{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

WorkedSolutions - Worked Solutions 1 enumerability Richard...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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. 2 Page 12 Exercise 2.1. If A , B and C are enumerable sets, the set of triples 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) is trivial. Now suppose that, for any enumerable sets A 1 , . . . , A n , the set 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. 3 Page 13 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 , and their 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
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
4 Page 14 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?). A is enumerable, and so we have an encoding 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 , . . . , a 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). 5 Page 25 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.
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern