Let \(E_1,E_2,\ldots\) be nonempty closed subsets of a complete metric space \((X,d)\) such that \[ E_{n+1}\subseteq E_n \] for every positive integer \(n\), and \[ \lim_{n\to\infty}\operatorname{diam}(E_n)=0, \qquad \operatorname{diam}(E)=\sup\{d(x,y):x,y\in E\}. \] Prove that \[ \bigcap_{n=1}^{\infty}E_n\ne\varnothing. \]
Solution
Choose \(x_n\in E_n\) for every \(n\).
1. The sequence \((x_n)\) is Cauchy.
Let \(\varepsilon>0\). Since \(\operatorname{diam}(E_n)\to0\), choose \(N\) such that \[ \operatorname{diam}(E_N)<\varepsilon. \] If \(m,n\geq N\), nestedness gives \[ x_m\in E_m\subseteq E_N, \qquad x_n\in E_n\subseteq E_N. \] Therefore \[ d(x_m,x_n) \leq \operatorname{diam}(E_N) < \varepsilon. \] Thus \((x_n)\) is Cauchy.
2. There exists \(x\in X\) such that \(x_n\to x\).
The metric space \(X\) is complete, and \((x_n)\) is Cauchy by step 1.
3. For every positive integer \(k\), \(x\in E_k\).
2. Since \(E_k\) is closed, it contains the limit \(x\).
4.\[ x\in\bigcap_{n=1}^{\infty}E_n. \]
This follows from step 3, since \(x\) belongs to every \(E_k\).
5. Q.E.D.
Step 4 exhibits an element of the required intersection.