Compact Set

Definition. Compact Set

A set $S \subset R$ is compact if every sequence in $S$ has a convergent sub-sequence whose limit is in $S$.

Theorem. $S$ is compact if and only if it is closed and bounded.

Proof:

1. $"\Rightarrow"$: Suppose $S$ is not bounded. Then there is a sequence $(a_n)$ is $S$, such that $|a_n| \ge n$, $\forall n \in N$. Now every sub-sequence in $(a_n)$ diverges. Therefore $S$ must be bounded. Next we show that $S$ contains all its limit points. Let $a$ be a limit point of $S$. Then there is a sequence $(a_n) \rightarrow a$. $(a_n)$ is a sub-sequence of itself and so $a \in S$. We have proved that $S$ is closed.
2. $"\Leftarrow"$: Assume $S$ is closed and bounded. Consider a sequence $(a_n)$ in $S$. By Weierstrass Theorem, there is a convergent sub-sequence $( a_{n_k} )$. But since $S$ is closed, $\lim_k a_{n_k} \in S$.

Theorem. $S$ is compact if and only if every open cover of $S$ has a finite sub-cover.

Proof:

1. $"\Rightarrow"$: If $S$ is compact, it is bounded. We can find a closed interval $I_0$ that contains $S$. Suppose $S$ can not be covered by a finite sub-cover. We bisect $I_0$ into two closed intervals of equal length. At least one of them can not be covered with finite sets. Denote this interval $I_1$. We continue the process, resulting in a series of intervals $I_1 \supset I_2 \supset I_3 \supset ...$. Define $J_n = I_n \cap S \quad \forall n \in N$. The construction of $I_n$ ensures $J_n \neq \emptyset \quad \forall n \in N$ and the length of $I_n$ approaches $0$ as $n \rightarrow \infty$. We claim that $\cap_n J_n \neq \emptyset$. Indeed, there is an element $a_n \in J_n$ for $n \in N$. It not difficult to see $(a_n)$ is a Cauchy sequence since $\forall m > n \ge N$, $a_n, a_m \in I_N$ and $|a_n - a_m| \le |I_N|$. Let $a = \lim_n a_n$. We conclude that $a \in J_n$ for $n \in N$. This is because $a_{n + 1}, a_{n + 2}..., \in J_n$ for all $n \in N$. It follows that $a \in \cap_n J_n$. Now $a$ must be covered by some open set $O_a$. On the other hand, the construction of $I_n$ ensures that there is an integer $N_a$ such that $I_{N_a}$ is small enough such that $I_{N_a} \subset O_a$. Contradicting the assumption that $I_{N_a}$ cannot be covered with finite sets.

2. $"\Leftarrow"$: Consider the collection of open balls $\{ B(O, 1),\ B(O, 2),\ B(O, 3), ...\}$ centered at $0$ with radii $1, 2, 3, ...$. It covers $S$ and has a finite sub-cover. It follows that $S$ is bounded by the ball with largest radius in the finite sub-cover. To show $S$ is closed, it suffices to prove it complement $S^c$ is open. Let $a \in S^c$. Take the collection of closed balls $\{ \bar B(a, 1),\ \bar B(a, 1/2),\ \bar B(a, 1/3), ...\}$ centered at $a$ with radii $1, 1/2, 1/3, ..., 1/n, ...$. Their complements $\{ \bar B(a, 1)^c,\ \bar B(a, 1/2)^c,\ \bar B(a, 1/3)^c, ...\}$ constitute an open cover of $S$. By assumption we can find a finite subset of them that covers $S$. Let $\bar B(a, \epsilon)^c$ be the one with the smallest radius in the finite subset collection. We learn that $B(a, \epsilon) \subset S^c$ and therefore $S^c$ is open.