WorkedSolutions - Worked Solutions 1 - enumerability...

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Worked Solutions 1 - enumerability Richard M c Kinley March 13, 2005 1 Introduction These are selected model answers to exercises from lecture handout 2. Each section corresponds to a slide from that handout. 2P a g e 1 2 Exercise 2.1. If A , B and C areenumerab lese ts ,these to ftr ip les A × B × C = { ( a, b, c ): a A, b Bc C } is enumerable. Proof. The product of two enumerable sets is enumerable. Writing A × B × C as ( A × B ) × C , we see that it is the cartesian product of two enumerable sets: A × B and C . Since A and B are enumerable, A × B is enumerable So we see that A × B × C is the product of two enumerable sets, and hence enumerable. Exercise 2.2. If A 1 ,...,A k are enumerable sets, then A 1 × ... × A k is enu- merable. Proof. By induction. The base case ( k =1)istriv ial . Now suppose that, for any enumerable sets A 1 n ,theset A 1 × × A n is enumerable. Then, given another enumerable set A n +1 , A 1 × × A n × A n +1 = B × A ,where B = A 1 × × A n is enumerable. B × A is the cartesian product of two enumerable sets, and so is enumerable. 3P a g e 1 3 Exercise 3.2. If B is enumerable and there is a total injective function A B , then A is enumerable. Proof. If B is enumerable, there is a coding of B - an total injective function f from B to N . We know we have a total injective function g : A B ,andthe ir composition f g is a function A N . This function is total and injective (you should check this, and in an exam we would expect you to prove it) and so A is enumerable. 1
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4P a g e 1 4 Exercise 4.1. The set Q + is enumerable. Proof. Let the function f : N × N Q + take ( a, b )to a b . This function is surjective,( for example, each rational number r has a form m n where m and n share no factors, and then ( m, n ) is a pre image for r) and we know that N × N is enumerable, (via Cantor’s zig-zag or an encoding c (( m, n )) = p n · q m where p and q are prime) so Q + is enumerable. Exercise 4.4. The set A of strings over an enumerable alphabet A is enumer- able. Proof. A is enumerable, and so we have an encoding c A : A N . We will give a total injective function A N × N . This will be enough to show that A is enumerable (Why?). Aisenumerab le ,andsowehaveanencod ing c A : A N . From a previous exercise, we know that for each n N there is an encoding c n : A n N . c n (( a 1 ,...,a n )) = c N ×···× N ( c A ( a 1 ) ,...c A ( a n )) Regarding each string s of length n as an n -tuple, we have a function c A : A N × N , c A ( a 1 a 2 a 3 ...a m )=( m, c m (( a 1 m )) This is injective and total, so we are done. (You should verify yourself that this function is injective and total, and in an exam you should prove these claims). 5P a g e 2 5 Exercise 5.2. Suppose that we have a programming language, such that every program describes a function N N . Show that there must be functions N N that are described by no program.
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This note was uploaded on 01/25/2011 for the course CSI 301 taught by Professor Mr.elienasr during the Spring '10 term at American University of Science & Tech.

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WorkedSolutions - Worked Solutions 1 - enumerability...

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