Show that \(X\) is countably compact if and only if every nested sequence \(C_1 \supset C_2 \supset \cdots\) of closed nonempty sets of \(X\) has a nonempty intersection.
Solution
1. Definition of Countable Compactness:
1.1. A topological space \(X\) is countably compact if every countable open cover of \(X\) has a finite subcover.
standard definition of countable compactness.
2. Forward direction (\(\implies\)): Countable compactness implies nested intersection property:
2.1. Let \(C_1 \supseteq C_2 \supseteq C_3 \supseteq \cdots\) be a nested sequence of non-empty closed subsets of \(X\).
setup.
2.2. Suppose for contradiction that \(\bigcap_{n=1}^\infty C_n = \emptyset\).
assumption for contradiction.
2.3. By De Morgan’s laws, the complements \(U_n = X \setminus C_n\) form a countable collection of open sets satisfying:\[ \bigcup_{n=1}^\infty U_n = \bigcup_{n=1}^\infty (X \setminus C_n) = X \setminus \left(\bigcap_{n=1}^\infty C_n\right) = X \setminus \emptyset = X. \]Thus \(\{U_n\}_{n=1}^\infty\) is a countable open cover of \(X\).
De Morgan’s laws.
2.4. Since \(X\) is countably compact, there exists a finite subcover: \(X = U_{n_1} \cup \cdots \cup U_{n_k}\).Let \(N = \max(n_1, \dots, n_k)\).Since the sequence \(\{C_n\}\) is nested (\(C_N \subseteq C_n\) for all \(n \le N\)), the complements are nested (\(U_n \subseteq U_N\) for all \(n \le N\)), so:\[ X = \bigcup_{j=1}^k U_{n_j} = U_N = X \setminus C_N. \]
monotonicity of nested sets.
2.5. Taking complements gives \(C_N = X \setminus U_N = \emptyset\), which contradicts the hypothesis that each \(C_n\) is non-empty.Thus \(\bigcap_{n=1}^\infty C_n \neq \emptyset\).
proof by contradiction.
3. Reverse direction (\(\impliedby\)): Nested intersection property implies countable compactness:
3.1. Let \(\{V_n\}_{n=1}^\infty\) be an arbitrary countable open cover of \(X\).
setup.
3.2. Suppose for contradiction that no finite subcollection covers \(X\).Then for every \(n \ge 1\), \(\bigcup_{k=1}^n V_k \neq X\).
assumption for contradiction.
3.3. Define \(C_n = X \setminus \bigcup_{k=1}^n V_k\).Each \(C_n\) is non-empty and closed (as the complement of a finite union of open sets).
2.
3.4. For each \(n \ge 1\):\[ C_{n+1} = X \setminus \left(\bigcup_{k=1}^{n+1} V_k\right) = \left(X \setminus \bigcup_{k=1}^n V_k\right) \setminus V_{n+1} = C_n \setminus V_{n+1} \subseteq C_n. \]Thus \(C_1 \supseteq C_2 \supseteq C_3 \supseteq \cdots\) is a nested sequence of non-empty closed sets.
monotonicity of unions.
3.5. By the nested intersection hypothesis, there exists \(x \in \bigcap_{n=1}^\infty C_n\).
hypothesis.
3.6. For this \(x\), \(x \in C_n = X \setminus \bigcup_{k=1}^n V_k\) for all \(n \ge 1\), which means \(x \notin V_k\) for all \(k \ge 1\).Thus \(x \notin \bigcup_{k=1}^\infty V_k = X\), a contradiction.
\(\{V_n\}\) covers \(X\).
3.7. Thus some finite subcollection must cover \(X\), so \(X\) is countably compact.
proof by contradiction.
4. Conclusion:\(X\) is countably compact if and only if every nested sequence of non-empty closed sets has non-empty intersection. Q.E.D.
2 and 3.