MATH527_HW2

# MATH527_HW2 - Serge Ballif MATH 527 Homework 2 Problem 1...

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Serge Ballif MATH 527 Homework 2 September 14, 2007 Problem 1. Consider a map h : ω ω , h ( x 1 , x 2 , x 3 , . . . ) = ( x 1 , 4 x 2 , 9 x 3 , . . . ) . (a) Is h continuous, when ω is given by the product topology? (b) Is h continuous, when ω is given by the box topology? (c) Is h continuous, when ω is given by the uniform topology? (a) Yes, h is continuous. To show this, we show that the inverse image under h of a basis set in ω is open in ω . An arbitrary basis set in ω has the form U = Q i =1 U i for intervals U i = ( a i , b i ) where the two equations a i = -∞ and b i = hold except for finitely many values of i . The calculation h - 1 ( { ( y 1 , y 2 , . . . , y i , . . . ) } ) = { ( y 1 , 1 4 y 2 , . . . , 1 i 2 y i , . . . ) } shows that h - 1 ( U ) = Q i =1 ( 1 i 2 a i , 1 i 2 b i ). This inverse image is also a product of open intervals of which only finitely many are not all of . Hence, h - 1 ( U ) is another basis element of ω . Thus the inverse image of a basis element is open under the map h . Therefore, h is continuous. (b) Yes, h is continuous. The inverse image under h of a basis set U = Q i =1 ( a i , b i ) is the basis set (and hence an open set) h - 1 ( U ) = Q i =1 ( 1 i 2 a i , 1 i 2 b i ). Hence, h is continuous. (c) No, h is not continuous. To show h is not continuous, we will show that the inverse image of an open ball does not contain an open ball (and hence is not open). An basis element of ω in the uniform topology is of the form B ¯ p ( x, ) = [ δ< Y i =1 ( x i - δ, x i + δ ) . The property h - 1 ( A α ) = h - 1 ( A α ) allows us to calculate h - 1 ( B ¯ p ( x, )) = [ δ< Y i =1 ( 1 i 2 ( x i - δ ) , 1 i 2 ( x i + δ ) ) .

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• Fall '07
• ROTMAN,REGINA
• Math, Topology, Topological space, inverse image, Serge Ballif

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