Let \(X\) be a compact space, and let \[A_1 \supseteq A_2 \supseteq \cdots \supseteq A_n \supseteq \cdots\] be a descending chain of non-empty closed subsets of \(X\). Show that their intersection \[\bigcap_{n=1}^{\infty} A_n\] is non-empty.
Solution
1. If \(\bigcap_{n=1}^{\infty} A_n = \varnothing\), then \(\{X \setminus A_n\}_{n\ge1}\) is an open cover of \(X\).
Each \(X\setminus A_n\) is open because \(A_n\) is closed. By De Morgan’s law, \(\bigcup_n (X \setminus A_n) = X \setminus \bigcap_n A_n = X\).
2. If \(\bigcap_{n=1}^{\infty} A_n = \varnothing\), then \(A_N=\varnothing\) for some \(N\).
By step 1 and compactness of \(X\), there are indices \(n_1,\ldots,n_k\) with \(X = \bigcup_{i=1}^k (X \setminus A_{n_i})\). By De Morgan’s law, \(\bigcap_{i=1}^k A_{n_i} = \varnothing\). Put \(N = \max(n_1, \ldots, n_k)\). Since the chain is descending, \(A_N\subseteq A_{n_i}\) for each \(i\), so \(A_N=\bigcap_{i=1}^k A_{n_i}=\varnothing\).
3. Q.E.D.
Each \(A_n\) is nonempty by hypothesis, so the conclusion of step 2 is false. Hence \(\bigcap_{n=1}^{\infty} A_n \neq \varnothing\).