Proving that a Closed Subset of a Complete Metric Space is Compact if and only if it is Totally Bounded

In the realm of metric spaces, the concept of compactness plays a crucial role in understanding the structure and properties of subsets. For a closed subset of a complete metric space, the equivalence between compactness and total boundedness is a fundamental theorem. This article delves into the proof of this equivalence, breaking down each step to provide a clear and comprehensive understanding.

Introduction to Metric Spaces and Compactness

A metric space is a set equipped with a distance function (metric) that defines the distance between any two points in the set. A subset of a metric space is said to be compact if every open cover of the subset has a finite subcover. On the other hand, a subset is totally bounded if, for every positive real number ε, there exists a finite number of open balls of radius ε that cover the subset.

Proving the Equivalence: From Compact to Totally Bounded

Let's begin by proving the first implication: if a closed subset A of a complete metric space X is compact, then it is totally bounded.

Implication 1: If A is compact then A is totally bounded.

By definition, every open cover of A has a finite subcover. To show that A is totally bounded, we need to show that for every ε > 0, there exists a finite set of points in A such that A can be covered by open balls of radius ε.

Consider an arbitrary ε > 0. Let the collection of open balls { B(x, ε) : x ∈ A } form an open cover of A. Since A is compact, there exists a finite subcover from this collection. Therefore, there exist points x_1, x_2, ..., x_n ∈ A such that:

A ? ∪_{i1}^n B(x_i, ε).

This shows that A can be covered by finitely many balls of radius ε, hence A is totally bounded.

Proving the Equivalence: From Totally Bounded to Compact

Now, let's prove the second implication: if A is totally bounded, then it is compact.

Implication 2: If A is totally bounded then A is compact.

Assume that A is totally bounded. We need to show that A is compact, i.e., every open cover of A has a finite subcover.

Since A is totally bounded, for every ε > 0, there exists a finite set of points x_1, x_2, ..., x_n ∈ A such that:

A ? ∪_{i1}^n B(x_i, ε).

Consider a sequence of open covers of A by balls of radius ε. In particular, let ε 1/n for n 1, 2, ....

For each n, since A is totally bounded, there exist points x_1^n, x_2^n, ..., x_{k_n}^n ∈ A such that:

A ? ∪_{i1}^{k_n} B(x_i^n, 1/n).

Now, since A is closed in a complete space, it is sequentially compact. Therefore, every sequence in A has a convergent subsequence whose limit lies in A.

If we take any open cover of A, we can find a suitable ε-net from the finite covering above. The finite covering can be adjusted to ensure that we can extract a finite subcover from the open cover.

This shows that every open cover of A has a finite subcover, hence A is compact.

Conclusion

Combining both implications, we conclude that a closed subset A of a complete metric space is compact if and only if it is totally bounded. This equivalence is a powerful tool in understanding the structure of subsets in complete metric spaces and is widely used in various areas of mathematics, including functional analysis and topology.

Related Keywords

Compact subset, complete metric space, totally bounded, proof, sequential compactness