# Equivalence of compactness in metric space

Lemma In a metric space ${(X,d)}$, if ${\{C_k\}_{k=1}^\infty \subset X}$ is a sequence of compact sets such that ${C_{k+1} \varsubsetneq C_k}$ for ${k \in {\mathbb N}}$ and ${\lim_{k \rightarrow \infty}}$ diam${C_k = 0}$. Then ${\cap_{k=1}^\infty C_k = \{c\}}$, where ${c}$ is a point in ${X}$.

Proof: First note that ${\cap_{k=1}^\infty C_k}$ cannot contain more than 1 points. If ${a,b \in \cap_{k=1}^\infty C_k}$, then ${a,b \in C_k \quad \forall k}$. Since ${\lim}$ diam${C_k}$=0, it forces ${a=b}$.

It remains to show ${\cap_{k=1}^\infty C_k}$ is nonempty. Assume it is empty, then ${\cup_{k=1}^\infty (X \setminus C_k ) = X \setminus (\cap_{k=1}^\infty C_k) = X}$. Since metric space is Hausdorff, ${C_k}$ is closed for all ${k}$ and ${\{ X \setminus C_k \}}$ is an open cover for ${C_1}$. Since ${C_1}$ is compact, there exists finite subcover such that ${C_1 \subset \cup_{i=1}^m (X\setminus C_{k_i}) = X \setminus C_{k_m}}$, which contradicts ${C_{k_m} \varsubsetneq C_1}$. $\Box$

Theorem In a metric space ${(X,d)}$ (no matter it has countable basis or not), let ${E}$ be a subset in ${X}$. The following statement are equivalent by assuming axiom of choice:

1. ${E}$ is compact;
2. ${E}$ is sequentially compact;
3. ${E}$ is complete and totally bounded.

I will prove ${(1) \Rightarrow (2)}$, ${(2) \Leftrightarrow (3)}$ and then ${(2) + (3) \Rightarrow (1)}$.

Proof: ${(1) \Rightarrow (2)}$:

Let ${\{a_n\}}$ be a sequence in ${E}$, we want to show that it has convergent subsequence. Denote by ${B(x,r)}$ the ball centered at ${x}$ of radius ${r}$. Then ${E \subset \cup_{x\in E} B(x,1)}$, by compactness ${E \subset \cup_{i=1}^m B(x_{i},1)}$. Therefore there must exist a ${B(x_i,1)}$ for some ${i}$ that contains infinite elements of ${\{a_n\}}$. Denote by ${A_1}$ the intersection of that ball and ${E}$. Use balls of radius ${\frac{1}{2}}$ to cover ${E}$, again there exists a ${B(x,{\frac{1}{2}})}$ such that ${A_2: = A_1 \cap B(x,{\frac{1}{2}})}$ contains infinite elements of ${\{a_n\}}$. Repeating this process we will have a sequence of sets ${\{A_k\}}$. Pick an element from ${\{a_n\}}$ in each ${A_k}$, we will have a cauchy subsequence, still denote by ${\{a_n\}}$. Since ${A_k \subset E}$ for all ${k}$, and ${E}$ is compact, then ${\bar{A_k}}$ is compact for all ${k}$. By the above lemma, ${\cap_{k=1}^\infty \bar{A_k} = \{a\}}$ for some ${a \in E}$. Therefore we know the cauchy subsequence ${\{a_n\}}$ converges to ${a}$.

${(2) \Rightarrow (3)}$:

If ${E}$ is sequentially compact, every Cauchy sequence in ${E}$ converges to a point in ${E}$, which means ${E}$ is complete. To show ${E}$ is also totally bounded, assume it is not, then there exists an ${\varepsilon > 0}$ such that ${E}$ cannot be covered by finitely many open balls of radius ${\varepsilon}$. Pick an ${x_1 \in E}$, and pick ${x_2 \in E}$ but not in ${B(x_1 , \varepsilon)}$. And then pick ${x_3 \in E}$ that is not in ${B(x_1, \varepsilon) \cup B(x_2,\varepsilon)}$. Since ${E}$ cannot be covered by finitely many balls of radius ${\varepsilon}$, we can find a sequence of open balls ${\{ B(x_i , \varepsilon) \}_{i \in {\mathbb N}}}$. Notice that

$\displaystyle \text{dist} (x_m , x_n) > \varepsilon \quad \text{ for any } m \neq n,$

the sequence ${\{ x_i \}}$ will not have convergent subsequence. So ${E}$ must be totally bounded.

${(3) \Rightarrow (2)}$:

If ${E}$ is complete and totally bounded. Given any sequence ${\{ x_i \}}$ in ${E}$, first we cover the set ${E}$ by finite open balls of radius ${1}$, there must be one ball, say ${B_1}$, contains infinite elements of the sequence. Then we cover the set by balls of radius ${\frac{1}{2}}$, again there must be one ball, say ${B_2}$, such that ${B_1 \cap B_2}$ contains infinite elements of the sequence. Keep doing the process we will have a Cauchy subsequence of ${\{ x_i \}}$, and it converges to some point in ${E}$ since ${E}$ is complete. Hence ${E}$ is sequentially compact.

${(2)+(3) \Rightarrow (1)}$:

Assume we have an open cover ${\{O_\alpha\}_{\alpha \in I}}$ for ${E}$, we want to show there exists a finite subcover. Since ${E}$ is totally bounded, it suffices to show there exists an ${\varepsilon >0}$ such that ${\forall x \in E, B(x, \varepsilon) \subset O_\alpha}$ for some ${\alpha \in I}$. This is because ${E}$ can be covered by finitely many ${B(x, \varepsilon)}$, and hence can be covered by finitely many ${O_\alpha}$. Now assume for all ${\varepsilon >0}$, there exists an ${x \in E}$ such that ${B(x,\varepsilon)}$ is not contained in any ${O_\alpha}$. Take ${\varepsilon = \frac{1}{n}}$ for ${n \in {\mathbb N}}$, we then have a sequence ${\{x_n\}}$ such that ${B(x_n, \frac{1}{n})}$ is not contained in any ${O_\alpha}$. By sequential compactness, there exists a convergent subsequence ${\{x_{n_k} \}}$ converges to a point ${x}$. Since ${\{O_\alpha\}}$ is an open cover, ${x \in O_{\alpha_x}}$ for some ${\alpha_x \in I}$. Since ${ O_{\alpha_x}}$ is open, if we take ${N}$ large such that ${\frac{1}{N} < \frac{1}{2} d(x, O_{\alpha_x})}$, ${B(x_N , \frac{1}{N}) \subset O_{\alpha_x}}$, which contradicts our assumption. Therefore there exists an ${\varepsilon >0}$ such that ${\forall x \in E, B(x, \varepsilon) \subset O_\alpha}$ for some ${\alpha \in I}$, and hence there is a finite subcover for ${E}$. $\Box$