Show that no two of the spaces \((0, 1)\), \((0, 1]\), and \([0, 1]\) are homeomorphic. [Hint: What happens if you remove a point from each of these spaces?]
Suppose that there exist imbeddings \(f: X \to Y\) and \(g: Y \to X\). Show by means of an example that \(X\) and \(Y\) need not be homeomorphic.
Show that \(\mathbb{R}^n\) and \(\mathbb{R}\) are not homeomorphic if \(n > 1\).
Solution
A point \(p\) of a connected interval is a cut point if deleting it disconnects the space. In \((0,1)\) every point is a cut point. In \((0,1]\), exactly one point, namely \(1\), is not a cut point. In \([0,1]\), exactly the two endpoints \(0,1\) are not cut points. A homeomorphism preserves the number of non-cut points, so no two of these three spaces are homeomorphic.
Let \[ X=(0,1),\qquad Y=[0,1]. \] The inclusion embeds \(X\) in \(Y\). The affine map \[ g:Y\to X,\qquad g(t)=\frac14+\frac t2 \] embeds \(Y\) in \(X\). But \(X\) and \(Y\) are not homeomorphic by part (a).
If \(n>1\), the punctured space \(\mathbb R^n-\{0\}\) is path connected: two points can be joined by a polygonal path avoiding the origin, using an intermediate point not lying on either line through the origin and one endpoint. By contrast, \[ \mathbb R-\{0\}=(-\infty,0)\cup(0,\infty) \] is disconnected. A homeomorphism \(\mathbb R^n\cong\mathbb R\) would restrict after deleting a point to a homeomorphism of these punctured spaces, impossible. Hence \(\mathbb R^n\not\cong\mathbb R\) for \(n>1\).