Let \(A_0\) be the closed interval \([0, 1]\) in \(\mathbb{R}\). Let \(A_1\) be the set obtained from \(A_0\) by deleting its “middle third” \((\tfrac{1}{3}, \tfrac{2}{3})\). Let \(A_2\) be the set obtained from \(A_1\) by deleting its “middle thirds” \((\tfrac{1}{9}, \tfrac{2}{9})\) and \((\tfrac{7}{9}, \tfrac{8}{9})\). In general, define \(A_n\) by the equation
\[ A_n = A_{n-1} - \bigcup_{k=0}^{\infty} \left( \frac{1+3k}{3^n}, \frac{2+3k}{3^n} \right). \]
The intersection
\[ C = \bigcap_{n \in \mathbb{Z}_+} A_n \]
is called the Cantor set; it is a subspace of \([0, 1]\).
Show that \(C\) is totally disconnected.
Show that \(C\) is compact.
Show that each set \(A_n\) is a union of finitely many disjoint closed intervals of length \(1/3^n\); and show that the end points of these intervals lie in \(C\).
Show that \(C\) has no isolated points.
Conclude that \(C\) is uncountable.
Solution
Write \(C\) for the Cantor set.
Let \(x<y\) be distinct points of \(C\). Choose \(n\) so large that \(3^{-n}<y-x\). The components of \(A_n\) are pairwise disjoint closed intervals of length \(3^{-n}\), so \(x\) and \(y\) cannot lie in the same component interval. Hence some deleted middle-third interval separates them. Therefore \(C\) admits a separation with \(x\) and \(y\) in different sides. Thus no connected subset of \(C\) contains two points, so \(C\) is totally disconnected.
Each \(A_n\) is closed in compact \([0,1]\), hence compact, and \[ C=\bigcap_{n\ge1}A_n \] is closed in \([0,1]\). Therefore \(C\) is compact.
Inductively, \(A_0=[0,1]\). At stage \(n\), each component interval of \(A_{n-1}\), of length \(3^{-(n-1)}\), has its open middle third deleted, leaving two disjoint closed intervals of length \(3^{-n}\). Hence \(A_n\) is the union of \(2^n\) pairwise disjoint closed intervals of length \(3^{-n}\). Endpoints are never deleted at later stages, so every endpoint belongs to every subsequent \(A_m\), hence lies in \(C\).
Fix \(x\in C\) and \(\varepsilon>0\). Choose \(n\) with \(3^{-n}<\varepsilon\), and let \(I_n\) be the component interval of \(A_n\) containing \(x\). One endpoint \(e_n\) of \(I_n\) is distinct from \(x\) unless \(x\) itself is that endpoint; in that case take the other endpoint. By (c), \(e_n\in C\), and \[ 0<|e_n-x|\le 3^{-n}<\varepsilon. \] Thus every neighborhood of \(x\) contains another point of \(C\), so \(C\) has no isolated points.
For each binary sequence \(\epsilon=(\epsilon_1,\epsilon_2,\dots)\in\{0,1\}^{\omega}\), choose recursively the left subinterval at stage \(n\) if \(\epsilon_n=0\) and the right subinterval if \(\epsilon_n=1\). This gives nested closed intervals \(I_n(\epsilon)\subset A_n\) with lengths \(3^{-n}\). Their intersection consists of exactly one point \(x_\epsilon\in C\). If \(\epsilon\ne\epsilon'\), at the first differing index the corresponding stage intervals are disjoint, so \(x_\epsilon\ne x_{\epsilon'}\). Hence \(\{0,1\}^{\omega}\) injects into \(C\). Since \(\{0,1\}^{\omega}\) is uncountable, \(C\) is uncountable.