An open subset $A \subseteq \mathbb{R}^{2}$ is not homeomorphic to an open subset of $\mathbb{R^{n}}$ for $n\geq 3$.

This question is from my General Topology class. We have seen up to part of algebraic topology.

My attempt:

Suppose a homeomorphism $f \colon X \to A$, where $X\subseteq \mathbb{R}^{n}$ is an open set. Then we can restrict this homeomorphism to an open ball contained in $X$ and, using some more homeomorphisms, suppose $X = \mathbb{R}^{n}$. Therefore we can also suppose $A \subseteq \mathbb{R}^{2}$ is an open simply connected set. Now, because of $\mathbb{R}^{n}-\{0\}$ is simply connected whenever $n\geq 3$, $f(\mathbb{R}^{n}-\{0\}) = A - \{f(0)\}$ is simply connected.

We want to deny that $A - \{f(0)\}$ is simply connected. In order to do that pick a $\delta > 0$ such that $B_{f(0)}[\delta] \subset A$. Then define the following function:

$$ \alpha \colon \mathbb{S}^{1} \to A - \{f(0)\}; ~ \alpha(z) = z\delta + f(0)$$

A set $X \subseteq \mathbb{R}^{n}$ is simply connected when every continuous function $p \colon \mathbb{S}^{1}\to X$ has a continuous extension $\overline{p}\colon \mathbb{B}^{2} \to X$.

I was not able to show that this function has not continuous extension to $\mathbb{B}^{2}$. Is it true? Any help would be appreciated but I would prefer help in my attempt.

Thank you.

  • $\begingroup$ I don't think you can reduce to the case of $A$ being open simply connected. If $A$ is an open annulus at the beginning, there is no way to reduce it to your case $\endgroup$ – Randall Jun 21 '18 at 2:15
  • $\begingroup$ It is not the same set, indeed. But this does not matter if we finnd a contradiction, no? $\endgroup$ – Oddone Jun 21 '18 at 2:20
  • $\begingroup$ Can you use the fact that $S^1$ is not simply connected? $\endgroup$ – arkeet Jun 21 '18 at 2:29
  • $\begingroup$ Yes I do, @arkeet $\endgroup$ – Oddone Jun 21 '18 at 2:32
  • $\begingroup$ I don't think you can assume that the open set in $\Bbb R^n$ is homeomorphic to $\Bbb R^n$. It could be an open shell and that's definitively not homeomorphic to $\Bbb R^n$. $\endgroup$ – Cameron Williams Jun 21 '18 at 2:58

In order to reduce the problem to a problem about simply-connectedness, the thing you need is that the local property "every point $x \in X$ has a neighbourhood $U$ such that $U \setminus \{x\}$ is simply connected" is a homeomorphism invariant of $X$. This property holds for any open set in $\mathbb{R}^n$ for $n \ge 3$ (since any point has a neighbourhood homeomorphic to $\mathbb{R}^n$), but not for $A \subseteq \mathbb{R}^2$, which can be proven as follows:

We want to show that if $U \subseteq \mathbb{R}^2$ is open and $a \in U$, then $U - \{a\}$ is not simply connected. There is $\delta > 0$ such that $U$ contains a closed ball $B_a(\delta)$, so in particular we have a map $$\alpha \colon \mathbb{S}^1 \to U - \{a\},\; \alpha(z) = \delta z + a.$$ The key is that this map has a left inverse $$\rho \colon U - \{a\} \to \mathbb{S}^1,\; \rho(x) = \frac{x-a}{|x-a|}$$ i.e. $\rho \circ \alpha = \operatorname{id}_{S^1}$. If $U - \{a\}$ were simply connected, then the map $\alpha$ would extend to $\bar\alpha \colon \mathbb{B}^2 \to U - \{a\}$. But then $$\rho \circ \bar\alpha \colon \mathbb{B}^2 \to \mathbb{S}^1$$ is a retraction, which is impossible since $\mathbb{S}^1$ is not simply connected.


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