Introduction to Proving Compactness in Metric Spaces

Metric spaces play a crucial role in modern mathematics, and understanding the properties of compact subsets is fundamental in analysis. This article will delve into the proof that in a metric space, a compact subset is both closed and bounded. We will provide detailed explanations and examples to help you grasp these concepts effectively. Understanding these ideas is crucial for anyone working with metric spaces.

What is a Metric Space?

A metric space is a mathematical structure used to define the notion of distance or metric between elements of a set. Formally, a metric space is a pair (X, d), where X is a set and d is a metric or distance function that defines the distance between any two points in X. The metric, d, is a function d: X × X → R that satisfies the following properties for all x, y, z in X:

Non-negativity: d(x, y) ≥ 0, and d(x, y) 0 if and only if x y. Symmetry: d(x, y) d(y, x). Triangle Inequality: d(x, z) ≤ d(x, y) d(y, z).

Compactness in Metric Spaces

A subset K of a metric space (X, d) is called compact if every open cover of K has a finite subcover. This definition might be initially abstract, but it's a powerful concept. Essentially, compactness ensures that certain types of limits and convergence properties are preserved, making compact sets particularly well-behaved.

Proving a Compact Subset is Closed and Bounded

Step 1: Proving that a Compact Subset is Closed

Let K be a compact subset of a metric space (X, d). To prove that K is closed, we need to show that the complement Kc is open. A set is open if every point in the set has a neighborhood that lies entirely within the set.

Take any point p in Kc. Since p is not in K, the distance from p to K must be positive. Let r min{d(p, x) | x in K}. Since x is not in K, the distance is strictly positive.

Consider the open ball Bp(r/2) centered at p with radius r/2. Every point in this ball is at a distance greater than r/2 from K. Thus, no point in this ball is in K, so the ball is contained in Kc.

Since every point in Kc has an open ball around it that lies in Kc, Kc is open. Therefore, K is closed.

Step 2: Proving that a Compact Subset is Bounded

To prove that a compact subset K is bounded, let x be any point in K (we can choose x as the reference point). We will use the compactness of K to construct a finite subcover that will help us establish the boundedness.

For each n in the natural numbers, define the open ball Bn,x {y : d(x, y)

Consider the collection of open balls {Bn,x} as an open cover of K. Since K is compact, there must exist a finite subcover. Let m be the largest integer such that Bm,x is in this finite subcover. Then, K is contained in Bm,x, which is a ball with radius 1/m centered at x.

Thus, K is bounded, as every point in K is within a fixed distance 1/m from x.

The Converse: Bounded and Closed Sets in Metric Spaces

The converse is not true in general; a set that is both closed and bounded in a metric space is not necessarily compact. However, in certain metric spaces, such as Euclidean spaces, bounded and closed sets are indeed compact. This is a consequence of the Heine-Borel Theorem in Euclidean spaces.

Conclusion

In summary, we have explored the proof that a compact subset of a metric space is both closed and bounded. This proof involves showing that the complement of a compact set is open and demonstrating that a compact set can be covered by a finite number of open balls. The concept of compactness is a fundamental tool in analysis, and understanding these ideas is crucial for advanced mathematical work.

References

For further reading and detailed proofs, refer to:

Rudin, W. (1976). Principles of Mathematical Analysis (3rd ed.). McGraw-Hill. Munkres, J. R. (2000). Topology (2nd ed.). Prentice Hall.