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
Suppose the metric space is compact.
Properties of complete metric spaces: If is a metric space and ,
- [1]: is a complete subspace. where is a closed set.
- [2]: is a totally bounded space where the closed set is compact.
- is a closed set in , so it is a complete space.
- Since the closed set is compact, it is totally bounded.
Suppose the metric space is a totally bounded space with completeness.
Pointwise compactness: being precompact (pointwise compact) means for all sequences defined in , there exists a convergent subsequence that converges to .
Proving that is sequentially compact, i.e., precompact, implies that , thus proving is compact, is the same as proving is sequentially compact.
Part 1.
Since is a totally bounded space, any can cover with a finite number of balls of radius .
Considering any sequence defined in , at least one of the finite balls covering must contain infinitely many points of .
Part 2.
Given a radius , at least one ball in the finite covering of contains infinitely many points of . Let’s call this ball and pick the set of indices that belong to included in it. Formally, it is as follows.
Part 3.
Following the method we used to select and , let’s pick and for . Since the method to define is the same as , all these sets are infinite, and we must reduce as converges there, thus the following holds.
Part 4.
Choose an , and for each , select such that holds. The feasibility of this selection is justified by Part 3. Consequently, for all , and belong to with a radius of . Hence, the subsequence of becomes a Cauchy sequence, and since is assumed to be complete, this Cauchy sequence must converge to . Therefore, is sequentially compact, and compact.
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Munkres. (2000). Topology(2nd Edition): p276. ↩︎