By Neil D. Jones
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Additional info for Computability Theory: An Introduction
Consequently for some fixed k0, g(x) = U(k0, x) for all x. This portion of the argument also applies in the current context. 8 arose from the equation U(k0,k0)=g(k0) = U(k0,k0)+l (1) which cannot be satisfied by any total function g, since it implies that 0 = 1. However in the current situation U is a partial function, so that Equation (1) can be satisfied by virtue of the fact that U(k0, k0) is undefined. 9 together imply the perhaps surprising fact that a set may be effectively enumerable, while a subset of it is not effectively enumerable.
It is not necessary to solve the halting problem in order to solve the printing problem, since it is easy to construct algorithms which print * and do not halt, or do not print * but do halt, or do both or neither. The following, however, is a valid argument which will be formally paraphrased in Chapter V. Suppose the printing problem were decidable. Then it could be used to solve the halting problem as follows : 1. Given an algorithm z and an input x, construct a new algorithm z' as follows : 2.
Show that the collection of all algebraic numbers is countable. Hint: Use the result of Exercise 2. 4. Show that if S and T are countable sets, then SuT and S x T are also countable. 5. Construct a unary function f(n) which enumerates the set of pairs (x, y) such that x, ye N. Give an explicit définition of/, rather than just an enumeration of its range. An Uncountable Set The proof of the following theorem uses a technique called diagonalization which is central to much of computability theory.
Computability Theory: An Introduction by Neil D. Jones