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The Equivalence between Compact Metric Spaces and Complete, Totally Bounded Spaces 📂Topology

The Equivalence between Compact Metric Spaces and Complete, Totally Bounded Spaces

Theorem 1

A metric space is compact if and only if it is complete and totally bounded.

Proof

()(\Rightarrow)

Suppose the metric space XX is compact.

Properties of complete metric spaces: If (X,d)(X,d) is a metric space and KXK \subset X,

  • XXX \subset X is a closed set in XX, so it is a complete space.
  • Since the closed set XXX \subset X is compact, it is totally bounded.

()(\Leftarrow)

Suppose the metric space XX is a totally bounded space with completeness.

Pointwise compactness: KXK \subset X being precompact (pointwise compact) means for all sequences {xn}K\left\{ x_{n} \right\} \subset K defined in KK, there exists a convergent subsequence {xn}{xn}\left\{ x_{n '} \right\} \subset \left\{ x_{n} \right\} that converges to xXx \in X.

Proving that XXX \subset X is sequentially compact, i.e., precompact, implies that X=X\overline{X} = X, thus proving XX is compact, is the same as proving XX is sequentially compact.


Part 1.

Since XX is a totally bounded space, any ε>0\varepsilon > 0 can cover XX with a finite number of balls of radius ε\varepsilon.

Considering any sequence {xn}nNX\left\{ x_{n} \right\}_{n \in \mathbb{N}} \subset X defined in XX, at least one of the finite balls covering XX must contain infinitely many points of {xn}nN\left\{ x_{n} \right\}_{n \in \mathbb{N}}.


Part 2. J1NJ_{1} \subset \mathbb{N}

Given a radius ε=1\varepsilon = 1, at least one ball in the finite covering of XX contains infinitely many points of {xn}nN\left\{ x_{n} \right\}_{n \in \mathbb{N}}. Let’s call this ball B1B_{1} and pick the set of indices J1J_{1} that belong to xnx_{n} included in it. Formally, it is as follows. J1:={nN:xnB1} J_{1} := \left\{ n \in \mathbb{N} : x_{n} \in B_{1} \right\}


Part 3. JkNJ_{k} \subset \mathbb{N}

Following the method we used to select B1B_{1} and J1J_{1}, let’s pick BkB_{k} and JkJ_{k} for ε=1k\varepsilon = {{ 1 } \over { k }}. Since the method to define JkJ_{k} is the same as J1J_{1}, all these sets are infinite, and we must reduce ε=1k\varepsilon = {{ 1 } \over { k }} as {xn}nN\left\{ x_{n} \right\}_{n \in \mathbb{N}} converges there, thus the following holds. J1J2 J_{1} \supset J_{2} \supset \cdots


Part 4.

Choose an n1J1n_{1} \in J_{1}, and for each Jk+1J_{k+1}, select nk+1Jk+1n_{k+1} \in J_{k+1} such that nk+1>nkn_{k+1} > n_{k} holds. The feasibility of this selection is justified by Part 3. Consequently, for all i,jki,j \ge k, xnix_{n_{i}} and xnjx_{n_{j}} belong to BkB_{k} with a radius of 1/k1/k. Hence, the subsequence {xnk}kN\left\{ x_{n_{k}} \right\}_{k \in \mathbb{N}} of {xk}kN\left\{ x_{k} \right\}_{k \in \mathbb{N}} becomes a Cauchy sequence, and since XX is assumed to be complete, this Cauchy sequence must converge to xXx \in X. Therefore, XXX \subseteq X is sequentially compact, and compact.


  1. Munkres. (2000). Topology(2nd Edition): p276. ↩︎