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# sol-05 - MATH 453 SOLUTIONS TO ASSIGNMENT 5 OCTOBER 8 2004...

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M  453 S   A  5 O  8, 2004 Exercise 4 from Section 18, page 111 We check that f is an imbedding; the argument for g is similar. f is clearly a bijection between X and f ( X ) = X × y 0 , so we need only show that it and its inverse are continuous. It follows from the deﬁnition of the subspace topology that the open sets of X × y 0 are of the form U × y 0 where U is open in X . Since f ( U ) = U × y 0 , it is immediate that both f and f - 1 are continuous. ± Exercise 5 from Section 18, page 111 The map f : [0 , 1] [ a , b ] given by f ( x ) = a + ( b - a ) x is clearly a homeomorphism that restricts to a homeomorphism from (0 , 1) to ( a , b ). ± Exercise 6 from Section 18, page 111 Deﬁne f : R R by f ( x ) = 0 if x is rational and f ( x ) = x otherwise. We will show that f is continuous only at 0 by using the sequence criterion for continuity. If { a i } is any sequence that converges to 0, then { f ( a i ) } is obtained from { a i } by replacing the rational a i ’s by 0, so it also converges to 0. Hence

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sol-05 - MATH 453 SOLUTIONS TO ASSIGNMENT 5 OCTOBER 8 2004...

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