The Cantor Intersection Theorem for Complete Metric Spaces
The Cantor Intersection Theorem for Complete Metric Spaces
Theorem 1 (The Cantor Intersection Theorem for Complete Metric Spaces): Let $X$ be a complete metric space. If $(x_n)_{n=1}^{\infty}$ is a sequence of points in $X$ and $(r_n)_{n=1}^{\infty}$ is a sequence of positive real numbers with $\displaystyle{\lim_{n \to \infty} r_n = 0}$ and such that $... \subseteq \overline{B}(x_{n+1}, r_{n+1}) \subseteq \overline{B}(x_n, r_n) \subseteq ... \subseteq \overline{B}(x_1, r_1)$ then the intersection of these closed balls in nonempty and more precisely, there exists a point $x \in X$ such that $\displaystyle{\bigcap_{n=1}^{\infty} \overline{B} (x_n, r_n) = \{ x \}}$. |
- Proof: We first show that the sequence $(x_n)_{n=1}^{\infty}$ given in the theorem is a Cauchy sequence. Let $\epsilon > 0$. Let $N \in \mathbb{N}$ be chosen such that $r_N < \epsilon$. Then if $m, n \geq N$ with $m \geq n$ we have that $\overline{B} (x_m, r_m) \subseteq \overline{B} (x_n, r_n)$ and so:
\begin{align} \quad d(x_m, x_n) \leq r_m < r_N < \epsilon \end{align}
- Therefore $(x_n)_{n=1}^{\infty}$ is indeed a Cauchy sequence in $X$. Since $X$ is a Cauchy sequence, $(x_n)_{n=1}^{\infty}$ converges to some $x \in X$. Since $(x_n)_{n \geq m} \subseteq \overline{B}(x_m, r_m)$ for every $m \in \mathbb{N}$ we have that:
\begin{align} \quad x \in \bigcap_{n=1}^{\infty} \overline{B}(x_n, r_n) \end{align}
- We now show that this intersection of closed balls contains only the point $x$. Suppose that $\displaystyle{x, y \in \bigcap_{n=1}^{\infty} \overline{B}(x_n, r_n)}$. Then for all $n \in \mathbb{N}$ we have that:
\begin{align} \quad d(x, y) \leq d(x, x_n) + d(x_n, y) < 2r_n \end{align}
- By taking the limit as $n \to \infty$ we see that $d(x, y) = 0$. So $x = y$. Hence:
\begin{align} \quad \bigcap_{n=1}^{\infty} \overline{B} (x_n, r_n) = \{ x \} \quad \blacksquare \end{align}