Show that \(\mathbb{R}^1\) and \(\mathbb{R}^n\) are not homeomorphic if \(n > 1\).
Show that \(\mathbb{R}^2\) and \(\mathbb{R}^n\) are not homeomorphic if \(n > 2\).
It is, in fact, true that \(\mathbb{R}^m\) and \(\mathbb{R}^n\) are not homeomorphic if \(n \neq m\), but the proof requires more advanced tools of algebraic topology.
Solution
Goal. Show \(\RR^1 \not\cong \RR^n\) for \(n > 1\) and \(\RR^2 \not\cong \RR^n\) for \(n > 2\).
1. (a) \(\RR^1 \not\cong \RR^n\) for \(n > 1\).
1.1. Removing a point from \(\RR^1\) gives a disconnected space.
\(\RR^1 \sm \theset{0} = (-\infty, 0) \cup (0, \infty)\) is disconnected.
1.2. Removing a point from \(\RR^n\) (\(n > 1\)) gives a connected space.
\(\RR^n \sm \theset{0}\) is path-connected (any two points can be joined by a path avoiding the origin).
1.3. Hence \(\RR^1 \not\cong \RR^n\).
connectedness is a homeomorphism invariant, and a homeomorphism would preserve the number of connected components after removing a point.
2. (b) \(\RR^2 \not\cong \RR^n\) for \(n > 2\).
2.1. \(\RR^2 \sm \theset{0}\) is not simply connected.
\(\RR^2 \sm \theset{0}\) deformation-retracts onto \(S^1\), so \(\pi_1(\RR^2 \sm \theset{0}) = \ZZ \neq 0\).
2.2. \(\RR^n \sm \theset{0}\) (\(n > 2\)) is simply connected.
\(\RR^n \sm \theset{0}\) deformation-retracts onto \(S^{n-1}\), and \(\pi_1(S^{n-1}) = 0\) for \(n - 1 \ge 2\).
2.3. Hence \(\RR^2 \not\cong \RR^n\).
\(\pi_1\) is a homeomorphism invariant, and the two spaces have different \(\pi_1\).
3. Q.E.D.
1 and 2 prove (a) and (b).