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Theorem [4.1.11] Let A and B be subsets of topological spaces X and Y respectively. Then Int(A × B) = Int(A) × Int(B). %3D Proof : Since Int(A) is an open set contained in A, and Int(B) is an open set contained in B, it follows that Int(A) x Int(B) is an open set in the product topology and is containd in A x B. Thus Int(A) x Int(B) S Int(A × B). Now we must prove that Int(A x B) C Int(A) x Int(B). H.W

Theorem [4.1.11] Let A and B be subsets of topological spaces X and Y respectively. Then Int(A × B) = Int(A) × Int(B). %3D Proof : Since Int(A) is an open set contained in A, and Int(B) is an open set contained in B, it follows that Int(A) x Int(B) is an open set in the product topology and is containd in A x B. Thus Int(A) x Int(B) S Int(A × B). Now we must prove that Int(A x B) C Int(A) x Int(B). H.W

Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter7: Distance And Approximation
Section7.4: The Singular Value Decomposition
Problem 54EQ
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Theorem [4.1.11] Let A and B be subsets of topological spaces X and Y respectively. Then Int(A × B) = Int(A) × Int(B). %3D Proof : Since Int(A) is an open set contained in A, and Int(B) is an open set contained in B, it follows that Int(A) x Int(B) is an open set in the product topology and is containd in A x B. Thus Int(A) x Int(B) S Int(A × B). Now we must prove that Int(A x B) C Int(A) x Int(B). H.W
Transcribed Image Text:Theorem [4.1.11] Let A and B be subsets of topological spaces X and Y respectively. Then Int(A × B) = Int(A) × Int(B). %3D Proof : Since Int(A) is an open set contained in A, and Int(B) is an open set contained in B, it follows that Int(A) x Int(B) is an open set in the product topology and is containd in A x B. Thus Int(A) x Int(B) S Int(A × B). Now we must prove that Int(A x B) C Int(A) x Int(B). H.W
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