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 \(Y = (a, c) \subset [0, 1]\) and \(U\) being some open set of \(C\). The subspace \(D\) may be described as

\[D = U \cap Y\]

Now, since \(D\) has more than one element, take distinct elements \(x, y \in D\) where \(x\) is the smaller element of the two and \(y\) is the larger of the two. Then, select a \(b \in [0, 1] - C\) such that \(a < b < c\).

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

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