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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 G�del 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},
107
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
� �
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 � = (�)G " M[G]. Furthermore, M[G] satisfies Extensionality
�
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 (", �!, �!, ").
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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