23.2. ARITHMETICAL DEFINABILITY AND FORCING

then the only way to expand N to get a model of F(G) is to take V as the denotation of G.

23.2 Arithmetical Definability and Forcing

We retain the terminology and notation of the preceding section. This entire section will be devoted to the proof of the following result.

23.3 Theorem (Addison’s theorem). The class of arithmetically definable sets of numbers is not an arithmetically definable class of sets.

The first notion we need is that of a condition, by which we mean a finite, consistent set of sentences of the language LG each either of the form Gm or Gm. The empty set is a condition. Other examples are {G17}, {G17, G59}, and

{G0, G1, G2, . . . , G999 999, G1 000 000}.

We use p, q, r as variables for conditions. We say a condition q extends or is an extension of a condition p if p is a subset of q. Thus every condition extends itself and extends .

Forcing is a relation between certain conditions and certain sentences of LG . We write p S to mean that condition p forces sentence S. The relation of forcing is inductively defined by the following five stipulations:

(1)If S is an atomic sentence of L, then p S if and only if N |= S.

(2)If t is a term of L and m is the denotation of t in N, then if S is the sentence Gt, then p S if and only if Gm is in p.

(3)If S is a disjunction (B C), then p S if and only if either p B or p C.

(4) If S is an existential quantification x B(x), then p S if and only if, for some n,

pB(n).

(5)If S is a negation B, then p S if and only if, for every q that extends p, it is not the case that q S.

The last clause bears repeating: a condition forces the negation of a sentence if and only if no extension forces the sentence. It follows that no condition forces some sentence and its negation, and also that either a condition forces the negation of a sentence or some extension forces the sentence. (It will soon be shown that if a condition forces a sentence, so does every extension of it.)

It follows from (2) and (5) that p Gm if and only if Gm is in p. For if Gm is not in p, then p {Gm} is an extension of p that forces Gm. So if p Gm, that is, if no extension of p forces Gm, then Gm must be in p. Conversely, if Gm is in p, then Gm is in no extension of p, and hence no extension of p forces Gm, and so p Gm.

Thus {G3} forces neither G11 nor G11, and so does not force (G11 G11). Thus a condition may imply a sentence without forcing it. (It will soon be seen that the inverse is also possible; that, for example, forces x Gx, even though it does not imply it, and does not force x Gx.)

23.4 Lemma. If p S and q extends p, then q S.

Proof: Suppose that p S and q extends p. The proof that q S is by induction on complexity of S. The atomic case has two subcases. If S is an atomic sentence of L, then since p S, S is true in N, and since S is true in N, q S. If S is an atomic sentence of form Gt, then since p S, Gm is in p, where m is the denotation of t in N, and since q extends p, Gm is also in q and q Gt. If S is (B C), then since p S, either p B or p C; so by the induction hypothesis, either q B or q C, and so q (B C). If S is x B(x), then since p S, we have p B(m) for some m; so by the induction hypothesis, q B(m) and q x B(x). Finally, if S isB, then since p S, no extension of p forces B; and then, since q is an extension of p, every extension of q is an extension of p, so no extension of q forces B, and so q B.

Two observations, not worthy of being called lemmas, follow directly from the preceding lemma. First, if p B, then p B; for any extension of p will force B, hence no extension of p will force B. Second, if p B and p C, then p (B C); for every extension of p will force both B and C, and so will force neither B nor C, and so will not force (B C).

A more complicated observation of the same kind may be recorded here for future reference, concerning the sentence

( ) ( ( B C) (B C))

( B C). Also p (B C), so p (B C). Hence by our second observation, p (*). Similarly, if p B and p C, then again p (*).

23.5 Lemma. If S is a sentence of L, then for every p, p S if and only if N |= S.

Proof: The proof again is by induction on the complexity of S. If S is atomic, the assertion of the lemma holds by the first clause in the definition of forcing. If S is (B C), then p S if and only if p B or p C, which by the induction hypothesis is so if and only N |= B or N |= C, which is to say, if and only if N |= (B C). If S is x B(x), the proof is similar. If S is B, then p S if and only if no extension of p forces B, which by the induction hypothesis is so if and only if it is not the case that N |= B, which is to say, if and only if N |= B.

Forcing is a curious relation. Since does not contain any sentence Gn, for no n does force Gn, and therefore does not force x Gx. But does force x Gx! For suppose some p forces x Gx. Let n be the least number such that Gn is not in p. Let q be p {Gn}. Then q is a condition, q extends p, and q forces Gn, so q forces x Gx. Contradiction. Thus no p forces x Gx, which is to say, no extension of forces x Gx, so forces x Gx.

We are going to need some more definitions. Let A be a set of numbers. First, we call a condition p A-correct if for any m, if Gm is in p, then m is in A, while if Gm is in p, then m is not in A. In other words, p is A-correct if and only if

denote A) is a model of p.

Further, say A FORCES S if some A-correct condition forces S. Note that the union of any two A-correct conditions is still a condition and is still A-correct. It follows that A cannot FORCE both S and S, since the union of an A-correct condition forcing S with one forcing S would force both, which is impossible.

Finally, we call A generic if for every sentence S of LG, either A FORCES S or A FORCES S. If this is so at least for every sentence S with at most n occurrences of logical operators, we call A n-generic. Thus a set is generic if and only if it is n-generic for all n.

The first fact about generic sets that we have to prove is that they exist.

23.6 Lemma. For any p, there is a generic set A such that p is A-correct.

Proof: Let S0, S1, S2, . . . be an enumeration of all sentences of LG . Let p0, p1, p2, . . . be an enumeration of all conditions. We inductively define a sequence q0, q1, q2, . . . of conditions, each an extension of those that come before it, as follows:

(0)q0 is p.

(1)If qi forces Si , then qi+1 is qi .

(2)If qi does not force Si , in which case there must be some q extending qi and forcing Si , then qi+1 is the first such q (in the enumeration p0, p1, p2, . . .).

Let A be the set of m such that Gm is in qi for some i.

We claim that p is A-correct and that A is generic. Since p = q0, and since for each i, either qi+1 Si or qi+1 Si , it will be enough to show that for each i, qi is A-correct. And since m is in A when Gm is in qi , it is enough to show that if Gm is in qi , then m is not in A. Well, suppose it were. Then Gm would be in q j for some j. Letting k = max(i, j), both Gm and Gm would be in qk , which is impossible. This contradiction completes the proof.

The next fact about generic sets relates FORCING and truth.

23.7 Lemma. Let S be a sentence of LG, and A a generic set. Then A FORCES S if and only if NAG |= S.

Proof: The proof will be yet another by induction on complexity, with five cases, one for each clause in the definition of forcing. We abbreviate ‘if and only if’ to ‘iff’.

Case 1. S is an atomic sentence of L. Then A FORCES S iff some A-correct p forces S, iff (by Lemma 23.5) N |= S, iff NAG |= S.

Case 2. S is an atomic sentence Gt. Let m be the denotation of t in N. Then A FORCES S iff some A-correct p forces Gt, iff Gm is in some A-correct p, iff m is in A, iff NAG |= Gt.

Case 3. S is (B C). Then A FORCES S iff some A-correct p forces (B C), iff some A-correct p forces B or forces C, iff either some A-correct p forces B or some A-correct p forces C, iff A FORCES B or A FORCES C, iff (by the induction hypothesis) NAG |= B or NAG |= C, iff NAG |= (B C).

292 ARITHMETICAL DEFINABILITY

Case 4. S is x B(x). Then A FORCES S iff some A-correct p forces x B(x), iff for some A-correct p there is an m such that p forces B(m), iff for some m there is an A-correct p such that p forces B(m), iff for some m, A forces B(m), iff (by the induction hypothesis) for some m, NAG |= B(m), iff NAG |= x B(x).

Case 5. S is B. No set FORCES both B and B. Since A is generic, A FORCES at least one of B or B. Hence A FORCES B iff it is not the case that A FORCES B, iff (by the induction hypothesis) not NAG |= B, iff NAG |= B.

The last fact about generic sets that we have to prove is that none of them is arithmetical.

23.8 Lemma. No generic set is arithmetical.

Proof: Suppose otherwise. Then there is a generic set A and a formula B(x) of L such that for every n, n is in A if and only if N |= B(n). So NAG |= x(Gx ↔ B(x)) or NAG |= x F(x), where F(x) is the following logical equivalent of (Gx ↔ B(x)):

( ( Gx B(x)) (Gx B(x))).

By Lemma 23.7, A FORCES x F(x), so some A-correct p forces x F(x), so for no q extending p and no n does q force F(n), which is to say

Let k be the least number such that neither Gk nor Gk is in p. Define a condition q extending p by letting q = p {Gk} if N |= B(k) and letting q = p {Gk} if N |= B(k). In the former case, q Gk, while by Lemma 23.5 q B(k). In the latter case, q Gk while by Lemma 23.5 q B(k). In either case, q (*) by our observations following Lemma 23.4, which is to say q F(n). Contradiction.

Suppose that at the beginning of the proof of Lemma 23.6, instead of enumerating all sentences we enumerate those sentences of complexity ≤ n (that is, having no more than n occurrences of logical operators). Then the proof would establish the existence of an n-generic set rather than of a generic set. Suppose that in the hypothesis of Lemma 23.7 we only assume the set A is n-generic rather than generic. Then the proof would establish the conclusion of Lemma 23.7 for sentences of complexity ≤ n, rather than for all sentences. But suppose that in the hypothesis of Lemma 23.8 we only assume the set A is n-generic rather than generic. Then the proof would break down entirely. And indeed, in contrast to Lemma 23.8, we have the following.

23.9 Lemma. For any n, there is an n-generic set A that is arithmetical.

Proof: The proof will be indicated only in outline. The idea is to carry out the construction in the proof of Lemma 23.6, starting from an enumeration of all sentences of complexity ≤ n, and with p = . It is necessary to show that, if code numbers are assigned in a suitable way, then various relations among code numbers connected with the construction will be arithmetical, with the result that the generic set constructed is arithmetical as well.

First note that, since we have seen in the preceding section that the set of code numbers of sentences of complexity ≤ n is recursive, the function enumerating the

elements of that set in increasing order is recursive. That is, if we enumerate the sentences S0, S1, S2, . . . in order of increasing code number, then the function taking us from i to the code number for Si will be recursive.

We also enumerate the conditions p0, p1, p2, . . . in order of increasing code number, where code numbers are assigned to finite sets of sentences—for that is what conditions are—as in section 15.2. As we observed in section 15.2, the relation ‘the sentence with code number i belongs to the set with code number s’ is recursive. Using this fact and the fact that the function taking m to the code number for Gm— essential the substitution function σ used in the preceding section—is recursive, it is not hard to show that the set of code numbers of conditions is recursive, and that the relation that holds between m and s if and only if s is the code number of a condition containing Gm is recursive. We also observed in section 15.2 that the relation ‘the set with code number s is a subset of the set with code number t’ is recursive. Hence the relation that holds between s and t if and only if they are code numbers of conditions p and q, with q an extension of p, is also recursive. Being recursive, the various functions and relations we have mentioned are all arithmetical.

We also need one more fact: that for each n, the relation that holds between i and s if and only if i is the code number of a sentence S of complexity ≤n and s is the code number of a condition p, and p forces S, is arithmetical. The proof is very similar to the proof in the preceding section that for each n the set Vn is arithmetical, and will be left to the reader.

Now returning to the construction of an n-generic set A, by the method of the proof of Lemma 23.6, we see that m is in A if and only if there exists a sequence s of conditions such that the following hold (for each i less than the length of the sequence):

(0)The 0th entry of the sequence is the empty condition

(1)If the ith entry of the sequence forces the negation of the ith sentence in the enumeration of sentences, then the (i + 1)st entry is the same as the ith.

(2)Otherwise, the (i + 1)st entry is a condition that extends the ith and that forces the ith sentence in the enumeration of sentences, and is such that no condition earlier in the enumeration of conditions (that is, no condition of smaller code number) does both these things.

(3)The sentence Gm belongs to the last entry of the sequence.

We can, of course, replace ‘there exists a sequence. . .’ by ‘there exists a code number for a sequence. . .’. When everything is thus reformulated in terms of code numbers, what we get is a logical compound of relations that we have noted in the preceding several paragraphs to be arithmetical. It follows that A itself is arithmetical.

At last we are in a position to prove Addison’s theorem.

Proof of Theorem 23.3: Suppose the theorem fails. Then there is a sentence S of LG such that for any set A, NAG |= S if and only if A is arithmetical. Let S be of complexity n. By Lemma 23.9 there exists an n-generic set A that is arithmetical. So NAG |= S. So by Lemma 23.7 (or rather, the version for n-generic sets and sentences of complexity ≤n, as in our remarks following Lemma 23.8), A FORCES S. So some