3
The Compactness Theorem (1st proof )
We review here the Henkin proof of the compactness theorem, as given in most logic classes; except that we abstract away all mention of syntactical proof. The interplay now is between finite satisfiability and existence of a model.Fix a language L.
Let ⌃ be a set of L-sentences. We write A ✏ ⌃ (A models ⌃, or A is a model of ⌃) if, for any 2⌃, A✏ .
AnL-sentence is said to bea logical consequence of ⌃ifA ✏⌃implies A ✏ for every L-structure A.
Notationally, it will be convenient to write ⌃✏ when is a logical conse-quence of some finite subset of ⌃. We will see later that in fact ⌃✏ i↵
is a logical consequence of ⌃.
written⌃✏ ,ifA ✏⌃impliesA✏ for everyL-structureA.For⌃infinite,
⌃✏ means that there is a finite ⌃0 ⇢⌃such that ⌃0 ✏ . 4
is calledlogically valid, written ✏ , if A ✏ for every L-structure A. A set ⌃ of L-sentences is said to be satisfiable if it has a model, i.e. there is an L-structure A such that A ✏ ⌃. ⌃ is said to be finitely satisfiable (f.s.) if any finite subset of ⌃ is satisfiable.
⌃ is said to be complete if, for any L-sentence , 2⌃ or¬ 2⌃.
Exercise 3.1. Let ↵,↵1, . . . ,↵n, , 1, . . . , n, be closed L-terms, P, f
L-symbols forn-ary predicate andn-ary function, correspondingly, and (v0, v1, . . . , vn) an L-formula with free variablesv0, v1, . . . , vn.Prove that
1. ↵l ✏ l↵;
2. ↵l , l ✏↵l ;
3. ✏↵l↵;
4. ↵1 l 1, . . . ,↵nl n, P(↵1, . . . ,↵n)|=P( 1, . . . , n);
5. ↵l ,↵1 l 1, . . . ,↵nl n, f(↵1, . . . ,↵n)l↵|= f( 1, . . . , n)l ;
4This definition will be convenient for the proof, and will be seen later to be equivalent
6. ( ,↵1, . . . ,↵n)|= 9v0 (v0,↵1, . . . ,↵n).
Definition 3.1. (1)-(5) are the axioms of equality. A binary relation satis-fying these laws is called a congruence.
A set of L-sentences⌃ is said to be deductively closedif
⌃✏ implies 2⌃.
Exercise 3.2. (i) If ⌃0 ✓⌃and ⌃0 ✏ then ⌃✏ ;
(ii) A complete f.s. ⌃ is deductively closed.
Proposition 1 (Lindenbaum’s Theorem). For any f.s. set ofL-sentences⌃
there is a complete f.s. set of L-sentences⌃# such that ⌃✓⌃#.
Proof Let
S ={⌃0 :⌃✓⌃0 a f.s. set of L-sentences }.
Clearly S satisfies the hypothesis of Zorn’s Lemma, so it contains a maximal element ⌃# say. This is complete for otherwise, say 2/ ⌃# and ¬ 2/ ⌃#. By maximality neither { }[⌃# nor {¬ }[⌃# is f.s.. Hence there exist finite S1 ✓ ⌃# and S2 ✓ ⌃# such that neither { }[S1 nor {¬ }[S2 is satisfiable. However, S1[S2 ✓⌃#,finite, so has a model, A say. But either A ✏ , soA ✏{ }[S1, orA ✏¬ , soA ✏{¬ }[S2, a contradiction.
Remark In class we saw another proof of Lindenbaum’s theorem, if the language is countable or more generally well-ordered. Namely, let { n} enu-merate all sentences of L. Define n recursively: if ⌃S{ 1, . . . , n 1, n} is consistent, let n = n; otherwise let n = ¬ n. One verifies easily that
⌃S{ n:n = 1,2, . . .} is complete and f.s.
Indeed the Henkin proof of compactness (or completeness) does not require the axiom of choice, provided the symbols of the language itself are well-ordered. This is an advantage over the ultraproduct proof in the next section.
Exercise 3.3. Let us allow 0-place relation symbols (Ri : i 2 I); they are also called propositional symbols. Assume there are no other symbols; so that a structure consists just of assigning a truth value to each Ri. Thus a structure is just an element of the I-fold product {0,1}I.
1. In this case, show that a complete f.s. set of sentences ⌃ determines a model M(⌃) of ⌃, simply by assigning 1 to Ri if Ri 2 ⌃ and 0 otherwise.
2. Define a topology on {0,1}I by letting a basic open set have the form B(i0, . . . , ik;⌫0, . . . ,⌫k) ={f :f(i0) = ⌫0, . . . , f(ik) =⌫k}
where i0, . . . , ik 2 I and ⌫0, . . . ,⌫k 2 {0,1}. Tychono↵’s theorem as-serts that this topology is compact. Prove this using this case of the compactness theorem.
A set ⌃ of L-sentences is said to be witnessing if for any sentence in ⌃ of the form 9v'(v) there is a closedL-term such that '( )2⌃.
Exercise 3.4. Any complete f.s. witnessing set of L-sentences has a closed L-term.
Hint: Consider the L-sentence9v v lv.
An L-structure A is called minimal if it has no proper substructure.
Exercise 3.5. Show that A is a minimal L-structure i↵ for every a 2 A there is a closed L-term such that A =a.
Let us say that a model of ⌃ iscanonical if it is minimal as an L-structure. 5
Proposition 2. For any complete, witnessing, f.s. set⌃ofL-sentences there is a canonical model A of ⌃.
Proof Let⇤ be the set of closed terms of L. This is nonempty by 3.4. For ↵, 2⇤ define ↵v i↵ ↵l 2⌃.
This is an equivalence relation by 3.1.1- 3.1.3. and 3.2.
For ↵2⇤, let ˜↵ denote the v-equivalence class containing ↵. Let A={↵˜: ↵2⇤}.
This will be the domain of our modelA.We want to define relations, functions and constants of L onA.
Let P be an n-ary relation symbol of L and ↵1, . . . ,↵n2⇤. Define h↵˜1, . . . ,↵˜ni 2PA i↵ P(↵1, . . . ,↵n)2⌃.
By 3.1.4 the definition does not depend on the choice of representatives in the v-classes.
For a unary function symbol f of L of arity m and ↵1, . . . ,↵m 2⇤ define fA(˜↵1, . . . ,↵m) = ˜˜ ⌧, where ⌧ =f(↵1, . . . ,↵m).
By 3.1.5 this is well-defined.
Finally, for a constant symbol, cA is just ˜c.
We now prove by induction on complexity of anL-formula '(v1, . . . , vn) that (⇤) A✏'(˜↵1, . . . ,↵n) i˜ ↵ '(↵1, . . . ,↵n)2⌃.
For atomic formulas we have this by definition. If '= ('1^'2) then
A ✏('1(˜↵1, . . . ,↵n)˜ ^'2(˜↵1, . . . ,↵n)) i˜ ↵A✏'1(˜↵1, . . . ,↵n) and˜ A✏'2(˜↵1, . . . ,↵n)˜ i↵ (by induction hypothesis) '1(↵1, . . . ,↵n),'2(↵1, . . . ,↵n) 2 ⌃ i↵ (by 3.2) ('1(↵1, . . . ,↵n)^'2(↵1, . . . ,↵n))2⌃. Which proves (*) in this case.
The case '=¬ is proved similarly. In case ' =9v
A ✏ 9v (v,↵1, . . . ,˜ ↵n) i˜ ↵ there is 2 ⇤ such that A ✏ ( ˜,↵1, . . . ,˜ ↵n) i˜ ↵
there is 2 ⇤ such that ( ,↵1, . . . ,↵n) 2 ⌃. The latter implies, by 3.1.6 and 3.2, that 9v (v,↵1, . . . ,↵n) 2 ⌃, and the converse holds because ⌃ is witnessing. This proves (*) for the formula and finishes the proof of (*) for all formulas.
Finally notice that (*) implies that A✏⌃.
We sometimes need to expand or reduce our language.
Let L be a language with non-logical symbols {Pi}i2I [{fj}j2J [{ck}k2K and L0 ✓ L with non-logical symbols {Pi}i
2I0 [{fj}j2J0 [{ck}k2K0 (I0 ✓ I, J0 ✓J, K0 ✓K). Let
A=hA;{PiA}i2I;{fjA}j2J;{cAk}k2Ki and
A0 =hA;{PiA}i2I0;{fjA}j2J0;{cAk}k2K0i.
Under these conditions we callA0 the L0-reduct of Aand, correspondingly,
A is an L-expansion of A0.
RemarkObviously, under the notations above for anL0-formula'(v1, . . . , vn) and a1, . . . , an 2A
Exercise 3.6. Let, for each i2N, ⌃i denote a set ofL sentences. Suppose
⌃0 ✓⌃1 ✓. . .⌃i. . . and each ⌃i is finitely satisfiable.
Then the union of the chain, Si2N⌃i, is finitely satisfiable.
Theorem 1(Compactness Theorem).Any finitely satisfiable set ofL-sentences
⌃ is satisfiable. Moreover, ⌃ has a model of cardinality less or equal to |L|.
Proof We introduce new languages Li and complete set of Li-sentences ⌃i (i = 0,1, . . .). Let L0 =L. By Lindenbaum’s Theorem there exists ⌃0 ◆ ⌃, a complete set of L0-sentences.
Given f.s. ⌃i in languageLi, introduce the new language Li+1 =Li[{c : a one variable Li-formula} and the new set of Li+1 sentences
⌃⇤i =⌃i[{(9v (v)! (c )) : a one variable Li-formula}. Claim. ⌃⇤
i is f.s. Indeed, for any finite S ✓ ⌃⇤i let S1 = S\⌃i and take a model A of S1 with domain A, which we assume well-ordered. Assign constants to symbols c as follows:
c = ⇢
the first element in (A) if (A)6=; the first element in A if (A) = ; . Denote the expanded structure A⇤. By the definition, for all (v),
A⇤ ✏9v (v)! (c ).So A⇤ ✏S. This proves the claim.
Let ⌃i+1 be a complete f.s. set of Li+1-sentences containing ⌃⇤i.
Take ⌃⇤ = Si2N⌃i. This is finitely satisfiable by 3.6. By construction one sees immediately that ⌃⇤ is also witnessing and complete set of sentences
in the language SLi = L +{new constants}. Proposition 2 gives us the canonical model, A⇤,of ⌃⇤.The reduct of A⇤ to language L is a model of ⌃.
The cardinality of the model we constructed is less or equal to |L| (see also Exercise 1.1).
As noted in the logic class, the contrapositive of the compactness theorem is the statement that logical consequence is intrinsically finitary:
Exercise 3.7. Show that a sentence is a logical consequence of some finite subset of a set ⌃ of sentences, if and only if it is a logical consequence of⌃. The following exercises draw some corollaries of compactness.
Exercise 3.8. Let T = T h(N,+,·,0,1). Let L0 ={+,·, c} be the language
obtained by adjoining a new constant symbol, T0 = T S{c 6= 0, c 6= 1, c 6= 1 + 1,· · ·}. Show that T0 has a model A0, and let A be the L-reduct of A0.
Show thatA is a model ofT, and is not minimalL-structure. Conclude that A, B are not isomorphic. This proves Skolem’s theorem, that the natural numbers are not characterized by their first-order theory.
Exercise 3.9. Show that there exists a model A of T h(R,+,·) containing an infinitesimal element, i.e. an element a such that for anyn 2N,A|= 0< na < 1.
Exercise 3.10. Assume the language has at least one constant symbol. Let
⌃ be a set of quantifier-free sentences. Assume ⌃ is satisfiable and that for any quantifier-free sentence , either 2 ⌃ or ¬ 2 ⌃. Show that there exists a unique minimalL-structure, up to isomorphism, which is a model of
⌃.
Exercise 3.11. Assume, for each n 2 N, that T has a model with at least n elements. Let be any set. Show thatT has a model Awhose universe A satisfies |A| | |. (Hint: introduce new constant symbols ci for i2 , and sentences ci 6=cj; use compactness.)
This was proved by Tarski, and called by him theupwardL¨owenheim-Skolem theorem.