Proving Cauchy Sequences Converge in Complete Metric Spaces

Understanding the behavior of sequences in metric spaces is crucial in analysis. One of the key properties is whether every Cauchy sequence converges. In this article, we delve into the proof that every Cauchy sequence converges in a complete metric space. We start with the definitions and then provide a detailed proof outline.

Definitions

A Cauchy sequence in a metric space Xsubd/sub is a sequence xn such that for every epsilon; 0, there exists an integer N such that for all m, n ≥ N, we have:

emdxm, xn/em epsilon;.

A sequence xn converges to a limit L in X if for every epsilon; 0, there exists an integer M such that for all n ≥ M, we have:

emdxn, L/em epsilon;.

A metric space Xsubd/sub is complete if every Cauchy sequence in X converges to a limit within X.

Proof Outline

To prove that every Cauchy sequence converges, we need to show that the space is complete. Here's a step-by-step guide to proving completeness:

Assume xn is a Cauchy sequence: Let xn be a Cauchy sequence in the metric space Xsubd/sub. Show that xn is bounded: Since xn is Cauchy, we can find some N such that for all m, n ≥ N, emdxm, xn/em 1. Choose epsilon; 1. Therefore, for n ≥ N, we have emdxn, xN/em 1. This implies that all xn for n ≥ N are within a distance of 1 from xN thus xn is bounded. Construct a limit point: Since xn is Cauchy, it has a subsequence that converges to some limit L in the completion of the metric space if it exists. The limit point L is in the completion of the metric space. Show convergence of the sequence: For any epsilon; 0, since xn is Cauchy, there exists N such that for all m, n ≥ N, emdxm, xn/em epsilon;. Thus for n ≥ N, we can show that emdxn, L/em epsilon;. This follows because for large enough n, xn gets arbitrarily close to the limit L.

Conclusion: Since the limit point L belongs to the metric space X, we conclude that the Cauchy sequence xn converges to L in X.

Example of a Non-Complete Space

Consider the space of rational numbers mathbb{Q} with the standard metric. This space is not complete. The sequence defined by the decimal expansion of , which is a Cauchy sequence in mathbb{Q}, does not converge in mathbb{Q} because sqrt{2} is not a rational number.

Summary

In summary, every Cauchy sequence converges in a complete metric space. The proof involves showing that the sequence is bounded, identifying a limit point in the completion of the space, and demonstrating that the sequence converges to that limit.