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Proof

(related to Theorem: Nested Closed Subset Theorem)

• Let $(X,d)$ be a metric space.
• We will first show that there is at most one point of intersection.
  • Because for the sequence of non-empty subsets of \(X\) with $$A_0\supset A_1\supset A_2\supset A_3\supset \ldots\quad ( * )$$ the corresponding sequence of diameters \((\operatorname{diam}(A_n))_{n\in\mathbb N}\) is converging against $0$, formally $$\lim_{n\to\infty}\operatorname{diam}(A_n)=0,\quad ( * * )$$ there are more than one point in the intersection $$\bigcap_{n=0}^\infty A_n$$.
  • Otherwise, let \(x\) and \(y\) be at least two points in the intersection.
  • Because $X$ is a metric space, it is also a Hausdorff space.
  • Therefore, because \(x\neq y\), we would have \(d(x,y) > 0\), contradicting $( * * )$.
  • Therefore, there is at least one point of intersection.
• It remains to be shown that such single point of intersection exists.
  • Choose a point \(x_n\in A_n\).
  • Because \[d(x_n,x_m)\le \operatorname{diam}(A_N)\quad\text{ for }n,m\ge N,\] the sequence $(x_n)_{n\in\mathbb N}$ is a Cauchy sequence in \(X\).
  • Because \(X\) is complete by hypothesis, the Cauchy sequence has a limit \(x\in X\).
  • Because \(x_n\in A_n\) for all \(n\in\mathbb N\), it follows from the characterization of closed sets by limits of convergent sequences that \(x \in A\).
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References

Bibliography

1. Forster Otto: "Analysis 2, Differentialrechnung im \(\mathbb R^n\), Gewöhnliche Differentialgleichungen", Vieweg Studium, 1984