Prove by any method you know that:
\(\mathbb R\) is not homeomorphic to \(\mathbb R^2\).
\(\mathbb R^2\) is not homeomorphic to \(\mathbb R^3\).
Solution
1. If \(h:X\to Y\) is a homeomorphism and \(x\in X\), then\[ h|_{X\setminus\{x\}}:X\setminus\{x\}\longrightarrow Y\setminus\{h(x)\} \]is a homeomorphism.
The restriction is bijective because \(h\) is bijective, and its inverse is the restriction of \(h^{-1}\) to \(Y\setminus\{h(x)\}\). Both restrictions are continuous.
2. The punctured line is disconnected, whereas the punctured plane is path connected.
For any \(a\in\RR\), \[ \RR\setminus\{a\}=(-\infty,a)\sqcup(a,\infty), \] a separation into two nonempty open subsets. Hence \(\RR\setminus\{a\}\) is disconnected.
For any \(b\in\RR^2\), translation by \(-b\) identifies \(\RR^2\setminus\{b\}\) with \(\RR^2\setminus\{0\}\). The latter is path connected: if \(x,y\ne0\), move radially from \(x\) to \(x/\|x\|\), follow an arc of the unit circle from \(x/\|x\|\) to \(y/\|y\|\), and then move radially to \(y\). All three pieces avoid the origin, so their concatenation is a path in \(\RR^2\setminus\{0\}\) from \(x\) to \(y\).
3. Therefore \(\RR\) is not homeomorphic to \(\RR^2\).
Suppose \(h:\RR\to\RR^2\) were a homeomorphism and choose \(a\in\RR\). By 1, \[ \RR\setminus\{a\}\cong\RR^2\setminus\{h(a)\}. \] The left side is disconnected by 2, while the right side is path connected and hence connected. Connectedness is preserved by homeomorphisms, a contradiction.
4. For \(n\ge2\), the punctured Euclidean space \(\RR^n\setminus\{0\}\) strongly deformation retracts onto \(S^{n-1}\).
Define \[ H:(\RR^n\setminus\{0\})\times[0,1]\longrightarrow\RR^n\setminus\{0\} \] by \[ H(x,t)=\left((1-t)+\frac{t}{\|x\|}\right)x. \] The scalar multiplying \(x\) is positive, so \(H(x,t)\ne0\). Moreover, \[ H(x,0)=x, \qquad H(x,1)=\frac{x}{\|x\|}\in S^{n-1}. \] If \(x\in S^{n-1}\), then \(\|x\|=1\) and \[ H(x,t)=x \] for all \(t\). Thus \(H\) is a strong deformation retraction onto \(S^{n-1}\).
5. The punctured plane has fundamental group \(\ZZ\), while punctured three-space has trivial fundamental group.
By 4 and invariance of the fundamental group under strong deformation retraction, \[ \pi_1(\RR^2\setminus\{0\})\cong\pi_1(S^1)\cong\ZZ \] and \[ \pi_1(\RR^3\setminus\{0\})\cong\pi_1(S^2)=0. \] Here \(\pi_1(S^1)\cong\ZZ\) is the standard computation from the universal covering map \(\RR\to S^1\), \(t\mapsto e^{2\pi i t}\). Also \(\pi_1(S^2)=0\) follows, for example, from Seifert–van Kampen applied to the cover of \(S^2\) by the complements of the north and south poles: both members are homeomorphic to \(\RR^2\) and hence simply connected, so their union has trivial fundamental group.
6. Therefore \(\RR^2\) is not homeomorphic to \(\RR^3\).
Suppose \(h:\RR^2\to\RR^3\) were a homeomorphism and choose \(a\in\RR^2\). By 1, \[ \RR^2\setminus\{a\}\cong\RR^3\setminus\{h(a)\}. \] Translations identify these spaces with \(\RR^2\setminus\{0\}\) and \(\RR^3\setminus\{0\}\), respectively. Thus a homeomorphism would induce an isomorphism of fundamental groups. But by 5 these groups are \[ \ZZ \quad\text{and}\quad 0, \] which are not isomorphic. This contradiction proves \[ \boxed{\RR^2\not\cong\RR^3}. \]