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V = L plus the Axiom of Choice plus the GCH. This proves the first part. The
proof of the second part is similar, starting with a model (A, E) that is not
necessarily transitive.
Remark 5.4.17. The previous theorem, due to Gödel 1939, is one of the most
significant achievements of axiomatic set theory. The second part is sometimes
described as a relative consistency result: ZFC + GCH is consistent relative to
ZF.
5.5 Forcing
Let M be a countable transitive model of ZFC. Let P = (P, d") be a partially
ordered set belonging to M. We say that p, q " P are compatible if there exists
r " P such that r d" p and r d" q. If p, q " P are incompatible, we write p ¥" q.
Definition 5.5.1. A filter on P is a set G †" P such that
1. p, q " G implies "r " G (r d" p, q);
2. p " G, p d" q imply q " G.
Definition 5.5.2. D †" P is dense open if
1. "p " P "q d" p (q " D);
2. "p " D "q d" p (q " D).
Definition 5.5.3. A filter G †" P is said to be M-generic if G )" D = " for all
dense open D †" P such that D " M.
Lemma 5.5.4. Given p " P we can find an M-generic filter G †" P such that
p " G.
Proof. Let {Dn | n " N} be an enumeration of {D " M | D dense open in P }.
Put p0 = p and, given pn, let pn+1 d" pn be such that pn+1 " Dn. It is easy to
verify that G = {q " P | "n (pn d" q)} is an M-generic filter.
Definition 5.5.5. Let G be an M-generic filter. We put
M[G] = {aG | a " M},
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where
aG = {bG | (p, b) " a for some p " G}.
Remark 5.5.6. Think of each a " M as a  name for aG " M[G]. We shall
show that M[G] is a countable transitive model of ZFC containing M.
Lemma 5.5.7. M[G] is a countable transitive set. We have M[G] ‡" M *" {G},
and Ord )" M[G] = Ord )" M.
Proof. It is obvious from the definition that M[G] is a countable transitive set.
For all a " M we have a = (a)G, where a = P × { | b " a}. We also have
Ù Ù
G = ( )G, where  = {(p, W) | p " P }. Thus M *" {G} †" M[G], and from this
it follows that Ord )" M †" Ord )" M[G]. On the other hand, for each a " M we
clearly have rank(a) e" rank(aG), hence Ord )" M ‡" Ord )" M[G].
A major result is:
Theorem 5.5.8. M[G] is a countable transitive model of ZFC.
Remark 5.5.9. The proof of Theorem 5.5.8 is long and employs a new method
known as forcing. However, some parts of the proof are easy and do not require
forcing.
For example, given a, b " M, put c = P × {a, b}, then cG = {aG, bG}. This
shows that M[G] satisfies the Pairing Axiom. Also, M[G] satisfies the Axiom of
Infinity because É = (É)G " M[G]. Furthermore, M[G] satisfies Extensionality
Ù
and Foundation automatically, because M[G] is a transitive set.
So far we have not used the assumption that G is an M-generic filter.
In order to prove the rest of Theorem 5.5.8, we now introduce the method
of forcing.
Definition 5.5.10. The forcing language consists of binary relation symbols "
and = plus constant symbols a for each a " M. Sentences of the forcing language
are of the form F (a1, . . . , an), where F (x1, . . . , xn) is a formula of L" with free
variables x1, . . . , xn, and a1, . . . , an " M. We have M[G] |= F (a1, . . . , an) if and
only if F (a1, . . . , an) is true in M[G], where quantifiers are interpreted as ranging
over M[G], and a1, . . . , an are interpreted as (a1)G, . . . , (an)G respectively.
Definition 5.5.11 (forcing). Let p " P and let F be a sentence of the forcing
language. We say that p F (read p forces F ) if and only if M[G] |= F for all
-
M-generic filters G such that p " G.
Lemma 5.5.12 (the extension lemma). If p F and q d" p, then q F .
- -
Proof. This is obvious, because q " G, q d" p imply p " G.
Lemma 5.5.13 (definability of forcing). For each formula F (x1, . . . , xn) there
"
is a formula F (p, x1, . . . , xn) such that, for all p " P and a1, . . . , an " M,
"
p F (a1, . . . , an) if and only if M |= F (p, a1, . . . , an).
-
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Lemma 5.5.14 (forcing equals truth). M[G] |= F if and only if "p " G (p F ).
-
Proof. We shall prove Lemmas 5.5.13 and 5.5.14 by simultaneous induction on
the number of connectives and quantifiers in F . We assume that F contains
only '", ¬ , and " (not (", Ò!, Ô!, ").
For the base step, we need to find formulas ""(p, x, y) and ="(p, x, y) of
L" defining the relations p a " b and p a = b, respectively, over M. The
- -
formulas "" and =" are defined by a rather complicated simultaneous transfinite
recursion within M. The properties
p a " b if and only if M |= ""(p, a, b)
-
and
p a = b if and only if M |= ="(p, a, b)
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