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 ${C_{k_m} \varsubsetneq C_1}$. $\Box$

TheoremĀ 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.