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
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3ae7bfd5-e7cb-40c6-b95d-6ac20a660c18%2Fb19cd02c-1fd5-48aa-b335-ff531b66cc5b%2Fh4mmvem_processed.jpeg&w=3840&q=75)
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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