hmw10s - 18.100B, Fall 2002, Homework 10 Due by Noon,...

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Unformatted text preview: 18.100B, Fall 2002, Homework 10 Due by Noon, Thursday December 5 Rudin: (1) Chapter 7, Problem 14 Solution. There is a function f as described, just set ) t 1 3 1 f ( t ) = 3( t 1 3 t 2 3 ) 3 1 2 t 1 3 and for instance f (2 t ) = f ( t ) for 1 t 2 and then f (2 k + t ) = f ( t ) for all k N , k = 0 , t [0 , 2] . This gives a continuous function. Consider (1) x ( t ) = 2 n f (3 2 n 1 t ) , y ( t ) = 2 n f (3 2 n t ) . n =1 n =1 The n th term in the series for (1) is bounded | 2 n f (3 2 n 1 t ) | 2 n . By Theorem 7.10, the series converges uniformly. Thus x ( t ) is continuous by Theorem 7.12. The same argument applies to y ( t ) so ( t ) = ( x ( t ) , y ( t )) is continuous by Theorem 4.10. Now, to see that is surjective, follow the hint. Each point in [0 , 1] has a dyadic decomposition x = 2 n b n , b n = or 1 . n =1 Indeed one can compute the successive b n , n = 1 , . . . , k to arrange that k x 2 n b n 2 k . Then choose b k +1 = or 1 depending on whether n =1 k k < 2 k 1 x 2 n b n or 2 k 1 x 2 n b n 2 k . Now we let n =1 n =1 the a 2 n 1 be these numbers for x and a 2 n those for y ....
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This note was uploaded on 12/16/2011 for the course STAT 5446 taught by Professor Frade during the Fall '09 term at FSU.

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hmw10s - 18.100B, Fall 2002, Homework 10 Due by Noon,...

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