[ Pobierz całość w formacie PDF ]
T such that F " Def(T ), we have
"a (a " T Ò! rng(F ¾!a) " T ).
Proof. Straightforward.
We shall now show that inaccessible cardinals give rise to transitive models
of ZFC. Recall that an inaccessible cardinal is a regular, uncountable, strong
limit cardinal. Recall also that we have identified cardinals with initial von
Neumann ordinals (cf. Sections 4.4.5 and 4.4.8).
Lemma 5.3.6. Let ´ be a limit ordinal > É. Then R´ is a transitive model of
all of the ZFC axioms except possibly the Replacement Scheme.
Proof. We apply Lemma 5.3.5. The Axioms of Extensionality and Foundation
hold in R´ because R´ is a transitive, pure, well-founded set. The Empty
Set, Power Set, Pairing, and Union Axioms and the Axiom of Choice and the
Comprehension Scheme hold in R´ because ´ is a limit ordinal. The Axiom of
Infinity holds in R´ because É " R´, since É
Lemma 5.3.7. An infinite cardinal » is regular if and only if, for all X †" »,
|X|
Proof. Straightforward.
Lemma 5.3.8. If » is an inaccessible cardinal, then
1. "x ((x †" R» '" |x|
2. "x (x " R» Ò! |x|
101
3. |R»| = ».
Proof. 1. Define Á : x ’! » by Á(u) = rank(u). Then |rngÁ| d" |x|
Since » is regular, it follows by the previous lemma that sup(rngÁ)
rngÁ †" ±
2. By transfinite induction on ±
If |R±| = º
cardinal. For limit ordinals ´
±
sup±
3. |R»| = sup±
Theorem 5.3.9. Let » be an inaccessible cardinal. Then R» is a transitive
model of ZFC.
Proof. Clearly » is a limit ordinal > É, hence by Lemma 5.3.6 we see that R»
satisfies all of the ZFC axioms except possibly the Replacement Scheme.
Let F : R» ’! R» and a " R» be given. Then rng(F ¾!a) †" R» and
|rng(F ¾!a)| d" |a|
cializing this to the case when F is definable over R», we see by Lemma 5.3.5
that the Replacement Scheme holds in R». This completes the proof.
Corollary 5.3.10. If there exists an inaccessible cardinal, then ZFC is consis-
tent.
Proof. Immediate from the theorem.
Exercise 5.3.11. A hereditarily finite set is a finite set x such that all elements
of x, elements of elements of x, . . . , are finite sets. Show that RÉ is the set of
all hereditarily finite, pure, well-founded sets. Show that RÉ is a model of all
of the axioms of ZFC except the Axiom of Infinity.
Exercise 5.3.12. Define E †" N by putting mEn if and only if 2m occurs in
1 k
the binary expansion of n, i.e., m = ni for some i where n = 2n + · · · + 2n .
1. Show that (N, E) (RÉ, "|RÉ).
=
2. Conclude that P †" Ék is definable over RÉ if and only if P is arithmetical.
Theorem 5.3.13. If there exists an inaccessible cardinal, then the existence of
an inaccessible cardinal is not a theorem of ZFC.
Proof. Assume that there exists an inaccessible cardinal. Let » be the smallest
inaccessible cardinal. By the previous theorem, R» is a model of ZFC. We claim
that R» also satisfies inaccessible cardinals do not exist . To see this, suppose
that R» satisfies there exists at least one inaccessible cardinal . Let º " R»
be such that R» satisfies º is an inaccessible cardinal . Then it is easy to see
that º is also an inaccessible cardinal in the real world. But clearly º
This contradicts the choice of ». Thus R» is a model of ZFC + inaccessible
cardinals do not exist .
102
The previous theorem shows that, if we assume only the axioms of ZFC,
then we cannot hope to prove the existence of inaccessible cardinals.
Exercise 5.3.14. Show that, if two or more inaccessible cardinals exist, then
the existence of two or more inaccessible cardinals is not a theorem of ZFC +
there exists at least one inaccessible cardinal .
Theorem 5.3.15. If there exists an inaccessible cardinal, then there exists a
countable, transitive model of ZFC.
Proof. Let » be an inaccessible cardinal. Then (R», "|R») is a model of ZFC.
By the Löwenheim-Skolem Theorem, there exists a countable set A †" R» such
that (A, "|A) is an elementary submodel of (R», "|R»). Thus (A, "|A) is a
countable, well-founded, extensional model of ZFC. By Theorem 5.3.2, (A, "|A)
is isomorphic to a transitive model (T, "|T ). Thus (T, "|T ) is a countable,
transitive model of ZFC.
Exercise 5.3.16. Let » be an inaccessible cardinal. Prove that there ex-
ists a limit ordinal ´
(R», "|R»).
5.4 Constructible Sets
Recall that, if T is any transitive, pure, well-founded set, Def(T ) is the set of all
subsets of T that are definable over T (i.e., over (T, "|T )) allowing parameters
from T .
Definition 5.4.1. By transfinite recursion we define L±, ± " Ord, as follows:
L0 = "
L±+1 = Def(L±)
[ Pobierz całość w formacie PDF ]