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I'm struggling for proving that $\mathbb{R}^{2}$ and $\mathbb{R}^{n}$ are not homeomorphic for $n \ge3.$ My approach is similar to proof of the state below.

$\mathbb{R}$ and $\mathbb{R}^{n}$ are not homeomorphic for n$\ge$2.

More specifically, suppose to the contrary that there is homeomorphism $h:\mathbb{R}^{2}\rightarrow \mathbb{R}^{n}$ . Since $\mathbb{R}^{2}-\mathbb{S}^{1}$ is NOT connected and it has separation $\{x\in \mathbb{R}^{2} \mid \left|x \right|<1\}$ and $\{x\in \mathbb{R}^{2} \mid \left| x \right| >1\}.$ I know intuitively that $\mathbb{R}^{n}-h \left( \mathbb{S}^{1}\right)$ is connected, but I cannot prove mathematically.

Thomas Andrews
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Mvaldi
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  • Do you know fundamental group? – Arctic Char Apr 16 '21 at 01:21
  • You can show that $\mathbb R^n\setminus {x}$ is simply connected for $n>2,$ and not for $n=2.$ – Thomas Andrews Apr 16 '21 at 01:23
  • @ArcticChar I know fundamental group. – Mvaldi Apr 16 '21 at 01:40
  • @ThomasAndrews I can use the fact that If a space is star convex, in other words, there is a point that all the other points can be connected by straight line segment, then it is simply connected. But If I remove {0} at $\mathbb{R}^{n}$, It is not star convex and therefore I must use different method. can you give me approach for this? – Mvaldi Apr 16 '21 at 01:44
  • I'll stick by the approach I sketched here: use dimension theory. – Henno Brandsma Apr 16 '21 at 06:29
  • I recommend also reading some closely related posts on this topic, particularly this one and it's various answers: https://math.stackexchange.com/questions/24873/elementary-proof-that-mathbbrn-is-not-homeomorphic-to-mathbbrm?rq=1 – Lee Mosher Apr 16 '21 at 12:39

2 Answers2

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Your approach works smoother if you flip the direction around.

Suppose you have a homeomorphism $f:\mathbb R^n\to \mathbb R^2$ and consider a fixed closed nonintersecting curve $\gamma$ in $\mathbb R^n$ that you can describe nicely. If you choose a nice straightforward circle, it will be easy to show that $\mathbb R^n\setminus\gamma$ is path-connected. On the other hand, $f(\gamma)$ is a Jordan curve in $\mathbb R^2$, so $\mathbb R^2\setminus f(\gamma)$ is not connected, a contradiction. (This assumes you can appeal to the Jordan curve theorem without proving it yourself).

Troposphere
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  • Oh! switching role of domain and range gives more convenient way of proof. I was struggling that image of circle is not planar curve in $\mathbb{R}^{n}$, but starting with planar curve in $\mathbb{R}^{n}$ and using the fact that continuous image of loop is also loop helps me to complete my proof. – Mvaldi Apr 16 '21 at 02:12
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Since you tagged this with algebraic topology, I will give you a very simple argument for this. Let $1\le m<n$ and suppose there was a homeomorphism $f\colon\mathbb{R}^m\to\mathbb{R}^n$. Removing any $x\in\mathbb{R}^m$ and the corresponding point $f(x)$ from their respective spaces, we have (still homeomorphic) punctured euclidean spaces. The first is homotopy equivalent to $S^{m-1}$, and the second to $S^{n-1}$. This gives us their homologies, $H_*(\mathbb{R}^m\setminus\{ x\}) \cong H_*(S^{m-1})$ and $H_*(\mathbb{R}^n\setminus\{ f(x)\})\cong H_*(S^{n-1})$.

We know that $n\ge2$, so $H_k(S^{n-1}) = 0$ if $k\neq 0, n-1$ and $H_0(S^{n-1})\cong \mathbb{Z} \cong H_{n-1}(S^{n-1})$. The same holds for $S^{m-1}$ (with $n$ replaced by $m$) if $m\ge2$, but if $m=1$, then $H_0(S^{m-1})\cong \mathbb{Z}\oplus\mathbb{Z}, H_k(S^{n-1})=0$ for $k\ge1$. Since the punctured spaces are homeomorphic to one another and homotopy equivalent to those spheres, the spheres are then homotopy equivalent to one another. This means that they have isomorphic homologies, but then $\mathbb{Z} \cong H_{n-1}(S^{n-1}) \cong H_{n-1}(S^{m-1}) = 0$, since $m\neq n$. This is an absurdity, and therefore the punctured spaces cannot be homeomorphic. Since the homeomorphism between them was obtained from a homeomorphism between the original euclidean spaces, such a homeomorphism cannot exist.

  • Ah.. I cannot use homology. – Mvaldi Apr 16 '21 at 01:34
  • Can you use homotopy? I can modify that to work for homotopy groups. The answer is probably going to be no, though. –  Apr 16 '21 at 01:37
  • I can use fundamental group and homotopy itself, but I did not learned about homotopy group. I have few tools of algebraic topology. – Mvaldi Apr 16 '21 at 01:38
  • I see. Then you can discard my ideas. –  Apr 16 '21 at 01:59
  • The original post is only about the case $m=2$, i.e. about the proof that $\mathbb R^2$ is not homeomorphic to $\mathbb R^n$ for $n \ge 3$. So one needs only the concept of "fundamental group", otherwise known as "first homotopy group". – Lee Mosher Apr 16 '21 at 11:40
  • @LeeMosher oh right! I forgot that completely. Good point. –  Apr 16 '21 at 12:26