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Carnegie Mellon University 21-300/21-600 Fall 2012 Assignment 6 Due October 17, 2012

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Carnegie Mellon University 21-300/21-600 Fall 2012 Assignment 6 Due October 17, 2012 Carnegie Mellon University 21-300/21-600 Fall 2012 Assignment 6 Due October 17, 2012 ¬ Let S be a first-order signature. A theory Γ is said to be com- plete if for every S-sentence ', either Γ j= ' or Γ j= :'. Prove that if Γ is complete and M is a model of Γ, then the set of logical consequences of Γ consists of all the sentences that are true in M. Solution. Suppose that Γ is complete and M is a model of Γ. We have to prove that f ' j M j= ' g = f ' j Γ j= ' g: Proof of ⊆: Suppose M j= '. Then Γ j= ', because other- wise, since Γ is complete, we would have Γ j= :', and then, since M is a model of Γ, we would also have M j= :', which is impossible. Proof of ⊇: Suppose Γ j= '. Then M j= ', because otherwise we would have M j= :', and then, since M is a model of Γ and Γ j= ', we would also have M j= ', which is impossible. ­ Let R be the structure (R; +; ·), where + and · are the usual addition and multiplication, respectively. p (a) Exhibit a formula that defines the set f 2g in R. Solution. Let '(x) be x2 = 1 + 1 ^ 9y(y2 = x). (b) Prove that every finite union of intervals in R whose end- points are algebraic numbers is definable in R. (Here, the intervals may be of any kind: open, closed, semiopen, bounded, unbounded, etc.) Solution. We show this in stages. First, let's observe that every rational number is definable. If m; n 2 N, the m rational number n is defined by the formulanx ¯ = m, and m the rational number − n is defined by the formulanx ¯ +m = 0. Next, if a0; : : : ; an are definable real numbers, the set of n roots of the polynomial a0 + a1x + ··· + anx is definable 1 by the formula n ^ n 9y1 ::: 9yn \yi = ai" ^ y0 + a1x + ··· + ynx = 0 : i=1 Thus, every algebraic number is definable. If a is a definable real number, the intervals (*) [a; 1); (a; 1) are definable, by the formulas 9y(\y = a" ^ y ≤ x) and 9y(\x = a" ^ y < x), respectively. Finally, notice that any finite union of intervals of any kind can be obtained from intervals the form (∗) by taking a finite number of unions, intersections, and complements. This concludes the proof because unions, intersections, and complements of definable sets are definable (we just use the boolean connectives _; ^; :, as needed). ® Prove that addition is not definable from multiplication in N, that is, prove that the set f (m; n; m + n) j m; n 2 N g is not definable in the structure (N; ·). [Hint: Start by finding an automorphism f : M ! M that switches two primes.] Solution. First we show that every permutation π of the set of primes can be extended to an automorphism fπ of (N; ·). (Recall that a permutation of a set A s a bijection from A onto A. ) Let π be a permutation of the set of primes, and define a function f : N ! N by letting fπ(1) = 1 and e1 ek e1 ek fπ(p1 ····· pk ) = π(p1) · : : : π(pk) ; e1 ek if p1; : : : ; pk are distinct primes, e1; : : : ; ek ≥ 0, and p1 ·· · ··pk ≥ 2. Notice that if π is any permutation of the set of primes, then fπ(mn) = fπ(m)fπ(n); for m; n 2 N: Thus, fπ is an automorphism of (N; ·) Let π be the permutation that switches two distinct primes, say p0 and p1, and leaves all other primes fixed. Then, fπ is an automorphism of (N; ·), but fπ is not an automorphism of (N; ·; +), because fπ(p0 + 1) = p1 + 1 6= p0 + 1 = fπ(p1 + 1): ¯ An expansion of a structure M is a structure that results from adding functions and/or relations and/or constants to M with- out changing the universe. Let S be a signature, and let M be an S-structure with uni- verse M. Let SM be the signature that results from adding to S a new constant symbol ca for each a 2 M, and let (M; a j a 2 M) denote structure that results from expanding M with one con- stant for each element of M. Notice that (M; a j a 2 M) is an SM -structure in a natural way: we just interpret each constant ca as the element a. Define ∆M as the set of all sentences of the form '(ca1 ; : : : ; can ) such that '(x1; : : : ; xn) is a quantifier-free S-formula, a1; : : : ; an 2 M, and (M; a j a 2 M) j= '(ca1 ; : : : ; can ): Prove that if N is any S-structure, then the following conditions are equivalent: (i) M is isomorphic to a substructure of N. (ii) Some expansion of N to the signature SM is a model of ∆M. Solution. Proof of (i))(ii): Let h be an isomorphism from M onto a substructure N0 of N. It follows by straightfor- ward induction on the complexity of ' that if '(x1; : : : ; xn) is a quantifier-free S-formula and a1; : : : ; an 2 M, M j= '[a1 : : : ; an] iff N0 j= '[a1 : : : ; an] iff N j= '[a1 : : : ; an]: Hence, (N; h(a) j a 2 M) j= ∆M: Proof of (ii))(ii): Suppose that some expansion of N to the signature SM is a model of ∆M. For each a 2 M, let h(a) be the interpretation of ca in this expansion. Then, if '(x1; : : : ; xn) is a quantifier-free S-formula and a1; : : : ; an 2 M, (*) M j= '[a1 : : : ; an] iff N j= '[h(a1) : : : ; h(an)]: The range of h is closed under the functions of N, because by (∗), if f is an n-ary function symbol of S, and and a1; : : : ; an 2 M, M N f (a1; : : : ; an) = b iff f (h(a1); : : : ; h(an)) = h(b): Let N0 be the structure of M that results from restricting all the functions and relations of M to the range of h. It follows by straightforward induction on the complexity of ' that if and a1; : : : ; an 2 M, (**) N j= '[h(a1) : : : ; h(an)] iff N0 j= '[h(a1) : : : ; h(an)]: By (∗) and (∗∗), M j= '[a1 : : : ; an] iff N0 j= '[h(a1) : : : ; h(an)]; for every quantifier − freeS−formula'(x1; : : : ; xn). Thus, h is an isomorphism from M onto N0. | A linear order (A; ≤) is said to be dense if (A; ≤) j= 8x8y[x < y ! 9z(x < z < y)]: (Here we are using some self-explanatory abbreviations; for ex- ample, x < y is written as an abbreviation of (x ≤ y ^ :(x = y)).) (a) Prove that any two countable dense linear orderings that have no maximum or minimum element are isomorphic. [Hint: Fix two such linear orderings (A; ≤) and (B; E). Fix an enumeration of A and an enumeration of B. Then construct your isomorphism inductively, going from A to B at the even stages of the induction, and from B to A at the odd ones. Prove formally that the function you constructed this way is indeed an isomorphism.] Solution. Let (A; ≤) and (B; E) be countable dense linear orderings that have no maximum or minimum element, and let A = f a0; a1;::: g B = f b0; b1;::: g be enumerations of A and B, respectively, without repeti- tion. By induction on k we shall define subsets Ak = f ai0 ; : : : aik g ⊆ A Bk = f bi0 ; : : : bik g ⊆ B such that the map aiu 7! biu ; for u 2 f0; : : : ; kg is an isomorphism from (Ak; ≤) onto (Bk; E). Assume that Ak−1 and Bk−1 have been already defined, in order to define Ak and Bk. We consider two cases: Case 1 : k is even. Let ik+1 be the least index such that ai+1 2= Ak. Now look now at how ak+1 sits in (A; ≤) with respect to the elements of Ak; since the order ≤ is linear and Ak is finite, we have only three possibilities: (i) ak is strictly above every element of Ak−1, (ii) ak is strictly below every element of Ak−1, (iii) There exist u; v 2 f0; : : : ; kg such that aiu ≤ ai+1 ≤ aiv and ai+1 is the only element of A that lies strictly between aiu and aiv . In case (i), we let bk+1 be an element of B that, with respect to the order E, is strictly above every element of Bk; such an element exists since B has no maximum element. In case (ii), we let bk+1 be an element of B that, with respect to the order E, is strictly below every element of Bk; such an element exists since B has no minimum element. In case (iii), we let bk+1 be an element of B that, with respect to the order E, sits strictly between biu and biv ; this is pos- sible since B has no minimum element. By construction, the map aiu 7! biu ; for u 2 f0; : : : ; k + 1g is an isomorphism from (Ak+1; ≤) onto (Bk+1; E). Case 2 : k is odd. We proceed as in Case 1, but the roles of (A; ≤) and (B; E) are exchanged. Let f be the map that results from iterating this process indefinitely. Then f is one-to-one by construction. At the even stages of the construction we made sure that every element of A is in the domain of f, and at the odd stages we guaranteed that every element of B is in the range of f. Thus, f is a bijection between A and B.
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