Show that \(\RR\) and \(\RR^2\) (with their usual topologies) are not homeomorphic.
Solution
1. For every \(c\in\RR\), the punctured line \(\RR\setminus\{c\}\) is disconnected.
It is the disjoint union \(\RR\setminus\{c\}=(-\infty,c)\sqcup(c,\infty)\). Both pieces are nonempty and open in the subspace \(\RR\setminus\{c\}\). Hence they form a separation.
2. For every \(a\in\RR^2\), the punctured plane \(\RR^2\setminus\{a\}\) is path-connected.
Translation by \(-a\) is a homeomorphism from \(\RR^2\setminus\{a\}\) to \(\RR^2\setminus\{0\}\), so it suffices to prove the latter is path-connected.
Take \(x,y\in\RR^2\setminus\{0\}\). Choose a point \(z\in\RR^2\setminus\bigl(\operatorname{span}(x)\cup\operatorname{span}(y)\bigr)\). Such a \(z\) exists because the union of two lines through the origin is not all of \(\RR^2\).
The line segment from \(x\) to \(z\) does not contain the origin: if \((1-t)x+tz=0\) for some \(0<t<1\), then \(z=-\frac{1-t}{t}x\in\operatorname{span}(x)\), contrary to the choice of \(z\). The same argument shows that the line segment from \(z\) to \(y\) avoids the origin. Concatenating these two line segments gives a path from \(x\) to \(y\) in \(\RR^2\setminus\{0\}\).
3. There is no homeomorphism \(h\colon\RR\to\RR^2\).
Assume such an \(h\) exists, choose \(c\in\RR\), and put \(a=h(c)\). The restriction \(h|_{\RR\setminus\{c\}}\colon\RR\setminus\{c\}\to\RR^2\setminus\{a\}\) is a bijection whose inverse is the corresponding restriction of \(h^{-1}\), so it is a homeomorphism. By step 1 its domain is disconnected, while by step 2 its codomain is path-connected and hence connected. A homeomorphic image of a disconnected space is disconnected, a contradiction.
4. Q.E.D.
Step 3 shows that \(\RR\) and \(\RR^2\) are not homeomorphic.