Proof. Suppose \(d(A,B) = K > 0\), then the equation \(K = \inf \big \{\big |a - b\big |: \, a\in A\, , \, b\in B\big \}\) implies that \(\big | a - b\big | \geq K\) for every pair of points \(a\in A\) and \(b\in B\).
If \(a\in A\), it is not an adherent point of \(B\), since the \(bd\big (a - 1/2 K\, , \, a + 1/2 K\big )\) of \(a\) contains not point of \(B\). Therefore \(A \cap \overline {B} = \emptyset \). Similarly \(\overline {A} \cap B = \emptyset \). Hence \(A\) and \(B\) are separated.

The condition is not necessary, since two sets \(A\) and \(B\) may be at zero distance apart, and yet be separated. For example, the sets \(x< 0\) and \(x> 0\) are separated but are at zero distance. But on the other hand, the sets \(x< 0\) and \(x\geq 0\) are disjoint but not separated.

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Proof. Since \(A\) and \(B\) are separated \(A \cap \overline {B} = \emptyset \). Thus \(d\big (A,\overline {B}\big ) > 0\). Taking \(d(A,B) = K > 0\), we have \(\big | a- b\big | = K > 0\) where \(a\in A\) and \(b\in \overline {B}\). Thus, for any point \(b\in \overline {B}\), it is not an adherent point of \(A\) and \(A_1\), since any \(nbd \big (b - 1/2 K \, , \, b+1/2 K\big )\) of \(b\) does not contain a point of \(A\). Consequently, \(d(A,\overline {B})= K > 0\). Therefore, \(A_1 \cap \overline {B}_1 = \emptyset \), since \(B_1\subset B \, \implies \, \overline {B}_1 \subset \overline {B}\).
Similarly, \(\overline {A}_1 \cap B_1 = \emptyset \). Hence, \(A_1\) and \(B_1\) are separated.

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Proof. Suppose \(F_1\) and \(F_2\) are closed subsets of \(\mathbb {R}\) which are separated. Then \(F_1\cap \overline {F}_2 = \emptyset = \overline {F}_1\cap F_2\) and \(F_1 = \overline {F}_1\hspace {0.2cm},\hspace {0.2cm} F_2 = \overline {F}_2\). This means that \(F_1\cap F_2 = \emptyset . \implies F_1\) and \(F_2\) are disjoints.
Conversely, suppose that \(F_1\) and \(F_2\) are closed and disjoint sets. Then \(F_1 = \overline {F}_1\hspace {0.2cm},\hspace {0.2cm} F_2 = \overline {F}_2\) and \(F_1\cap F_2 =\emptyset = F_1\cap \overline {F}_2 = \emptyset \). Similarly \(\overline {F}_1 \cap F_2 = \emptyset \). Hence \(F_1\) and \(F_2\) are separated sets.

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Proof. Suppose that \(A\) and \(B\) are not empty. For if \(A = \emptyset \) then \(B = O\), and \(A\) and \(B\) are both open. Suppose \(x\) is any point of \(A\). Then there exist a \(nbd \big (x - r\, , \, x + r\big )\) of \(x\) consisting entirely of points of \(O\). Since \(x\) is not a limit point of \(B\), there exist a \(nbd \big ( x-r_1\, , \, x + r_1\big )\) of \(x\) , where \(r_1 \leq r\), which contains no point of \(B\) and so consists entirely of points of \(A\). Every point of \(A\) is thus an interior point of \(A\). Hence \(A\) is open similarly for \(B\).

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Proof. Since \(F= A\cap B\) is closed, we have \(A\cup B = \overline {A\cup B} = \overline {A} \cup \overline {B}\). Hence, \[\overline {A} = \overline {A} \cap \big (\overline {A}\cup \overline {B}\big ) = \overline {A} \cap \big ( A\cup B) = \big (\overline {A} \cap A\big ) \cup \big ( \overline {A} \cup B\big ) = A\cup \emptyset = A\] Thus, \(A\) is closed. Similarly \(B\) is closed.

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Proof. Suppose \(\, \exists \, x, \, y \in A\) and some \(z \in (x,y)\ni z\not \in A\). Then \(A = P \cup Q'\) where \( P = A \cap \big (-\infty \, ,\, z\big )\) and \(Q = A\cap \big (z\, , \, \infty \big )\). Since \(x\in P\) and \(y \in Q\, ,\, P\) and \(Q\) are non-empty. Since \(P\subset \big (-\infty \, ,\, z\big )\) and \(Q\subset \big (z\, , \,\infty \big )\), they are separated. Hence \(A\) is not connected, being a union of two non-empty separated sets. Hence \(z \in A\).
Conversely, suppose that \(A\) is not connected. Then there are non-empty separated sets \(P\) and \(Q\) such that \(P\cup Q = A\). Take \(x\in P \) and \(y \in Q\) and assume (without loss of generality) that \(x<y\). Define \(z = \sup \big (P\cap \big [x,y\big ]\big )\). Then \(z\in P\) and hence \(z\not \in Q\). In particular, \(x\leq z\leq y\). If \(z\not \in P\), it follows that \(x < z < y\) and \(z\in A\). If \(z \in P\), then \(z\not \in Q\), hence there exist \(z_1\ni z < z_1 < y\) and \(z_1 \not \in Q\). Then \(x < z_1 < y\) and \(z_1\not \in A\). This contradicts that \(A\) is not connected. Therefore \(A\) is connected.

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Proof. Suppose to the contrary that \(A\) is an interval. Then \(\, \exists \, a, \, b\in A\) with \(a< b\) and \(c\in \big (a,b\big ) \ni c\not \in A\). Let \(G_c = A \cap \big (-\infty \, , \, c\big )\) and \(H_c = A \cap \big ( c\, ,\, \infty \big )\). Note that both \(H_c\) and \(G_c\) are non-empty. Also since \(c\not \in A\hspace {0.2cm},\hspace {0.2cm} A = G_c\cup H_c\). Since \(\big (-\infty \, , \, c\big ) \cap \big [c\, ,\, \infty \big ) = \emptyset \) and \(\big (-\infty \, ,\, c\big ] \cap \big [c\, , \infty \big ) = \emptyset \) then \(G_c\) and \(H_c\) are separated. Thus, \(A\) is not connected. Therefore, if \(A\) is connected, then \(A\) is an interval.
Conversely, \(A\) is not connected and write \(A = G\cup H\) where \(G\) and \(H\) are non-empty separated sets. Without loss of generality, let \(a\in G\, ,\, b\in H\) and \(a< b\). Note that \(G\cap [a,b]\) is bounded so that \(c = \sup G \cap [a,b]\) exists in \(\mathbb {R}\) Then \(c\in \overline {G \cap [a,b]}\subset \overline {G}\cap \overline {[a,b]}\). In other words \(c\in \overline {G}\). Since \(\overline {G} \cap H = \emptyset , \, c\not \in H\). If \(c\not \in G\) then \(c\not \in A\), and so \(A\) is not connected by the previous theorem. Hence, \(A\) is not an interval, by the previous theorem. Therefore, if \(A\) is an interval then \(A\) is connected.

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Proof. Let \(A\) and \(B\) be two connected sets which are not separated. If \(A\cup B\) is not connected, \(A\cup B = C \cup D\) where \(C\) and \(D\) are non-empty separated sets. Then \(A = A\cap (A \cup B) = A\cap (C\cup D) = (A\cap C) \cup (A\cap D)\) which expresses \(A\) as the union of two separated sets, which is impossible since \(A\) is connected, unless one of \(A\cap C\) and \(A\cap D\) is empty. Suppose that \(S\cap D\, B\subseteq C\) or \(B\subseteq D\). But this gives \(A \cap \overline {B} \subseteq C \cap \overline {D} = \emptyset \hspace {0.2cm},\hspace {0.2cm} \overline {A}\cap B \subseteq \overline {C} \cap D = \emptyset \). which is not possible. Hence \(A\cup B\) is connected.

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Proof. Let \(A\) and \(B\) be two connected sets, whose intersection \(A\cap B\) is not empty since \(A\cap B \subseteq A\cap \overline {B}\) and \(A \cap B \subseteq \overline {A} \cap B\), the result is an immediate consequence of the theorem.

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