Some_sols_Homework1_typeset - 1 Solution to Problem 2....

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1 Solution to Problem 2. Define X 0 = X - g ( Y ) and Y 0 = Y - f ( X ). For any x X 0 , consider the sequence x, g ( f ( x )) ,g ( f ( g ( x )))) ,.... Define A X to be all elements of such sequences. y Y 0 , consider the sequence g ( y ) ( f ( g ( y )) ( f ( g ( f ( g ( y )))) Define B X to be all elements of such sequences. Finally let C = X - ( A B ). Then A B = / ± , A C = / ± , B C = / ± , and of course X = A B C . Define h ( x )= f ( x )i f x A , g - 1 ( x f x B , f ( x f x C . Then h : X Y is bijective. It is kind of fun to apply this to the case X =[ - 1 , 1], Y =( - 1 , 1) and f ( x x/ 2 ( y y . In that case, (1) X 0 = {- 1 , 1 } , and A = 1 , ± 1 / 2 , ± 1 / 4 ,... } .On A , we set h ( x x/ 2; (2) Y 0 - 1 , 1) - [ - 1 / 2 , 12], and B - 1 , 1) - ( A ∪{ 0 } ). On B , we set h ( x x ; (3) C = { 0 } , and h (0)=0. . There are many entertaining cases, like setting up a bijective correspondance between R and R 2 . Note that this construction is completely constructive. Suppose that instead of supposing f and g injective, we had supposed them surjective. I have been asking around, more particularly our logician Justin Moore, and it turns out
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Some_sols_Homework1_typeset - 1 Solution to Problem 2....

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