solutionshw13-554-2010

# solutionshw13-554-2010 - x-f y | ≤ | x-y | Let now x n...

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Solutions homework 13. Page 86 Problem 7. From 0 f ( x ) x it follows that f is continuous at 0. If x > 0 and x rational, let x n x with x n irrational. Then f ( x n ) = x n x 6 = 0 = f ( x ). Hence f is discontinuous at rational x > 0. Similarly, If x > 0 and x irrational, let x n x with x n rational. Then f ( x n ) = 0 0 6 = x = f ( x ). Page 91 Problem 6. From the triangle inequality it follows that | x - a | ≤ | x - y | + | y - a | . This implies f ( x ) = inf {| x - a | ; a A } ≤ inf {| x - y | + | y - a | : a A } = | x - y | + f ( y ). Hence f ( x ) - f ( y ) ≤ | x - y | . Interchanging x and y we get that f ( y ) - f ( x ) ≤ | y - x | = | x - y | . Hence | f
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Unformatted text preview: ( x )-f ( y ) | ≤ | x-y | . Let now x n → x . Then | f ( x )-f ( x n ) | ≤ | x-x n | → 0, so f ( x n ) → f ( x ). Hence f is continuous. Page 91 Problem 10. Let x ∈ R . Then from Q = R it follows that there exists r n ∈ Q such that r n → x . Then f ( r n ) → f ( x ), g ( r n ) → g ( r ) and f ( r n ) = g ( r n ) imply that f ( x ) = g ( x ). 1...
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