Proof. Suppose \(d(A,B) = K > 0\). Then \(| a - b| \geq K\) for every \(a\in A\) and \(b\in B\). So if \(a\in A\), the neighbourhood \(\left (a - \frac {K}{2}, a + \frac {K}{2}\right )\) of \(a\) contains no point of \(B\), and by Theorem 1.2.15(iv), \(a\notin \overline {B}\). Therefore \(A \cap \overline {B} = \emptyset \). Similarly \(\overline {A} \cap B = \emptyset \), and \(A\) and \(B\) are separated.

The condition is not necessary: the sets \((-\infty ,0)\) and \((0,\infty )\) are separated, since neither closure meets the other set, but their distance is \(0\). □

Proof. Since \(B_1\subseteq B\), we have \(\overline {B}_1 \subseteq \overline {B}\) (every limit point of \(B_1\) is a limit point of \(B\)). Hence \[A_1 \cap \overline {B}_1 \subseteq A \cap \overline {B} = \emptyset .\] Similarly \(\overline {A}_1 \cap B_1 \subseteq \overline {A}\cap B = \emptyset \). Hence \(A_1\) and \(B_1\) are separated. □

Proof. Let \(F_1\) and \(F_2\) be closed, so that \(\overline {F}_1 = F_1\) and \(\overline {F}_2 = F_2\). Then \[F_1\cap \overline {F}_2 = F_1\cap F_2 \qquad \text {and}\qquad \overline {F}_1\cap F_2 = F_1\cap F_2 ,\] so the two conditions for being separated both reduce to \(F_1\cap F_2 = \emptyset \). □

Proof. Let \(x \in A\). Since \(O\) is open, there is an \(r>0\) with \((x - r, x + r)\subseteq O\). Since \(x\notin \overline {B}\), there is an \(r_1\) with \(0<r_1 \leq r\) such that \((x - r_1, x + r_1)\) contains no point of \(B\). This interval lies in \(O = A\cup B\), so it lies in \(A\). Thus every point of \(A\) is an interior point of \(A\), and \(A\) is open. Similarly \(B\) is open. □

Proof. Since \(A\subseteq F\) and \(F\) is closed, \(\overline {A}\subseteq F = A\cup B\) by Theorem 1.2.15(ii). But \(\overline {A}\cap B = \emptyset \), so \(\overline {A} \subseteq A\). Hence \(\overline {A} = A\) and \(A\) is closed. Similarly \(B\) is closed. □

Proof. Suppose there are \(x, y \in A\) and \(z \in (x,y)\) with \(z\notin A\). Let \(P = A \cap (-\infty , z)\) and \(Q = A\cap (z, \infty )\). Since \(z\notin A\), \(A = P \cup Q\), and \(P\) and \(Q\) are non-empty because \(x\in P\) and \(y \in Q\). Since \(\overline {P}\subseteq (-\infty ,z]\) and \(\overline {Q}\subseteq [z,\infty )\), neither set meets the closure of the other, so \(P\) and \(Q\) are separated. Hence \(A\) is not connected. This proves that a connected set has the stated property.

Conversely, suppose \(A\) has the property but is not connected, say \(A = P\cup Q\) with \(P\) and \(Q\) non-empty and separated. Take \(x\in P\) and \(y \in Q\), and assume without loss of generality that \(x<y\). Let \(z = \sup (P\cap [x,y])\). Then \(x\leq z\leq y\), so \(z\in A\) by the property, and \(z\in \overline {P}\), so \(z\notin Q\) (as \(\overline {P}\cap Q = \emptyset \)). Hence \(z\in P\), and in particular \(z\neq y\), so \(z<y\). Since \(z \in P\) and \(P\cap \overline {Q} = \emptyset \), there is a \(\delta >0\) such that \((z-\delta , z+\delta )\) contains no point of \(Q\). Choose \(w\) with \(z<w<\min \{z+\delta , y\}\). Then \(w\in A\) by the property and \(w\notin Q\), so \(w\in P\cap [x,y]\) with \(w>z\), contradicting the choice of \(z\). Hence \(A\) is connected. □

Proof. A subset of \(\mathbb {R}\) is an interval (of one of the nine kinds in Section 1.1, or empty, or a single point) exactly when it contains every point lying between two of its points. By Theorem 1.6.10, this is exactly the condition for \(A\) to be connected. □

Proof. Let \(A\) and \(B\) be connected sets which are not separated, and suppose \(A\cup B = C \cup D\), where \(C\) and \(D\) are non-empty separated sets. Then \[A = (A\cap C) \cup (A\cap D),\] and \(A\cap C\) and \(A\cap D\) are separated, being subsets of \(C\) and \(D\). Since \(A\) is connected, one of them is empty, so \(A\subseteq C\) or \(A\subseteq D\). Likewise \(B\subseteq C\) or \(B\subseteq D\).

If \(A\) and \(B\) lay in the same set, say \(C\), then \(D\) would be empty, which is not the case. So one of them lies in \(C\) and the other in \(D\), say \(A\subseteq C\) and \(B\subseteq D\). Then \[A \cap \overline {B} \subseteq C \cap \overline {D} = \emptyset \qquad \text {and}\qquad \overline {A}\cap B \subseteq \overline {C} \cap D = \emptyset ,\] so \(A\) and \(B\) are separated, contrary to hypothesis. Hence \(A\cup B\) is connected. □

Proof. Let \(A\) and \(B\) be connected with \(A\cap B\neq \emptyset \). Since \(A\cap B \subseteq A\cap \overline {B}\), the set \(A\cap \overline {B}\) is non-empty, so \(A\) and \(B\) are not separated. The result follows from the theorem. □

Solution. (a) \(\overline {(0,1)} = [0,1]\) does not meet \((1,2)\), and \(\overline {(1,2)} = [1,2]\) does not meet \((0,1)\). So the sets are separated, and their union \((0,1)\cup (1,2)\) is disconnected. (It omits \(1\), so it is not an interval.)

(b) \(\overline {[0,1)} = [0,1]\) meets \([1,2]\) in the point \(1\), so the sets are not separated. Both are connected, so their union \([0,2]\) is connected by the theorem on unions of non-separated connected sets.

(c) The sets are closed and disjoint, hence separated, and the union is disconnected.

Solution. Let \(P = \mathbb {Q}\cap (-\infty ,\sqrt {2})\) and \(Q = \mathbb {Q}\cap (\sqrt {2},\infty )\). Since \(\sqrt {2}\notin \mathbb {Q}\), \(\mathbb {Q} = P\cup Q\). Both sets are non-empty (\(0\in P\), \(2\in Q\)), and \(\overline {P}\subseteq (-\infty ,\sqrt {2}]\) does not meet \(Q\), while \(\overline {Q}\subseteq [\sqrt {2},\infty )\) does not meet \(P\). So \(\mathbb {Q}\) is disconnected.

For the irrationals, cut at the rational number \(0\) instead: \(\mathbb {Q}^C = \bigl (\mathbb {Q}^C\cap (-\infty ,0)\bigr )\cup \bigl (\mathbb {Q}^C\cap (0,\infty )\bigr )\), and the same argument applies. Both conclusions also follow from Theorem 1.6.10, since neither set is an interval.

Solution. Suppose \(B = C\cup D\) with \(C\) and \(D\) non-empty and separated. Then \(A = (A\cap C)\cup (A\cap D)\) with \(A\cap C\) and \(A\cap D\) separated, so, \(A\) being connected, one of them is empty; say \(A\subseteq C\). Then \(B\subseteq \overline {A}\subseteq \overline {C}\), so \(D = B\cap D\subseteq \overline {C}\cap D = \emptyset \), a contradiction. Hence \(B\) is connected.

If \(I\) is an interval, it is connected, so \(\overline {I}\) is connected and therefore an interval. For example \(\overline {(a,b)} = [a,b]\).