Theorem. Let \(X\) be a compact Hausdorff space. Let \(\mathcal{A}\) be a collection of closed connected subsets of \(X\) that is simply ordered by proper inclusion. Then
\[ Y = \bigcap_{A \in \mathcal{A}} A \]
is connected. [Hint: If \(C \cup D\) is a separation of \(Y\), choose disjoint open sets \(U\) and \(V\) of \(X\) containing \(C\) and \(D\), respectively, and show that
\[ \bigcap_{A \in \mathcal{A}} (A - (U \cup V)) \]
is not empty.]
Solution
Goal: Prove that the intersection \(Y = \bigcap_{A \in \mathcal{A}} A\) of a nested (simply ordered by inclusion) family \(\mathcal{A}\) of closed connected subsets in a compact Hausdorff space \(X\) is non-empty and connected.
1. \(Y\) is non-empty and compact:Proof:
1.1. Each \(A \in \mathcal{A}\) is closed in \(X\), hence compact.
1.2. Since \(\mathcal{A}\) is simply ordered by inclusion, any finite subcollection \(\{A_1, \dots, A_k\} \subset \mathcal{A}\) has a minimal element \(A_0 = \min_i A_i \in \mathcal{A}\), so \(\bigcap_{i=1}^k A_i = A_0 \neq \emptyset\).
1.3. Thus \(\mathcal{A}\) satisfies the finite intersection property.
1.4. By compactness of \(X\), \(Y = \bigcap_{A \in \mathcal{A}} A\) is non-empty, closed, and compact.
2. Separation assumption and neighborhood separation:Suppose for contradiction that \(Y\) is not connected.Proof:
2.1. There exists a separation \(Y = C \cup D\) into disjoint, non-empty closed subsets \(C, D \subset Y\).
2.2. Since \(Y\) is closed in \(X\), \(C\) and \(D\) are disjoint closed subsets of the compact Hausdorff space \(X\).
2.3. Every compact Hausdorff space is normal (\(T_4\)).
2.4. By normality of \(X\), there exist disjoint open sets \(U, V \subset X\) such that \(C \subset U\) and \(D \subset V\).
2.5. Then \(Y = C \cup D \subset U \cup V\) and \(U \cap V = \emptyset\).
3. Finiteness reduction on \(A \setminus (U \cup V)\):There exists a member \(A_0 \in \mathcal{A}\) such that \(A_0 \subset U \cup V\).Proof:
3.1. For each \(A \in \mathcal{A}\), define \(F_A = A \setminus (U \cup V) = A \cap (X \setminus (U \cup V))\).
3.2. Since \(X \setminus (U \cup V)\) is closed in \(X\), each \(F_A\) is a closed (hence compact) subset of \(X\).
3.3. The total intersection is:\[\bigcap_{A \in \mathcal{A}} F_A = \left(\bigcap_{A \in \mathcal{A}} A\right) \setminus (U \cup V) = Y \setminus (U \cup V) = \emptyset.\]
3.4. By compactness of \(X\), the family of closed sets \(\{F_A\}_{A \in \mathcal{A}}\) cannot satisfy the finite intersection property, so there exists a finite subfamily \(\{A_1, \dots, A_k\} \subset \mathcal{A}\) with \(\bigcap_{i=1}^k F_{A_i} = \emptyset\).
3.5. Since \(\mathcal{A}\) is totally ordered by inclusion, choose \(A_0 = \min\{A_1, \dots, A_k\} \in \mathcal{A}\).
3.6. Then \(F_{A_0} = \bigcap_{i=1}^k F_{A_i} = \emptyset\), which means \(A_0 \setminus (U \cup V) = \emptyset\), i.e., \(A_0 \subset U \cup V\).
4. Derivation of contradiction:Proof:
4.1. Since \(A_0 \subset U \cup V\) and \(U \cap V = \emptyset\), we can write \(A_0 = (A_0 \cap U) \cup (A_0 \cap V)\) as a union of disjoint open sets in the subspace topology of \(A_0\).
4.2. Furthermore:\[A_0 \cap U \supseteq Y \cap U \supseteq C \neq \emptyset \quad \text{and} \quad A_0 \cap V \supseteq Y \cap V \supseteq D \neq \emptyset.\]
4.3. Thus \(A_0 = (A_0 \cap U) \cup (A_0 \cap V)\) constitutes a separation of \(A_0\) into two non-empty disjoint open sets.
4.4. This strictly contradicts the hypothesis that \(A_0 \in \mathcal{A}\) is connected.
5. Conclusion:No separation of \(Y\) can exist, so \(Y = \bigcap_{A \in \mathcal{A}} A\) is connected. Q.E.D.