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Enumerability and Diagonalization

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Enumerability and Diagonalization 5 Enumerability and Diagonalization 5.1 Introduction 5.2 Equinumerous Sets 5.3 Countable and Uncountable Sets 5.4 Enumerating Programs 5.5 Abstract Families of Functions 5.6 Exercises and Supplements 5.7 Bibliographical Comments 5.1 Introduction One of the purposes of this chapter is to present a formal approach to the intuitive notion of the size of a set. The facts discussed here play a fundamental role in computer science; they serve to make precise the idea that there are more functions on the set of natural numbers than programs (in any programming language), and therefore, there exist functions that cannot be computed by any program in that programming language. A focal point of this chapter is a proof method originating in set theory: the diagonalization method. Developments in theoretical computer science have made diagonalization an essential instrument for proving negative and limitative results in this field. We discuss diagonalization both in its original set-theoretical context and in the context of computability theory. 5.2 Equinumerous Sets It follows from Corollary 4.2.7 that there is a bijection f : {ao, ... ,am-I} --+ {bo, ... , bp-tl if and only if m = p, that is, if and only if the two sets involved have the same number of elements. In this section, our main objective is to extend this idea to arbitrary sets. Definition 5.2.1 Two sets M and P are called equinumerous, denoted by M", P, if there exists a bijection f : M --+ P. It is easy to verify the following properties for any sets M, P, and Q: 339 P. A. Fejer et al., Mathematical Foundations of Computer Science © Springer-Verlag New York, Inc. 1991 340 5. Enumerability and Diagonalization 1. M,....M; 2. if M,.... P, then p,.... M; and 3. if M ,.... P and p,.... Q, then M ,.... Q. The first property follows immediately from the fact that the mapping 1M : M --+ M is a bijection. Further , if I : M --+ P is a bijection, then 1- 1 : P --+ M is also a bijection; therefore, if M ,.... P then, P ,.... M. Finally, if I : M --+ P and 9 : P --+ Q are two bijections, then gl : M --+ Q is also a bijection. Example 5.2.2 Consider the set E of all even natural numbers. We have E C N. Nevertheless, we also have N ,.... E because of the fact that the mapping I : N --+ E defined by I(n) = 2n for every nE N is a bijection. Example 5.2.3 The set Z of integers is equinumerous with N. Consider the mapping 9 : N --+ Z defined by { n/2 if n is even g(n) = -(n + 1)/2 if n is odd. It is not difficult to see that 9 is a bijection, which proves that Z ,.... N. Example 5.2.4 Any two open intervals of the set of real numbers are equinumerous. Indeed, consider two intervals (a, b) and (c, d), where a < b and e< d. Define g: (a,b) --+ (e,d) by d - e be - ad g(x) -b-x b . = -a + -a It is easy to see that 9 is a bijection between (a,b) and (c,d). Of course, this is true even if (a, b) ~ (c, d) since we did not use the relative positions of the two intervals in our argument. Furthermore, we can prove that R,.... (a, b) for any open interval (a, b). It suffices to consider the mapping tan : (-t, t) --+ R, which is a bijection, to obtain (-t, t) ,.... R. On the other hand, any open interval (a,b) is equinumerous with (-t, t), which gives R,.... (a, b). The following theorem gives some evidence that two equinumerous sets do in fact have the same size. Theorem 5.2.5 Let M and P be sets such that M,.... P. Then, either M and P are both finite or they are both infinite. Proof: If one of M and Pisfinite and M ,.... P, then it follows from Theorem 2.4.10 that the other of these two sets is also finite. I In our next definition, we try to make precise the ideas of one set being at least as large as or strictly larger than another. 5.2. Equinumerous Sets 341 Definition 5.2.6 Let M and P be two sets. M is dominated by P, written M ~ P, if there exists an injection f : M --+ P. M is strictly dominated by P, written M ~ P, if M ~ P and M is not equinumerous with P. Example 5.2.7 For any finite set M, we have M ~ N since if M has n elements, M = {xQ, ... , xn-d, the mapping f : M --+ N defined by f(Xä) = i for 0 ~ i ~ n - 1 is injective. Theorem 5.2.8 For every M, P, Q, R, Ml , ... , Mb Pl , ••• , Pk , we have 1. M ~ M; 2. if M ~ P and P ~ Q, then M ~ Q; 3. if M ~ P, then M ~ P; 4. if M '" P, then M ~ P; 6. if M ~ P, Q'" M, and R", P, then Q ~ R. Proof. The argument consists of direct applications of the definition of ~, and it is left to the reader. I The connection between finiteness and domination Is given in the next theorem. Theorem 5.2.9 Let M and P be sets such that M ~ P. Then, if P is finite, so is M, and if M is infinite, so is P. Proof: Since M ~ P, we have M '" K for some subset K of P. If P is finite, then by Theorem 4.2.11 so is K, but then by Theorem 5.2.5, M is finite. The second part of the theorem follows immediately from the first. I The next theorem shows the connection between cardinality and domi­ nation for finite sets. Theorem 5.2.10 Let M and P be finite sets. Then, M ~ P if and only if IMI ~ IPI and M ~ P if and only if IMI < IPI. Proof: The theorem follows immediately from Corollary 4.2.9. I Theorem 5.2.8 implies that if M '" P then M ~ P and P ~ M. By the previous theorem, the converse of this statement is true if M and P are finite sets. We now show that the converse is true in general. This supports our intuition that M ~ P me ans that P is at least as big as M and M '" P means that M and P have the same size. Theorem 5.2.11 {Schröder-Bernstein Theorem} For all sets M and P, if M ~ P and P ~ M, then M '" P. 342 5. Enumerability and Diagonalization Prool. According to the hypothesis of the theorem, we have two injective mappings f: M --+ P and g: P --+ M. Define ~ : 1'(M) --+ 1'(M) by ~(K) = M -g(P- f(K)). This mapping is monotonie since K 1 ~ K 2 implies f(Kt} ~ f(K2 ) (because of Theo­ rem 2.5.6); hence, P - f(K2) ~ P - f(Kd, which gives g(P - f(K2)) ~ g(P- f(Kd). This, in turn, implies M -g(P- f(K1 )) ~ M -g(P- f(K2 )); hence, ~(Kt) ~ ~(K2). Let Ko be a subset of M such that ~(Ko) = Ko. Such a subset exists because of Theorem 3.9.10. For this set, we have K o = ~(Ko) = M - g(P - f(Ko)). (5.1) Define now a mapping h : M --+ P by h(a) = { f(a) if a E Ko g-l(a) if a E M - K o. Since 9 is an injection, g-l is a function and if a E M - Ko, then a E g(P - f(Ko)), because of (5.1), so g-l(a) is defined. Thus, h is well­ defined. The mapping h is onto. In order to justify this claim, let b E P. If bE f(Ko), then there is a E Ko such that b = f(a). In this case, we have h(a) = f(a) = b. If b f/. f(Ko), then bE P - f(Ko) implies a = g(b) f/. Ko by (5.1), so b = h(a) and this proves that every element of P is the image (by h) of an element of M. The mapping h is injective. Suppose that h(a) = h(a') = b. If both a, a' E Ko, then we have f(a) = b = f(a') so a = a', because of the fact that fis injective. Ifneither a nor a' are in K o, then we have a = g(b) = a'. Finally, suppose that a E Ko and a' f/. Ko. Then, b = h(a) = f(a), and a' = g(h(a')) = g(b), so a' = g(f(a)) E g(f(Ko)). Together with a' E M - Ko = g(P - f(Ko)), this contradicts the injectivity of g. Thus, h is a bijection and M "" P. I Theorem 5.2.12 If M, P, and Q are sets such that either M --( P:5 Q or M :5 P --( Q, then M --( Q. Prool: Suppose that M --( P :5 Q. Then, M :5 P :5 Q, so by Theo­ rem 5.2.8, M :5 Q. Since P :5 Q, there must be an injection 9 : P --+ Q. If M "" Q, say, f : Q --+ M is a bijection, then fg is an injection from P to M, so P :5 M. In combination with M :5 P, we get from Theorem 5.2.11 that M "" P, contradicting M --( P. Thus, we do not have M "" Q, and hence, M --( Q.
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