sols7 - M114S, Solutions to HW #7 x7.10. The fact that U o...

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Unformatted text preview: M114S, Solutions to HW #7 x7.10. The fact that U o V is a linear ordering is quite trivial. If negationslash = X U o V , let y = min V { y V | ( x U )[( x,y ) X ] } , x = min U { x U | ( x,y ) X } and check easily that ( x ,y ) is the least element of X . x7.11. We use the Transfinite Recursion Theorem 7.24. Let U be well ordered, and let E = { , 1 } . Let flip = { (0 , 1) , (1 , 0) } be the function which flips 0 and 1. Let h : ( U E ) U E be given by h ( g,x ) = braceleftBigg flip ( f ( y )) , if x = S ( y ) for some (unique) y, , otherwise . Let Parity : U E satisfy the recursion Parity( x ) = h (Parity seg ( x ) ,x ) for all x U . Thats all thats to it: Parity is exactly the function we want, by an easy induction on U . x7.12. Suppose not and let y be the least element of U which falsifies the claim in the problem. If y = 0 or y is a limit point, then y = S ( y ) by the definition of the function S n , which contradicts its choice, and hence y is the successor of some u in U , i.e., y = S ( u ). Now u < y , and so, by the choice of y , u = S n ( x ) where x is 0, or a limit point; hence y = S n +1 ( x ) with the same x . To prove the uniqueness of this representation, suppose that....
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This homework help was uploaded on 04/16/2008 for the course MATH 114S taught by Professor Moschovakis during the Winter '08 term at UCLA.

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sols7 - M114S, Solutions to HW #7 x7.10. The fact that U o...

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