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Show that a finite union of compact subspaces is compact.

Show that a finite union of compact subspaces is compact.

Elementary Linear Algebra (MindTap Course List)
Elementary Linear Algebra (MindTap Course List)
8th Edition
ISBN:9781305658004
Author:Ron Larson
Publisher:Ron Larson
Chapter4: Vector Spaces
Section4.3: Subspaces Of Vector Spaces
Problem 56E: Give an example showing that the union of two subspaces of a vector space V is not necessarily a...
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**Prove that a Finite Union of Compact Subspaces is Compact** In this section, we will demonstrate that the union of a finite number of compact subspaces results in a compact space. This is an important concept in topology, as compactness is a key property that extends many fundamental results from finite-dimensional spaces to more general settings. Let's delve into the proof: 1. **Definition Review**: Recall that a subspace is compact if every open cover has a finite subcover. 2. **Finite Union Consideration**: Suppose \( K_1, K_2, \ldots, K_n \) are compact subspaces of a topological space. We need to prove that the union \( K = K_1 \cup K_2 \cup \ldots \cup K_n \) is compact. 3. **Proof Strategy**: To demonstrate that \( K \) is compact, consider an open cover \( \{ U_\alpha \}_{\alpha \in A} \) of \( K \). Since \( K = \bigcup_{i=1}^n K_i \), each \( K_i \) must be covered by \( \{ U_\alpha \}_{\alpha \in A} \). 4. **Application of Compactness**: For each \( K_i \), because \( K_i \) is compact, there exists a finite subcover \( \{ U_{\alpha_{i1}}, U_{\alpha_{i2}}, \ldots, U_{\alpha_{im_i}} \} \). 5. **Combine Subcovers**: Consider the union of all these finite subcovers from each \( K_i \). This union will be a finite collection since there are a finite number of sets \( K_i \). 6. **Conclusion**: The combined finite subcovers provide a finite subcover for \( K \) itself. Hence, \( K \) is compact. Understanding this result is fundamental in mathematical analysis and topology, where the compactness of a space is often a key assumption for theorems concerning continuity, convergence, and more.
Transcribed Image Text:**Prove that a Finite Union of Compact Subspaces is Compact** In this section, we will demonstrate that the union of a finite number of compact subspaces results in a compact space. This is an important concept in topology, as compactness is a key property that extends many fundamental results from finite-dimensional spaces to more general settings. Let's delve into the proof: 1. **Definition Review**: Recall that a subspace is compact if every open cover has a finite subcover. 2. **Finite Union Consideration**: Suppose \( K_1, K_2, \ldots, K_n \) are compact subspaces of a topological space. We need to prove that the union \( K = K_1 \cup K_2 \cup \ldots \cup K_n \) is compact. 3. **Proof Strategy**: To demonstrate that \( K \) is compact, consider an open cover \( \{ U_\alpha \}_{\alpha \in A} \) of \( K \). Since \( K = \bigcup_{i=1}^n K_i \), each \( K_i \) must be covered by \( \{ U_\alpha \}_{\alpha \in A} \). 4. **Application of Compactness**: For each \( K_i \), because \( K_i \) is compact, there exists a finite subcover \( \{ U_{\alpha_{i1}}, U_{\alpha_{i2}}, \ldots, U_{\alpha_{im_i}} \} \). 5. **Combine Subcovers**: Consider the union of all these finite subcovers from each \( K_i \). This union will be a finite collection since there are a finite number of sets \( K_i \). 6. **Conclusion**: The combined finite subcovers provide a finite subcover for \( K \) itself. Hence, \( K \) is compact. Understanding this result is fundamental in mathematical analysis and topology, where the compactness of a space is often a key assumption for theorems concerning continuity, convergence, and more.
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