Exercises in Set Theory: Classical Independence Results
Mar 20, 2013
I worked on Chapter 13 and 14 of Thomas Jech’s book ‘‘Set Theory’’.
Doing the exercises from these chapters gave me the opportunity to come back
to the ‘‘classical’’ results about the independence of the Axiom of Choice
and (Generalized) Continuum Hypothesis by Kurt Gödel and Paul Cohen.
It’s funny to note that it’s easier to prove that AC holds in
the definition by ordinal induction provides the
well-ordering of the class of contructible sets) than to prove that GCH
holds in (you rely on AC in and on the technical condensation
lemma). Actually, I believe Gödel found his proof for AC
one or two years after the one for GCH.
On the other hand, it is easy to make GCH fails (just add
Cohen reals by Forcing) but more difficult to make AC fails
(e.g. AC is preserved by Forcing). This can be interpreted as AC being
more ‘‘natural’’ than GCH.
After reading the chapters again and now I analyzed in details
the claims, I’m now convinced about the correctness of the proof.
There are only two points I
didn’t verify precisely about the Forcing method (namely that all axioms of
predicate calculus and rules of inference are compatible
with the Forcing method ; that the Forcing/Generic Model theorems can
be transported from the Boolean Algebra case to the general case) but
these do not seem too difficult. Here are some notes about claims that were
not obvious to me
at the first reading. As usual, I hope they might be useful to
the readers of that blog:
In the first page of chapter 13, it is claimed that
for any set , and
. The first statement
is always true because
is by definition. However, the second
statement can only be true if is transitive
(since that implies ).
Indeed, if is transitive then for all
we have and since is we
If moreover we consider then
is also transitive. Hence the transitivity of the
can still be shown by ordinal induction.
The proof of lemma 13.7 can not be done exactly by induction on
the complexity of , as suggested. For example to prove (ii)
for , we would consider
and would like to say that is . Nevertheless,
we can not
deduce that from the induction hypothesis. Hence the right thing to do is
to prove the lemma for first and deduce the
lemma for (and ). Then
we can proceed by induction.
In the proof of theorem 13.18, it is mentioned that the assumption
implies that if then . To show that,
we consider the least ordinal such that
. In particular, is not limit
( and if for some limit
then there is such that ) and we can
write it . We have
so there is a formula and
elements such that
. Hence . Moreover
by minimality of , so
by (ii) we have and by (i) .
In lemma 14.18, we have expressions that seem
ill-defined for example where .
This happens in other places, like lemma 14.17 or definition 14.27.
is to understand that the functions are extended by 0. Indeed, for any
and similarly we get . Then we can use the inequality
page 207 () to
replace by its extension .
In lemma 14.23, the inequality
seems obvious but I don’t believe that it can be proved so easily at
that point. For example the proof from chapter 2 requires at least the
Separation axiom and the formulation from chapter 10 is based
on the Axiom of Regularity. To solve that issue, it seems to me that
the lemma should be moved after the proof that axioms of ZFC are valid
in . This is not an issue since lemma 14.23 is only used much
later in lemma 14.31.
Many details could be added to the proof of theorem 14.24, but
let’s just mention Powerset. For any , some
is defined and satisfies
(this follows from the
definitions, using the Boolean inequality
Since moreover we get
In theorem 14.34, we prove that any regular in
remains regular in (the hard case is really
uncountable and this assumption is implicitely used later
to say that is bounded).
It may not be obvious why this is enough. First recall that
for any ordinal ,
any (regular) cardinal in is a (regular) cardinal in .
Next we have,
to ‘‘ and have the same regular cardinals’’. Similarly,
we can prove that
is equivalent to
‘‘ and have the same cardinals’’.
The proof of theorem 14.34 shows that
‘‘ and have the same regular cardinals’’ and so
to complete the proof, it is enough to show that
and assume that there is such that
. Consider the least such .
By minimality of , and so
is a cardinal in . is actually regular in .
Otherwise, suppose and
such that . By minimality of ,
we have and
, a contradiction. Finally, we get
. This is again a contradiction and so