solutionshw9-554-2010

solutionshw9-554-2010 - Solutions homework 9 Page 60...

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Solutions homework 9. Page 60 Problem 2. : (i) Let x y with x, y I J . Then I s an interval implies that [ x, y ] I and similarly [ x, y ] J . Hence [ x, y ] I J . It follows therefore from Theorem 4.1.4 that I J is an interval. (ii) False, take e.g. I = [0 , 1) and J = [1 , 2]. (iii) True. Let I = ( a 1 , b 1 ) and J = ( a 2 , b 2 ). Let a = min { a 1 , a 2 } . Then a = inf( I J ), Similarly if b = max { b 1 , b 2 } , then b = sup( I J ). If I J is an interval, it is equal to ( a, b ). Assume a = a 1 . If a 2 = a 1 , then obviously I J 6 = . If a 2 > b 1 , then I J = ( b 1 , a 2 ) is non-empty. If a 2 b 1 , then a 2 ( a, b ) \ ( I J ), which is impossible. (iv) True. Let I = [ a 1 , b 1 ] and J = [ a 2 , b 2 ]. Let a = min { a 1 , a 2 } . Then a = inf( I J ), Similarly if b = max { b 1 , b 2 } , then b = sup( I J ). If I J is an interval, it is equal to [ a, b ]. Assume a = a 1 . If a 2 = a 1 , then obviously I J 6 = . If a 2 b 1 , then I J = [ b 1 , a 2 ] is non-empty. If a 2 < b 1 , then a 2 + b 1 2 ( a, b ) \ ( I J ), which is impossible.
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