# 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.

# Kolmogorov-Riesz theorem

I will show an ${L^1}$ version of Arzela-Ascoli theorem, which is called Kolmogorov-Riesz theorem, by using a fact in topology as following.

Definition In a metric space, a set is totally bounded, if for any fixed ${\varepsilon>0}$, the set can be covered by finitely many open balls of radius ${\varepsilon}$.

Lemma In a metric space, a set ${E}$ is compact if and only if ${E}$ is complete and totally bounded.

Proof of the lemma can be find here.

Theorem 1 (Kolmogorov Riesz Theorem (finite measure domain)) Let ${\mathcal{F}}$ be a bounded set of ${L^1({\mathbb R}^n)}$ functions, Assume that ${\forall \varepsilon >0}$, there exists a ${\delta >0}$ such that

$\displaystyle \| f(\cdot + y) - f \|_{L^1({\mathbb R}^n)} \le \varepsilon$

whenever ${|y| < \delta}$, for all ${f \in \mathcal{F}}$. Then ${\mathcal{F}_{| \Omega}}$ has a compact closure in ${L^1(\Omega)}$ for any finite measure set ${\Omega \subset {\mathbb R}^n}$.