r}{2}}(p) \subset \overline{B_{\frac{r}{2}(p)}}$, $S^2_{\frac{r}{2}}(p)$ is shpere of $\mathbb{R}^3$ with center p radius $\frac{r}{2}$.

This implies that $S^2_{\frac{r}{2}}(p) \subset \mathbb{R}^n$. Consider $f_1:S^2_{\frac{r}{2}}(p) \to V=f\left( S^2_{\frac{r}{2}}(p)\right)\subset\mathbb{R}^2$. $f_1$ is homeomorphism. In the other hand, $S^2_{\frac{r}{2}}(p)$ homeomorphism with $S^2$ be unit shpere. So there exists $f_2:S^2\to V$ homeomorphism.

Then I think I use Theorem Borsuk- Ulam, to show conditration.

The idea of proof is right? And help me to explain why $S^2_{\frac{r}{2}}(p) \subset \overline{B_{\frac{r}{2}(p)}}$.

Sunflower
  • 47

1 Answers1

0

As there are many examples of not empty open sets being homeomorphic, the title problem is to give an example of
two not empty open sets that are not homeomorphic.

An example is the interior of the unit sphere and it's exterior.
They are not homeomorphic because:
one is simply connected, the other is not;
one is contractible, the other is not;
one has a compact closure, the other does not.

William Elliot
  • 17,598