Compact Sets in Metric Spaces are Complete
Compact Sets in Metric Spaces are Complete
Recall from the Complete Metric Spaces page that a metric space $(M, d)$ is said to be complete if every Cauchy sequence in $M$ converges in $M$. Furthermore, if $S \subseteq M$ then we said that $S$ is complete if every Cauchy sequence in $S$ converges in $S$.
We will now look at an important theorem which says that if $S \subseteq M$ and if $S$ is a compact set then $S$ is also complete.
| Theorem 1: Let $(M, d)$ be a metric space and let $S \subseteq M$. If $S$ is compact then $S$ is complete. |
- Proof: Suppose that $S \subseteq M$ is a compact set. Let $(x_n)_{n=1}^{\infty}$ be a Cauchy sequence in $S$, and let $X \subseteq S$ be the set of terms in this sequence.
- Suppose that $X$ is finite. Then clearly $(x_n)_{n=1}^{\infty}$ must converge to some $p \in X$ since by the definition of a Cauchy sequence we must have that for every $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ then $d(x_m, x_n) < \epsilon$, i.e., given any $\epsilon > 0$ there exists a tail of the sequence $(x_n)_{n=1}^{\infty}$ such that the distance between any terms in the tail of the sequence is less than $\epsilon$.
- Instead suppose that $X$ is an infinite set. Since $S$ is a compact set we by the Every Infinite Subset of a Compact Set in a Metric Space Contains an Accumulation Point page that $X$ contains an accumulation point $p \in X$. We will show that $(x_n)_{n=1}^{\infty}$ converges to $p$.
- Since $(x_n)_{n=1}^{\infty}$ is a Cauchy sequence we have that for $\epsilon_1 = \frac{\epsilon}{2} > 0$ there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ then:
\begin{align} \quad d(x_m, x_n) < \epsilon_1 = \frac{\epsilon}{2} \end{align}
- Furthermore, since $p$ is an accumulation point of $X$ we have that for all $r > 0$ that $B(p, r) \cap X \setminus \{ x \} \neq \emptyset$, so for $r = \epsilon_1 = \frac{\epsilon}{2} > 0$ there exists an $x_m \in X$ with $m \geq N$ such that:
\begin{align} \quad x_m \in B(p, \epsilon_1) \cap X \setminus \{ x \} \neq \emptyset \end{align}
- Hence $d(x_m, p) \leq \epsilon_1 = \frac{\epsilon}{2}$. By the triangle inequality, we have that if $n \geq N$ then:
\begin{align} \quad d(x_n, p) \leq d(x_n, x_m) + d(x_m, p) < \epsilon_1 + \epsilon_1 = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}
- Therefore $(x_n)_{n=1}^{\infty}$ converges to $p \in X \subseteq S$, and so every Cauchy sequence of $S$ converges in $S$, so $S$ is complete. $\blacksquare$