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

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