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

x Let S be a ﬁrst-order signature. A theory Γ is said to be complete if for every S-sentence ϕ, either Γ |= ϕ or Γ |= ¬ϕ. 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

{ ϕ | M |= ϕ } = { ϕ | Γ |= ϕ }.

y Let R be the structure (R, +, ·), where + and · are the usual addition and multiplication, respectively. √ (a) Exhibit a formula that deﬁnes the set { 2} in R.

Solution. Let ϕ(x) be x2 = 1 + 1 ∧ ∃y(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 ∈ 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 ∃y1 ... ∃yn “yi = ai” ∧ y0 + a1x + ··· + ynx = 0 . i=1 Thus, every algebraic number is definable. If a is a definable real number, the intervals (*) [a, ∞), (a, ∞) are definable, by the formulas ∃y(“y = a” ∧ y ≤ x) and ∃y(“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). z Prove that addition is not definable from multiplication in N, that is, prove that the set { (m, n, m + n) | m, n ∈ N } 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 ∈ 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 without changing the universe. Let S be a signature, and let M be an S-structure with universe M. Let SM be the signature that results from adding to S a new constant symbol ca for each a ∈ M, and let (M, a | a ∈ M) denote structure that results from expanding M with one constant for each element of M. Notice that (M, a | a ∈ 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 ϕ(ca1 , . . . , can ) such that ϕ(x1, . . . , xn) is a quantiﬁer-free S-formula, a1, . . . , an ∈ M, and

(M, a | a ∈ M) |= ϕ(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 straightforward induction on the complexity of ϕ that if ϕ(x1, . . . , xn) is a quantiﬁer-free S-formula and a1, . . . , an ∈ M,

M |= ϕ[a1 . . . , an] iﬀ N0 |= ϕ[a1 . . . , an] iﬀ N |= ϕ[a1 . . . , an]. Hence,

(N, h(a) | a ∈ M) |= ∆M. Proof of (ii)⇒(ii): Suppose that some expansion of N to the signature SM is a model of ∆M. For each a ∈ M, let h(a) be the interpretation of ca in this expansion. Then, if ϕ(x1, . . . , xn) is a quantiﬁer-free S-formula and a1, . . . , an ∈ M,

(*) M |= ϕ[a1 . . . , an] iﬀ N |= ϕ[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 ∈ 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 ϕ that if and a1, . . . , an ∈ M,

(**) N |= ϕ[h(a1) . . . , h(an)] iﬀ N0 |= ϕ[h(a1) . . . , h(an)]. By (∗) and (∗∗),

M |= ϕ[a1 . . . , an] iﬀ N0 |= ϕ[h(a1) . . . , h(an)],

for every quantifier − freeS−formulaϕ(x1, . . . , xn). Thus, h is an isomorphism from M onto N0. | A linear order (A, ≤) is said to be dense if (A, ≤) |= ∀x∀y[x < y → ∃z(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 = { a0, a1,... }

B = { b0, b1,... } be enumerations of A and B, respectively, without repeti- tion. By induction on k we shall deﬁne subsets

Ak = { ai0 , . . . aik } ⊆ A

Bk = { bi0 , . . . bik } ⊆ B such that the map

aiu 7→ biu , for u ∈ {0, . . . , k}

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 ∈/ 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 ∈ {0, . . . , k} 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 ∈ {0, . . . , k + 1}

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. (b) (For 21-600) Prove that every countable dense linear order must be isomorphic to one of the following four structures: (i) (Q ∩ (0, 1), ≤), (ii) (Q ∩ [0, 1), ≤), (iii) (Q ∩ (0, 1], ≤), (iv) (Q ∩ [0, 1], ≤). Solution. Letl M = (M, E) be a countably infinite dense linear order, and let M0 be the subset of M that results from removing the maximum and minimum elements of M, if they exist. By part (a), the structure (M0, E) is isomorphic to (Q ∩ (0, 1), ≤). Fix an isomorphism h between these two structures. If M has a minimum but not a maximum, then h can be extended to an isomorphism between M and (Q ∩ [0, 1), ≤). If M has a maximum but not a minimum, h can be extended to an isomorphism between M and (Q ∩ (0, 1], ≤). Finally, if M has both maximum and minimum elements, the isomorphism can be extended to an isomorphism between M and (Q ∩ [0, 1], ≤). } (For 21-600) Prove that if M is a finite structure and N ≡ M, then M and N are isomorphic. Note. This is an exercise in writing things well, so I won’t write a full solution. The main point is that if S is a finite signature and M is an S-structure with finite universe, then one can write a first-order sentence ϕ that describes M completely, in the sense that every S-structure that satisfies ϕ is isomorphic to M. The sentence ϕ is of the form

∃x1 ... ∃xn[“the universe consists of {x1, . . . , xn}” ∧ θ(x1, . . . , xn)].

Where θ(x1, . . . , xn) is a quantifier-free formula that specifies the graph of each function and relation of the structure and specifies, for each constant c in the signature, which of x1, . . . , xn equals cM.