MIT6_042JS10_lec06_sol

MIT6_042JS10_lec06_sol - Massachusetts Institute of...

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Unformatted text preview: Massachusetts Institute of Technology 6.042J/18.062J, Spring 10 : Mathematics for Computer Science February 16 Prof. Albert R. Meyer revised February 9, 2010, 1094 minutes Solutions to In-Class Problems Week 3, Tue. Problem 1. Lemma 4.9.4 . Let A be a set and b / A . If A is infinite, then there is a bijection from A { b } to A . Proof. Heres how to define the bijection: since A is infinite, it certainly has at least one element; call it a . But since A is infinite, it has at least two elements, and one of them must not be equal to a ; call this new element a 1 . But since A is infinite, it has at least three elements, one of which must not equal a or a 1 ; call this new element a 2 . Continuing in the way, we conclude that there is an infinite sequence a ,a 1 ,a 2 ,...,a n ,... of different elements of A . Now we can define a bijection f : A { b } A : f ( b ) ::= a , f ( a n ) ::= a n +1 for n N , f ( a ) ::= a for a A { b,a ,a 1 ,... } . (a) Several students felt the proof of Lemma 4.9.4 was worrisome, if not circular. What do you think? Solution. There is no solution for this discussion problem, since it depends on what seems bothersome. It may be bothersome that the proof asserts that f is bijection without spelling out a proof. But the bijection property really does follow directly from definition of f , so it shouldnt be much burden for a bothered reader to fill in such a proof. Another possibly bothersome point is that the proof assumes that if a set is infinite, it must have more than n elements, for every nonnegative integer n . But really thats the definition of infinity: a set is finite iff it has n elements for some nonnegative integer,...
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MIT6_042JS10_lec06_sol - Massachusetts Institute of...

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