This is exercise 6 on page 178 in Chapter 27 of James R. Munkers’ Topology 2nd Edition textbook.

Question

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” \((\frac{1}{3}, \frac{2}{3})\). Let \(A_2\) be the set obtained from \(A_1\) by deleting its “middle thirds” \((\frac{1}{9}, \frac{2}{9})\) and \((\frac{7}{9}, \frac{8}{9})\). In general, define \(A_n\) by the equation

\[A_n = A_{n - 1} - \bigcup_{k = 0}^{\infty}(\frac{1 + 3k}{3^n}, \frac{2 + 3k}{3^n}).\]

The intersection

\[C = \bigcup_{n \in \mathbb{Z}_{+}}A_n\]

is called the Cantor Set; it is a subspace of \([0, 1]\).

Proof

a) Show that \(C\) is totally disconnected

It suffices to prove the conditional: If \(D\) is a connected subspace of \(C\), then \(D\) is a one-point set.

The contrapositive of the conditional above is: If \(D\) is a subspace of \(C\) with more than one element, then \(D\) has a separation. We will prove the contrapositive.

To begin, suppose \(D\) is a subspace of \(C\) with two or more elements. Next, take any interval \((a, c) \subset [0, 1]\) and \(U\) being some open set of \(C\). The subspace \(D\) may be described as

\[D = U \cap (a, c)\]

Now, since \(D\) has more than one element, take distinct elements \(x, y \in D\) in which \(x < y\).

To show we may find two disjoint sets whose union is \(D\), it suffices to show there exists some \(b \in (a, c)\) such that \(b \notin C\). Because \((a, c)\) is a fixed interval, there exists some \(n \in \mathbb{N}\) such that some \(I \subset A_n\) is fully contained within \((a, c)\), that is \(I \subset (a, c)\). Next, take \(A_{n+1}\). By definition of the Cantor Set, the “middle thirds” of \(I\) are intersected out. These “middle thirds” contain points within \((a, c)\) and not within \(C\). Take some \(b\) from a “middle third” to satisfy the condition imposed in the first sentence of this paragraph.

Because \(x, y\) are distinct, we find two open, disjoint sets \(X = (a, b) \cap U\) and \(Y = (b, c) \cap U\) where \(X \cup Y = D\). Thus, proving the contraposition, completing part a.

b) Show that \(C\) is compact

Because each \(A_i\) for \(i \in \mathbb{Z}_{+}\) can be written as a finite union of closed sets in \([0, 1]\), the infinite intersection of these sets are also closed. Thus, \(C\) is closed in \([0, 1]\). Now, using Theorem 26.2, the subspace \(C\) is compact as \([0, 1]\) is also compact. This concludes part b of the proof.

\[\tag*{$\blacksquare$}\]

Extras

link to the mathjax LaTeX specification: https://treeofmath.github.io/tex-commands-in-mathjax/TeXSyntax.htm