Closed Subset Of Compact Set Is Compact

Closed Subset of a Compact Set is Compact: A Proof and Explanation

This article proves the fundamental topological theorem stating that a closed subset of a compact set is itself compact. This theorem is crucial in analysis and topology, providing a powerful tool for establishing compactness in various contexts. We will provide a clear, concise proof and discuss its implications. Understanding this theorem helps solidify your understanding of compactness and its relationship with closed sets.

What is Compactness?

Before diving into the proof, let's refresh our understanding of compactness. A set is considered compact if every open cover of the set has a finite subcover. In simpler terms, you can always find a finite number of open sets from any collection that completely covers the original set. This is a significant property, often implying boundedness and "completeness" in a certain sense.

What is a Closed Set?

A closed set is a set that contains all its limit points. A limit point is a point that can be arbitrarily closely approximated by points within the set. Equivalently, a set is closed if its complement is open.

The Theorem: Closed Subset of a Compact Set is Compact

Theorem: Let K be a compact set, and let C be a closed subset of K. Then C is compact.

Proof:

1. Consider an arbitrary open cover of C: Let {Uα} be an open cover of C, meaning that every point in C is contained in at least one Uα.

2. Extend the cover to include K: Since C is a subset of K, we can extend the open cover {Uα} to a cover of K by adding the complement of C, denoted by C^c. The complement is open because C is closed. Therefore, {Uα} ∪ {C^c} is an open cover of K.

3. Finite Subcover for K: Since K is compact, there exists a finite subcover of K from {Uα} ∪ {C^c}. Let's denote this finite subcover as {Uα₁, Uα₂, ..., Uα_n, C^c} (some of the Uα_i might be absent if C^c covers part of K).

4. Finite Subcover for C: Note that the set C is disjoint from C^c. Therefore, every point in C must be contained within one of the Uα_i from the finite subcover. This means {Uα₁, Uα₂, ..., Uα_n} is a finite subcover of C.

5. Conclusion: We have shown that any open cover of C has a finite subcover. Therefore, C is compact.

Implications and Applications

This seemingly simple theorem has profound implications in various areas of mathematics. For instance, it allows us to easily demonstrate the compactness of certain sets by identifying them as closed subsets of known compact sets. This significantly simplifies many proofs and arguments within analysis, topology, and related fields. The theorem plays a critical role in proofs concerning continuous functions on compact sets, the Heine-Borel theorem, and many other essential results.

Further Exploration

This proof provides a solid foundation for understanding the relationship between compactness and closed sets. Exploring counterexamples (e.g., showing that a closed subset of a non-compact set need not be compact) can further deepen your understanding of these concepts. Delving into the different characterizations of compactness in metric spaces will offer additional insight into the power and versatility of this fundamental topological property.