Let \(X\) be a metric space and let \(T_1\supseteq T_2\supseteq\cdots\) be nonempty closed subsets of \(X\). If \(T_1\) is compact, show that \[ \bigcap_{n=1}^{\infty}T_n\ne\varnothing. \]
Solution
1. Suppose for contradiction that\[ \bigcap_{n=1}^{\infty}T_n=\varnothing. \]Then\[ \{X\setminus T_n:n\ge1\} \]is an open cover of \(T_1\).
Each \(T_n\) is closed, so \(X\setminus T_n\) is open. The assumed empty intersection implies, by De Morgan’s law, \[ X = X\setminus\bigcap_{n=1}^{\infty}T_n = \bigcup_{n=1}^{\infty}(X\setminus T_n). \] In particular these open sets cover \(T_1\).
2. There is an integer \(N\ge1\) such that\[ T_1\subseteq X\setminus T_N. \]
Since \(T_1\) is compact, step 1 has a finite subcover. Thus there are indices \[ n_1,\ldots,n_k \] such that \[ T_1 \subseteq \bigcup_{j=1}^k(X\setminus T_{n_j}). \] Let \[ N\coloneqq\max\{n_1,\ldots,n_k\}. \] The nesting \[ T_1\supseteq T_2\supseteq\cdots \] implies \[ X\setminus T_{n_j} \subseteq X\setminus T_N \] for every \(j\). Hence the whole finite union is contained in \(X\setminus T_N\), proving the claim.
3. Step 2 contradicts the nonemptiness of \(T_N\).
Because the sets are nested, \[ T_N\subseteq T_1. \] Step 2 says that every point of \(T_1\) lies outside \(T_N\). Therefore \[ T_N \subseteq T_1\cap(X\setminus T_N) = \varnothing. \] Thus \(T_N=\varnothing\), contrary to the hypothesis that every \(T_n\) is nonempty.
4. Therefore\[ \boxed{ \bigcap_{n=1}^{\infty}T_n\ne\varnothing. } \]
The contrary assumption in step 1 leads to the contradiction in step 3.
5. Q.E.D.
Step 4 is the required conclusion.