# 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.

Github: ### References

#### Bibliography

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