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.
Best Answer
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).