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2.4.  Connected Sets

We now focus on the fact that the Cantor set has gaps at every scale: given any \(x\in C\) and any \(r>0\), no matter how small \(r\) is, the neighborhood \((x-r,x+r)\) contains some open interval that does not belong to \(C\). Thus, if we zoom in around any point of the Cantor set, the set never starts looking like a solid interval. To express the property of a set forming one global piece, we introduce connectedness.

Definition 2.4.1. Let \(M\) be a metric space, and let \(A,B\subseteq M\). The two sets \(A\) and \(B\) are said to be separated if both \(A\cap \overline B\) and \(\overline A\cap B\) are empty, that is, if neither set contains a closure point of the other.

Note that separated sets are of course disjoint, but disjoint sets need not be separated.

Definition 2.4.2. Let \(M\) be a metric space, and let \(A\subseteq M\). \(A\) is said to be connected if it is not a union of two nonempty separated sets. And, \(A\) is said to be disconnected if it is not connected.

Theorem 2.4.3. Let \(M\) be a metric space, and let \(A\subseteq M\). Then, \(A\) is disconnected if and only if there are two nonempty sets \(U\) and \(V\) open in \(A\) such that \(A = U\,\dot\cup\, V\).

Proof. Suppose first that \(A\) is disconnected. Then there exist two nonempty separated sets \(B\) and \(C\) such that \(A=B\,\dot\cup\,C\). Define \(U=A\setminus \overline C\) and \(V=A\setminus\overline B\). Since \(M\setminus \overline C\) and \(M\setminus\overline B\) are open, Theorem 2.1.18 shows that the sets \(U\) and \(V\) are open in \(A\). Since \(B\) and \(C\) are separated, \(B\cap\overline C=\varnothing\), so \(B\subseteq U\). If \(x\in U\), then \(x\in A=B\cup C\) and \(x\notin\overline C\). Thus \(x\notin C\), so \(x\in B\). Thus \(U=B\). Similarly, \(V=C\). Therefore, \(A=U\,\dot\cup\,V\), where \(U\) and \(V\) are nonempty and open in \(A\). Conversely, suppose that \(U\) and \(V\) are nonempty sets open in \(A\) such that \(A=U\,\dot\cup\,V\). Since \(U=A\setminus V\) and \(V=A\setminus U\), both \(U\) and \(V\) are also closed in \(A\). Hence \(\overline U^{\raise{-0.5em}{A}}=U\) and \(\overline V^{\raise{-0.5em}{A}}=V\). Therefore we have

\[U\cap\overline V^{\raise{-0.5em}{A}} =U\cap V =\varnothing.\]

Since \(U\subseteq A\) and \(\overline V^{\raise{-0.5em}{A}} = \overline V \cap A\) by Theorem 2.1.19, it follows that

\[U \cap \overline V = (U\cap A) \cap \overline V = U \cap (A\cap \overline V) = U\cap\overline V^{\raise{-0.5em}{A}} = \varnothing.\]

Similarly, \(\overline U\cap V=\varnothing\). Thus \(U\) and \(V\) are separated. Since they are nonempty and \(A=U\,\dot\cup\,V\), the set \(A\) is disconnected.\(\square\)

On the real line \(\mathbb R\), a gap is the only possible obstruction to connectedness. A set is therefore connected precisely when it contains every point lying between any two of its points.

Theorem 2.4.4. A subset \(A\) of \(\mathbb R\) is connected if and only if it has the following property:

\[x,y\in A \ \ \ \text{and} \ \ \ x<z<y \ \Longrightarrow \ z\in A.\]

Proof. Suppose \(x,y\in A\), \(x<z<y\), and \(z\notin A\). Define

\[U = (-\infty, z) \cap A \quad \text{and} \quad V = (z,\infty) \cap A.\]

Then, \(A = U\,\dot\cup\,V\). Since \(x\in U\) and \(y\in V\), \(U\) and \(V\) are nonempty. By Theorem 2.1.18, \(U\) and \(V\) are open in \(A\). Theorem 2.4.3 now shows that \(A\) is disconnected. We now prove the converse. Suppose, for contradiction, that \(A\) is disconnected and has the stated property. Then there are nonempty separated sets \(A_1\) and \(A_2\) such that \(A = A_1\cup A_2\). Let \(a_1\in A_1\) and \(a_2\in A_2\), and assume, without loss of generality, that \(a_1<a_2\). By the property, we have

\[[a_1,a_2]\subseteq A.\]

As we move from \(a_1\) to \(a_2\), there must be a boundary between the two sets. To locate this boundary, define

\[\alpha = \sup {\left(A_1\cap [a_1,a_2]\right)}.\]

For all \(x\in A\), we have \(x\in A_1\) or \(x\in A_2\). Thus \(\alpha\in [x,y]\subseteq A\) implies that \(\alpha\in A_1\) or \(\alpha \in A_2\). If \(\alpha \notin A_1\), then \(\alpha \in A_2\). Since \(\alpha \in \overline{A_1}\), it follows that \(\overline {A_1} \cap A_2\ne \varnothing\). This contradicts the separatedness. If \(\alpha \in A_1\), then \(\alpha<a_2\) because \(a_2\in A_2\). Since \(\alpha\) is the supremum, \((\alpha,a_2]\nsubseteq A_1\), so that \((\alpha,a_2]\subseteq A_2\). Therefore \(\alpha \in \overline {A_2}\), so that \(A_1\cap \overline{A_2}\ne \varnothing\). The result is a contradiction once again. Therefore \(A\) is connected.\(\square\)

References

  1. Rudin, W. (1976). Principles of mathematical analysis (3rd ed.). McGraw-Hill Education.