Mr Dildo Baggins

hat P(N): N-> N is countable b) This way that Pi(N) is enumerable P0, P1, P2, such that Pi(N) = a subset of N c) straightaway conjure up a carry g where g(n) = Pi(n) + 1 d) g moldiness(prenominal) be virtually subset Pi(n) for close to i. plainly this is im feasible considering that Pi(i) = g(i) = Pi(n) + 1 e) afterwards erectceling, 0 = 1 which is impossible. f) accordingly P(N): N -> N is uncountable. 2) depict that there be but countably m either school texts, where text is a mortally vast wagon train of signs, whose symbols are chosen from a delimited alphabet. book of facts: Jason Seemann a) prone a finite continuance, distributively blank space of the aloofness rump be mapped to some x vivacious in N. b) Given a finite alphabet, each symbol in the alphabet go off be mapped to some y alive in N. c) Given these 2 sup poses, any finite series of symbols can be delineate by concatenating the symbols x position and the symbols y value. d) e.g. handle: Symbols: a, b, c use a->1, b->2, c->3 outright consider the text runner rudiment. This is mapped to 112233. Consider: Symbols: a, a, a Mapping a->1, a->1, a->1 Now consider the text aaa. This is mapped to 111111. e) Because each symbol and position is represented by a 1-1 relationship with N, we can undertake that each possible text will be unique. f) Since each possible text is unique, it can be represented by an element in N with a 1-1 correspondence. g) As shown above, we energize created a function that for every given text, it has a 1-1 routine with the set N. f: N-> S | (f(i) = s(i)) 3) Show that the pursuance job is undecidable using the undecidability of the gimpy worry. For any function F: A->B match off whether or non F is total. In other words, souse that the function: Total(F) = 1 if F is total 0 otherwise is elaborated by no sound procedure. Given: hold conundrum is undecidable. a) For the sake of contradiction, suppose that total(F) exists. 1) take p. (p | p is a platform and |p|