Forcing and absoluteness as means to prove
theorems
Matteo Viale
Dipartimento di Matematica
Universit`a di Torino
13 June 2012
Freiburg
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FORCING
Forcing was introduced in 1963 by Paul Cohen to prove the independence
with respect to ZFC of the continuumhypothesis,the ﬁrst in the list of 23
Hilbert’s problems.
It soon emerged that forcing is a very powerful tool to prove independence
results in ZFC and also in many other branches of pure mathematics,for
example:
Group theory (Whitehead’s problem) Shelah 1974,
General topology:Todorˇcevi ´c and Moore’s results on the Sspace
and the Lspace problems,
Functional analysis:many results of Todor ˇcevi ´c on Banach spaces,
Operator algebras:Farah’s works on the automorphisms of the Calkin
algebra which develops on Shelah and Veli ˇckovi ´c’s analysis of the
automorphismgroup of P(!)=FIN,
...............
I’msurely forgetting many fundamental results and contributions,
hopefully
by people not in this room.
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FORCING
Forcing can be seen as a “computable” function:
(M;B) 7!M
B
M is a (transitive) model of (a large enough fragment of) ZFC.
B 2 M is a boolean algebra which M models to be complete.
(being a boolean algebra is an absolute property of transitive models
while being complete varies fromone transitive model to another)
M
B
is a boolean valued model of ZFC and is a deﬁnable class in M.
There is a deﬁnable in M injective map
i
B
:M!M
B
i
B
:M!M
B
a 7!
a
such that M will be naturally identiﬁed with
i
B
[M] M
B
.
Truth in M
B
depends on
I
The ﬁrst order theory of M.
I
The combinatorial properties that M gives to B.
Truth in M
B
is (almost) deﬁnable in M with parameter B.
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The forcing relation
To evaluate the semantics of M
B
one introduces a (deﬁnable in M)
boolean evaluation of formulas:
Deﬁnition
Given
1
;:::;
n
2 M
B
,
(x
1
;:::;x
n
) formula in the language of set theory in the free
variables x
1
;:::;x
n
,
a boolean value ~(
1
;:::;
n
)
B
in B is assigned as follows:
~ ^
B
= ~
B
^
B
~
B
,
~:
B
=:
B
~
B
,
~8x(x)
B
=
V
B
f~()
B
: 2 M
B
g.
What is difﬁcult is to deﬁne ~
1
2
2
B
and ~
1
=
2
B
.
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What holds in M
B
Theorem(Cohen)
Assume M is a (transitive) model of ZFC and B 2 M is a complete boolean
algebra in M.Then
M j= (~
B
= 1
B
)
for any axiom of ZFC.
What else holds in M
B
depends on the choice of B and the ﬁrst order
properties of M.
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Baire’s category theorem and forcing
Given a boolean algebra B,B
+
is the set of its positive elements
(i.e.B
+
= B n f0
B
g).
G B
+
is a ﬁlter on B if:
for all p 2 G and q p,q 2 G,
1
B
2 G,
for all p;q 2 G,p ^ q 2 G.
D B
+
is dense in B if for all p 2 P,there is q 2 D such that q p.
A B
+
is open in B if whenever p 2 A and q p,q 2 A as well.
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Baire’s category theorem and forcing
FA
(B) holds if for all family fD
: < g of dense open subsets of B,there
is a ﬁlter G which meets all these dense sets.
Theorem(Baire’s category theorem)
For all boolean algebras B,FA
@
0
(B) holds.
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Towards Cohen’s forcing Theorem
Fromnow on,we assume V exists and is the “true” universe of sets.Else
(if one does not want to be platonist) one has to reformulate everything
with more care.
Let (M;2) 2 V be a model of a large enough fragment of ZFC.
Typically:
M H
or M V
for some large enough regular cardinal or some
big enough ordinal ,
M =
N
[N] where N H
(N V
) and
N
is the transitive collapse of
the structure (N;2\N
2
).
Let B 2 M be such that M models B is a boolean algebra.
Deﬁnition
G B is an Mgeneric ﬁlter for B if
G\D\M is nonempty for all D 2 M dense open subset of B.
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Cohen’s forcing Theorem
Theorem(Cohen’s forcing Theorem)
Assume:
M 2 V is a transitive model of ZFC of size at most ,
B 2 M is a boolean algebra such that FA
(B) holds in V,
G 2 V is an Mgeneric ﬁlter.
Then there is a surjective map
G
:M
B
!M[G] such that:
M[G] is transitive and
G
(
a) = a for all a 2 M.
(M[G];2) j= (
G
(
1
);:::;
G
(
n
))
if and only if
~(
1
;:::;
n
)
B
2 G.
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Cohen’s absoluteness Lemma
Corollary
Assume that:
(x;r) is a
0
formula with real parameter r.
B 2 V is a Boolean algebra such that
V j= (1
B
= ~9x(x;
r)
B
):
Then L(R) j= 9x(x;r).
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Proof:
Assume B 2 V is a Boolean algebra such that
V j= (1
B
= ~9x(x;
r)
B
):
To simplify matters assume there is an inaccessible such that B 2 V
(redundant assumption).
Then V
j= ZFC and
V
j= (1
B
= ~9x(x;r)
B
):
Pick N V
countable such that B 2 N.
Let M =
N
[N] and Q =
N
(B).Notice that r 2 P(!) and
N
(!) =!,
Thus
N
(r) = r.
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Proof continued
Since
N
:N!M is an isomorphismand Q =
N
(B),
N
(r) = r,
M j= (1
Q
= ~9x(x;r)
Q
):
Now M is countable and transitive,Q 2 M and FA
@
0
(Q) holds in V.
Thus there is G 2 V which is an Mgeneric ﬁlter for Q.
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Proof continued
By Cohen’s forcing Theoremwe can deﬁne
G
:M
Q
!M[G] surjective
such that
G
(
a) = a for all a 2 M,
M[G] is transitive,
M[G] j= iff ~
Q
2 G.
In particular since
G
(
ir) = r
M[G] j= 9x(x;r):
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Proof continued
Thus there is a 2 M[G] such that M[G] j= (a;r).
Since M[G] is countable and transitive,M[G] 2 L(R) and M[G] L(R),
thus a;r 2 L(R).
Since (a;r) is a
0
formula with parameters in M[G] L(R):
M[G] j= (a;r) () L(R) j= (a;r):
In particular a witnesses that L(R) j= 9x(x;r).
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Cohen’s absoluteness Lemma reformulated
Actually if one doesn’t want to commit to any philosophical position on the
onthology of sets Cohen’s absoluteness Lemma can be formulated as
follows:
Corollary (Cohen)
Let T be any ﬁrst order theory which extends ZFC and (x;r) be a
0
formula with a parameter r such that T`r !.
TFAE:
T`9x(x;r).
T`There exists a boolean algebra B such that 1
B
= ~9x(x;
r)
B
.
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Woodin’s absoluteness
Theorem(Woodin)
Assume there are class many Woodin cardinals in V,then for every
formula with real parameters:
L(R)
V
j=
if and only if there exists a boolean algebra B 2 V such that
V j= (1
B
= ~
L(R
(r)
B
):
Notice that we had to relativize the formulas to L(R) to obtain the
absoluteness results.
This is an unavoidable consequence of the fact that formulas which are not
0
are neither upward absolute nor downward absolute between transitive
structures.
(We needed the upward absoluteness of (a;r) to conclude the proof of
Cohen’s absoluteness.)
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Effects of large cardinals on the theory of L(R)
If one investigates with care Woodin’s proof,the assumption that V is
transitive is redundant.
In particular Woodin actually proved:
Theorem(Woodin)
Let T be any theory which extends ZFC+there are class many Woodin
cardinals and r be a parameter such that T`r !.Then for any formula
(x) TFAE:
T`[L(R) j= (r)],
T`There is a boolean algebra B such that 1
B
= ~
L(R)
(
r)
B
.
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What can be said about L(P()) and forcing
absoluteness?
If @
1
there are (provably in ZFC) boolean algebras B such that FA
(B)
is false.
Nonetheless we can prove on the same lines the following:
Corollary (Generalized Cohen’s absoluteness)
Assume that:
(x;p) is a
0
formula with parameter p 2 P().
B 2 V is a Boolean algebra such that
V j= (1
B
= ~9x(x;p)
B
):
FA
(B) holds in V.
Then L(P()) j= 9x(x;p).
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When does some B satisfy FA
(B)?
Without further assumptions,the generalized version of Cohen’s
absoluteness lemma is not so useful.
Whether a boolean algebra B 2 V satisfy FA
(B) for some @
1
is very
much dependent on the particular choice of set theoretic universe V we
work in.
We need more assumption on V.
We have a very natural candidate for such an assumption,subsumed by
the following slogan:
FA
(B) holds for all B for which we cannot prove that it fails.
Notice that for = @
0
this slogan is trivially true:
in ZFC we can prove that FA
@
0
(B) holds for all B.
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Martin’s maximum makes the slogan true for = @
1
A complete Boolean algebra B is locally stationary set preserving if there
is some ﬁxed b 2 B
+
such that
b
B
~
S is stationary in !
1
B
for every S which is a stationary subset of!
1
.
Shelah proved that if B is not locally stationary set preserving then
FA
@
1
(B) fails.
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Martin’s maximum makes the slogan true for = @
1
,
continued
Foreman,Magidor,Shelah proved the following:
Theorem
Assume there is a supercompact cardinal in V.Then there is a boolean
algebra B 2 V such that
1
B
= ~FA
@
1
(B) holds for all B which are locally stationary set preserving
B
Let us abbreviate by MM(Martin’s maximum) the assertion that
FA
@
1
(B) holds for all B which are locally stationary set preserving.
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Why MM (actually MM
++
) is useful.
The model V
B
produced by Foreman,Magidor,Shelah shows the
consistency of an enhanced version of MMwhich we call MM
++
.
MM
++
is a natural statement stronger than MMbut not so simple to
formulate.
MM(MM
++
) is extremely powerful,a variety of problems in set theory and
in general mathematics have a solution if we assume MM
++
(often much
less sufﬁce):
the continuumproblem,
the singular cardinal problem,
the Suslin problem,
all mathematical problems mentioned in the ﬁrst slides,
many,many other problems
the majority of which can be formalized by
2
properties of H
@
2
have a solution assuming MM.
Why?
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Most likely because of forcing absoluteness!
At least this is the case for all problems which can be formulated as
2
questions on H
!
2
.Among which:
the continuumproblem,
the Souslin hypothesis,
the ﬁve element basis for uncountable linear orders,
the Kurepa hypothesis,
......
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Theorem(V.)
Assume T is a theory which extends ZFC+MM
++
+there are class many
Woodin cardinals.
For any
2
formula (p) in a parameter p such that
T`p !
1
the following are equivalent:
T`
H
!
2
(p),
T`There is a boolean algebra B which is stationary set preserving
and such that 1
B
= ~BMM
B
and 1
B
= ~
H
!
2
(p)
B
.
(BMMis a weakening of MMand follows fromMM
++
).
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Conclusions
First conclusion:we cannot use stationary set preserving forcings to
prove independence results for the
2
theory of H
!
2
in models of MM
++
+
large cardinals.
We can just use these forcings to prove implications fromMM
++
.
Eventually we could use forcings which are not stationary set preserving to
prove independence results for the theory MM
++
+ large cardinals.
It is not known if this can be the case,one line of attack could be to
investigate the relation between Woodin’s axiom () and MM
++
.
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Second conclusion:the above theoremmakes MM
++
a very appealing
axiomto extend ZFC in conjunction with large cardinals:
it is simple to state and assert a maximality principle for the cardinal @
1
which provably holds at @
0
(i.e.that an enhanced version of the Baire’s
category theoremFA
@
1
(B) holds for the largest conceivable class of
boolean algebras B).
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Third conclusion:to make this axiommore appealing we should show
that this maximality property does not hold just for @
0
and @
1
but can be
formulated and proved consistent simultaneously for all regular
uncountable cardinals .
This is what I’mcurrently working on...
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I should also say that I cut out many other absoluteness results of Woodin
about CH
and on axiom (),
a variant of BMMwhich does not follow from
MMand whose relation with MM
++
is unknown.
Thanks for your patience and attention.
BUT:
don’t leave immediately!
I may go on on the blackboard with a
second (harder) part.........
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