Nested Sphere Theorem

Theorem
Let $M = \left({A, d}\right)$ be a complete metric space.

Let $\left\langle{S_n}\right\rangle$ be a sequence of closed balls in $M$ defined by:


 * $S_n = B^-_{\rho_n} \left({x_n}\right)$

where $\rho_n \to 0$ as $n \to \infty$ and:


 * $S_1 \supseteq S_2 \supseteq \cdots \supseteq S_n \supseteq \cdots$

Then there exists $x \in A$ such that:


 * $\displaystyle \bigcap_{n \mathop = 1}^\infty S_n = \left\{{x}\right\}$

Proof
Let $S_n = B^-_{\rho_n} \left({x_n}\right)$ be the closed ball of radius $\rho_n$ about the point $x_n$.

That is, let $S_n = \left\{{x \in A: d \left({x_n, x}\right) \le \rho_n}\right\}$.

Then the sequence $\left\langle{x_n}\right\rangle$ forms a Cauchy sequence:


 * $d \left({x_n, x_{n+p}}\right) < \rho_n$

for any $p \ge 0$ since $S_{n+p} \subseteq S_n$.

However, $\rho_n \to 0$ as $n \to \infty$ and therefore $d \left({x_n, x_{n+p}}\right) \to 0$ as $n \to \infty$ for any $p \ge 0$.

Since the space $M$ is complete, there exists $x \in X$ such that $x_n \to x$ as $n \to \infty$.

Then since the subsequence $\left\langle{x_k}\right\rangle_{k = n}^\infty$ is contained entirely in $S_n$ and converges to $x$, $x$ is a member of the closure of $S_n$ by Closure of Subset of Metric Space by Convergent Sequence.

Since $S_n$ is closed, by Closed Set Equals its Closure, we have:
 * $\forall n \in \N: x \in S_n$

Hence:
 * $\displaystyle x \in \bigcap_{n \mathop = 1}^\infty S_n$

Suppose now that $y \ne x$.

Then it follows that $d \left({x, y}\right) > 0$, and hence that for some $n$:


 * $d \left({x, y}\right) > 2 \rho_n$

Since $x, x_n \in S_n$, it then follows that:


 * $d \left({x, x_n}\right) \le \rho_n$

Now:

From the above, it follows that:


 * $d \left({x_n, y}\right) > \rho_n$

so that $y \notin S_n$, and consequently:


 * $\displaystyle y \notin \bigcap_{i \mathop = 1}^\infty S_n$

Hence:


 * $\displaystyle \bigcap_{n \mathop = 1}^\infty S_n = \left\{{x}\right\}$