independence of the continuum hypothesis - Department of
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INDEPENDENCE OF THE CONTINUUM HYPOTHESIS
CAPSTONE
MATT LUTHER
1
INDEPENDENCE OF THE CONTINUUM HYPOTHESIS 2
1. Introduction
This paper will summarize many of the ideas from logic and set theory that areneeded in order to follow Paul Cohen's proof that the Continuum Hypothesis is notimplied by the axioms of Zermelo-Fraenkel set theory with Choice. We will thentrace the majority of his proof, in which we will see the technique of forcing andhow it is used to build a model in which the Continuum Hypothesis fails.
2. History
Georg Cantor began development of set theory in the 1870s while investigatingtrigonometric series and the structure of the real numbers. Cantor used the exis-tence of one-to-one correspondences between the elements of sets to compare theirsizes, or cardinalities. His ļæ½rst major results regarding the comparison of cardinali-ties of sets came in 1874 when he published a paper containing a proof that the setof algebraic numbers could be put in one-to-one correspondence with the naturalnumbers and another proof that there was no such correspondence between the realnumbers and the naturals. The latter of these proofs established for the ļæ½rst timethat there are inļæ½nite sets of diļæ½erent sizes.
Over the next two decades, Cantor expanded his theory, and by the 1890s had de-veloped the transļæ½nite cardinal and ordinal numbers, which serve as representativesfor the diļæ½erent sizes and order types of sets, respectively. Among his publicationswere proofs that the cardinality of the reals is the same as the cardinality of thepower set of the naturals, P(N), and that for any set X, its power set P(X) has alarger cardinality. Cantor proved that by taking successive power sets, one couldalways ļæ½nd higher cardinality sets. However, it was unclear whether or not thecardinality of P(X) is the next cardinal, Īŗ+, whenever the cardinality of X is someinļæ½nite cardinal Īŗ. That is, it was unclear whether the power set operation alwaysgives the next inļæ½nite cardinal, or if there might be some cardinal between |X| and|P(X)|. He proposed that the cardinality of the reals |R|, and equivalently |P(N)|,was indeed the next smallest cardinal after the size of N, but he was unable to provethis. This claim is known as the Continuum Hypothesis.
The transļæ½nite cardinals begin at the cardinality of the naturals, which is de-noted by the cardinal number āµ0. Sets with cardinality āµ0 are called countable.Assuming the Axiom of Choice, we have that the next smallest cardinal after āµ0
is the cardinality of the set of all ordinals with cardinality less than or equal toāµ0, that is, the set of all at-most countable ordinals. This next smallest cardinal isdenoted āµ1. We also know that the size of P(X) for any set X is the cardinality of
the set of functions from X to the set {0, 1}, which is 2|X|. Thus we can write theContinuum Hypothesis (CH) quite succinctly as
āµ1 = 2āµ0
This can be read as stating that the next cardinality after āµ0 is the size of thepower set of any set with cardinality āµ0. It can also be read as saying that theset of all ordinals with cardinality ā¤ āµ0 is the same size as the collection of allsubsets of a set with cardinality āµ0. More tangibly, this claims that the size of theset of reals is the next smallest cardinality after the size of the naturals. Cantormade a general version of this claim, reasonably named the Generalized Continuum
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Hypothesis (GCH), which is written as
āµĪ±+1 = 2āµĪ± for all ordinals Ī±
This says that the next cardinality after |X| is always the size of the power set,|P(X)|. In the absence of the Axiom of Choice, we do not know what the smallestcardinal after āµĪ± is, so the GCH would be stated as follows.
Given an inļæ½nite cardinal Īŗ, there exists no cardinal Ī» such that Īŗ < Ī» < 2Īŗ
In the years after Cantor's development of what would later became knownas naive set theory, work was done to axiomatize the intuitive idea of sets. In1908, Ernst Zermelo put forth his axiomatic set theory, which eventually becameZermelo-Fraenkel set theory with the Axiom of Choice. Zermelo-Fraenkel set theoryis commonly abbreviated as ZF when treated without the Axiom of Choice, andabbreviated ZFC when the Axiom of Choice is included. We will abbreviate theAxiom of Choice as AC when thought of as a statement on its own. The majorwork regarding the Continuum Hypothesis following Cantor's investigations wouldlater all take place within ZF and ZFC.
Little progress was made regarding the truth of the Continuum Hypothesis until1938 when Kurt Gƶdel established that if the set ZF of statements that serve asthe axioms of ZF set theory is consistent, then there is a model of ZF in which theGeneralized Continuum Hypothesis and the Axiom of Choice are both true. As aresult of Gƶdel's completeness theorem from logic, the existence of such a modelimplies that neither GCH nor CH can be disproven in either ZF or ZFC set theory,provided that ZF set theory is consistent. This however did not demonstrate thateither form of the Continuum Hypothesis is true or provable, just that they cannotbe disproven in these theories.
The question of the Continuum Hypothesis again went without much progressuntil 1963 when Paul Cohen showed that if the set of statements ZF which makeup the axioms of ZF set theory is consistent, then there is a model of ZF in whichthe Generalized Continuum Hypothesis fails but the Axiom of Choice holds. Forthe same reason that Gƶdel's result shows that Ā¬GCH cannot be proven, Cohen'sresult shows that CH cannot be proven in ZFC set theory either.
The end result of Gƶdel and Cohen's work is that regardless of whether we areworking in ZF or ZFC set theory, both the Continuum Hypothesis and GeneralizedContinuum Hypothesis cannot be proven or disproven. Statements such as thesewhich cannot be proven true or false in some theory are called independent of thetheory.
3. Models, Consistency, and Independence
We will begin with several concepts from mathematical logic which will allow usto better understand exactly what it means for the Continuum Hypothesis to beindependent and what the proofs by Gƶdel and Cohen actually say.
Deļæ½nition 3.1. A theory is a set of sentences in a formal language.
For example, we may have a theory TO intended to represent strictly orderedsets which has only the following two statements as elements.
ā¢ The Axiom of Asymmetry: there are no a, b such that a < b and b < a.ā¢ The Axiom of Transitivity: for all a, b, c, if a < b and b < c, then a < c.
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Deļæ½nition 3.2. A model for a theory T is a structure that tells how to interpretthe non-logical symbols of the language that T is in, so that every statement in thetheory T is true.
Models provide a domain for variables to range over and provide deļæ½nitions for therelations like the < sign that appears in our above sentences. To continue with ourexample theory TO, consider the structure M where we let our variables range overthe set {0, 1, 2} and we deļæ½ne < as the usual less-than relation on integers. We cansee that M is a model for TO because
ā¢ The Axiom of Asymmetry holds: for any two numbers selected from {0, 1, 2},at least one of them is not less than the other.
ā¢ The Axiom of Transitivity holds: for any a, b, c in {0, 1, 2}, if a is less thanb and b is less than c, then a is less than c.
However, if we consider the structure N where the variables range over the sameset {0, 1, 2} but the relation < is deļæ½ned by a < b if and only if b = a + 1, thenwe see that N is not a model for TO because the Axiom of Transitivity is false; wehave 0 < 1 and 1 < 2, yet 0 < 2 is false since 2 6= 0 + 1.
Deļæ½nition 3.3. We say that a statement S is provable in a theory T if thereexists a proof of S which may contain the statements in T as assumptions. Thatis, S is provable in T if there exists a ļæ½nite sequence of statements (S1, S2, ..., Sn)such that each statement Si is either a statement in T or follows from the previousstatements by some rule of logic.
The following theorem relates the provable statements of a theory to the modelsof that theory.
Theorem 3.4. (The Soundness Theorem): If a statement S is provable in atheory T , then S is true in every model of T .
Simply put, this says that if we can deduce a statement from a set of assump-tions, then that statement is true in every structure satisfying those assumptions.Equivalently, this theorem tells us that if we have a theory T and a model for it inwhich some statement S is false, then there cannot be a proof of S in T .
Now, given a statement and a theory, we might wonder if we can prove thatstatement in that theory. Consider again our theory TO of strictly ordered sets. Ifwe were to propose an additional axiom for our theory, we would be interested inwhether or not the axiom was already provable in TO or if it conļæ½icted with theaxioms we already have in the theory. Suppose the following axiom is suggested.
ā¢ The Axiom of Trichotomy: for all a, b, we have a < b or a = b or b < a.
By the Soundness Theorem, if we could prove the Axiom of Trichotomy in TO, thenthe axiom would have to be true in any model of TO. But, consider the structureM1 where variables range over the set S = {{}, {0}, {1}, {0, 1}} and our relation< is taken to be the proper subset relation ā between sets, as indicated in thediagram below.
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{1}ā
EEEE
EEEE
{}
ā AAAA
AAAA
ā}}}}}}}}
{0, 1}
{0}ā
yyyyyyyy
This structure is a model for TO, because no two of the elements in S are propersubsets of each other, and the subset relation is transitive. However, the Axiomof Trichotomy fails in M1, because we have {0} and {1} in S, but {0} 6ā {1},{0} 6= {1}, and {1} 6ā {0}. Thus, because the Axiom of Trichotomy fails in amodel of TO, the Soundness Theorem tells us that the Axiom of Trichotomy is notprovable in TO.
Since the axiom is not provable in TO, and we have seen that it fails in atleast one model, we might wonder if we could disprove it. That is, we would beinterested in whether we could deduce the negation of the Axiom of Trichotomyin our theory. However, if we consider again our earlier model M where variablesrange over {0, 1, 2} and our relation is simply the usual < from the integers, wesee that the Axiom of Trichotomy holds. The Soundness Theorem now tells usthat the negation of the axiom is not provable in TO, because if the negation wereprovable, then the axiom would have to be false in every model, but here we havea counterexample.
Deļæ½nition 3.5. We say that a theory T is consistent if there is no statement Ssuch that both S and its negation, Ā¬S, are provable in T . That is, T is consistentif no contradiction can be deduced from it. We extend this notion and say that astatement S is consistent with a theory T if the theory T āŖ {S} is consistent.
Another theorem establishes the relationship between models and consistency.
Theorem 3.6. (The Completeness Theorem): A theory is consistent if andonly if it has a model.
Thus we can quickly see from our above examples that both the Axiom of Tri-chotomy and its negation are consistent with the theory of strictly ordered sets,because we have found a model for both TOāŖ{the Axiom of Totality} and TOāŖ{thenegation of the Axiom of Totality}. This leads us to our next deļæ½nition.
Deļæ½nition 3.7. We call a statement S independent of a theory T when both Sis consistent with T and Ā¬S is consistent with T .
Equivalently, in light of the above theorem, we can say that a statement S isindependent of a theory T if and only if there exists a model MS for T āŖ{S} and amodel MĀ¬S for T āŖ{Ā¬S}. Remembering that having a model in which a statementfails means that the statement cannot be proven from the theory, we can state thisin yet one more equivalent way: a statement is independent of a theory if and onlyif it cannot be proven or disproven from the theory.
So, we can now see exactly what is meant by the independence of the ContinuumHypothesis. It is a statement CH such that both ZFCāŖ{CH} and ZFCāŖ{Ā¬CH}are consistent, and as such, neither Ā¬CH nor CH is provable in ZFC.
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Gƶdel's consistency proof builds a model of ZF āŖ {GCH, AC}, that is, a modelwhere all of the axioms of ZF set theory hold, the Generalized Continuum Hypothe-sis is true, and the Axiom of Choice is true. This establishes both the consistency ofZF āŖ{GCH} and ZFC āŖ{GCH}, and thus that GCH and CH cannot be provenfalse in either ZF or ZFC set theory. Cohen's model establishes the consistency ofZF āŖ {Ā¬CH, AC}, and thus that CH cannot be proven in ZF or ZFC set theory.Since GCH implies CH, this additionally means that GCH cannot be proven inthese theories either.
Both Gƶdel and Cohen assume the existence of some model of ZF at the startof their proofs, and as such, both of their proofs are contingent on the consistencyof ZF . However, note that if ZF is not consistent, then it contains a contradictionand thus any statement is provable in it anyway.
4. Some Concepts from Set Theory
Before we can begin ļæ½nding the models we need, we require a few deļæ½nitions andproperties relating to ordinal numbers.
Deļæ½nition. A relation < orders a set S if
(1) For all x, y in S, exactly one of the following holds.
x = y, x < y, y < x
(2) If x < y and y < z, then x < z.
Deļæ½nition. A relation < well-orders a set S if < orders S and for any nonemptysubset B ā S, there exists an x ā B such that for any y ā B, we have either x = yor x < y.
A relation thus well-orders a set if every non-empty subset has a least elementwith respect to that relation. We say that a set S is well-ordered if S is a setwith a relation < such that < well-orders S.
A familiar example of a well-ordered set is the set of natural numbers N ={0, 1, 2, ...} under its usual ordering. Given a well-ordered set S we can see thatany nonempty subset X ā S is also well-ordered, since if Y is a nonempty subsetof X, then Y is a nonempty subset of S, and so Y has a least element since S iswell-ordered. Thus, any nonempty subset of the natural numbers will also serve asan example.
Deļæ½nition. A set x is transitive if z ā y and y ā x implies that z ā x.
A simple example of a transitive set is the set x = {ā , {ā }}. The only elementsof this set are y0 = ā and y1 = {ā }. To see that x is a transitive set, we need tocheck that every element of yo is in x and that every element of y1 is in x. We cansee that y0 has no elements, and we see that the only element z ā y1 is the set ā ,which is an element of x.
Deļæ½nition. An ordinal is a set Ī± which is transitive and well-ordered by the setmembership relation ā. We denote that a set Ī± is an ordinal by On(Ī±).
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In set theory, we develop natural numbers by induction in the following way.
0 = ā 1 = {ā }2 = {ā , {ā }}3 = {ā , {ā }, {ā , {ā }}}
and in general
n+ 1 = n āŖ {n}
Since each successive number contains all of the elements of the previous number,we see that each n is transitive. This development also ensures that each naturalnumber is a set well-ordered by ā. Thus, each natural number is an ordinal. Thisconstruction causes ā to act identically like the usual less-than relation betweennatural numbers, and so when the entire set N = {0, 1, 2, ...} is considered, we cansee that it is also an ordinal.
In a similar way to how the cardinal numbers generalize the natural numbersand convey information about size, the ordinals generalize the natural numbers andconvey information about order. Sets have the same cardinality when there is aone-to-one, onto function between them. Two sets with order relations, (X,<X)and (Y,<Y ), have the same order type when there is a one-to-one, onto function fbetween them such that x1 <X x2 if and only if f(x1) <Y f(x2). One example of thedistinction between these ideas can be seen by comparing the cardinalities and ordertypes of the natural and rational numbers. Both N and Q have the same cardinality,because there is a one-to-one correspondence between these sets. However, theyhave diļæ½erent order types when considered under their usual orderings, becauseany one-to-one, onto function f between N and Q which preserves order as requiredwould have to assert the existence of some n ā N such that 1 < n < 2, since 1 < 2implies f(1) < f(2), but f(1) and f(2) are distinct rational numbers, and so thereis some rational between them that is the output for some n ā N. This rationalf(n) is such that f(1) < f(n) < f(2), and so we must have 1 < n < 2, which isimpossible. This example demonstrates that there are distinct ordinals with thesame cardinality.
The ordinals serve as representatives for the order types of sets, similarly to howcardinals serve as representatives for the sizes of sets. An important thing for us toknow as we move forward is that we can perform transļæ½nite induction on ordinals.This lets us extend the idea of strong induction on the integers to any well-orderedset, including inļæ½nite sets with cardinality larger than āµ0. The models we considerwill involve inductively building up structures by starting at 0, and then worryingabout what happens at some ordinal Ī± when every earlier ordinal Ī² < Ī± has beentaken care of.
5. Concepts from Gƶdel's Model
In order to carry out his consistency proof, Gƶdel builds a model featuring onlycertain kinds of sets that arise in a particular way. The ideas and theorems involvedin Gƶdel's proof play a signiļæ½cant role in Cohen's work.
The Axiom of Replacement in ZF set theory allows very liberal use of propertiesto deļæ½ne new sets. These properties can do things such as range over collectionsof sets including even the set that is currently being deļæ½ned. Gƶdel's idea was
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essentially to restrict the properties that are acceptable for this axiom so that theyrange only over the sets that were already deļæ½ned. By starting out with the emptyset, and iteratively building up new sets in this way, we end up with what are calledthe constructible sets.
Deļæ½nition 5.1. Let A be a formula, and let X be a set. We denote by AX theformula A with all of its variables which are bound by a quantiļæ½er restricted sothat they range only over the set X. We say that AX is the formula A restricted
to X.
What the above deļæ½nition says is that, starting with the formula A that containssome quantiļæ½ers such as ļæ½for all xļæ½ and ļæ½there exists yļæ½ in it, AX is the formula onegets when A is changed so that each quantiļæ½er is instead read as ļæ½for all x ā Xļæ½and ļæ½there exists y ā Xļæ½.
Deļæ½nition 5.2. Let X be a set. We deļæ½ne X ā² as the union of X and the set ofall sets Y for which there is some formula A(z, t1, ..., tk) in ZF such that for sometā1, ..., t
āk in X, we have Y = {z ā X | AX(z, tā1, ..., t
āk).
This deļæ½nition gives us that X ā² contains all of the elements of X, along with allof the sets Y that can be built by using a formula which has variables ranging onlyover X. This is the set of everything that can be built from X using a formularestricted to X.
Deļæ½nition 5.3. Given an ordinal Ī±, deļæ½neMĪ± byM0 = ā andMĪ± = (āĪ²<Ī±MĪ²)ā².
Note that in the above deļæ½nition, the set MĪ± is of the form X ā². We start withM0 = ā , then iteratively make these MĪ± which consist of all of the elements ofthe previous sets along with all of the new sets that can be created using formulasrestricted to the previous sets. This deļæ½nition ensures that each MĪ± is transitive,since it continues to include every element of MĪ± into all future sets, each of whichis only adding subsets of the previous sets.
Deļæ½nition 5.4. A set x is constructible if there is an ordinal Ī± such that x āMĪ±.We say that x is constructed at Ī± or at stage Ī± when this is the case.
So, the constructible sets are the sets gained in this method of iteratively buildingup new sets from the empty set using only formulas restricted to the sets whichwere already constructed at a previous stage.
Deļæ½nition 5.5. Given an ordinal Ī±, we deļæ½ne XĪ± asāĪ²<Ī±MĪ² . We say that a set
x is constructed before stage Ī± if we have x ā XĪ±.
Note that we could now also write MĪ± = (XĪ±)ā² for each ordinal Ī±. The XĪ±
are the sets of elements that exist by the time stage Ī± of the construction is takingplace, butXĪ± does not yet include the new elements that will be added at this stage.The XĪ± collect up all the existing elements by the time we get to Ī±, the MĪ± collectup the elements, and additionally include the new sets that can be constructed.We deļæ½ne these terms separately so that we can more easily speak about when anelement is constructed. Note, however, that elements which are constructed beforeĪ± are also in a sense ļæ½reconstructedļæ½ at Ī±, because we have XĪ± ā MĪ±. Thesephrases do not necessarily refer to the minimal Ī± for which x begins to appear.
Notation 5.6. We denote by L the class of all constructible sets, and we denote byV the class of all sets.
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Gƶdel's proof that ZF āŖ {GCH} is consistent if ZF is consistent comes as theresult of three theorems. The ļæ½rst theorem is the veriļæ½cation that L satisļæ½es theaxioms of ZF set theory and thus is a model of ZF . The next theorem veriļæ½esthat L additionally satisļæ½es the statement V = L; that is, when the relativizedconstructible sets are built inside of the actual class of constructible sets, the resultis the same as when the constructible sets are built in ZF set theory as a whole.The ļæ½nal theorem establishes that (V = L) ā (AC & GCH) holds within ZFset theory. Taken together, these theorems establish that the Axiom of Choiceand the Generalized Continuum Hypothesis hold within L, and thus they are bothconsistent with the axioms of ZF set theory.
Before we move on to developing Cohen's model, there are a couple more deļæ½-nitions we will need which relate to constructibility.
Deļæ½nition 5.7. The transitive closure of a setX is the setānXn whereX0 = X
and Xn+1 =āXn.
That is, the transitive closure of X is the union over all successive unions startingat X, then
āX, then
ā(āX), and so on. This ensures that the transitive closure
of a set is actually transitive by adding the necessary elements.
Deļæ½nition 5.8. Given a set y, we deļæ½ne MĪ±(y) inductively by setting M0(y) asthe transitive closure of y, andMĪ±(y) = (
āĪ²<Ī±MĪ²(y))ā². Likewise, we deļæ½ne XĪ±(y)
asāĪ²<Ī±MĪ²(y). We say that an element x āMĪ±(y) is constructed from y at Ī±,
and similarly we say that an element x ā XĪ±(y) is constructed from y before Ī±.
The above deļæ½nition just generalizes the earlier deļæ½nitions about constructibilityby allowing the process to begin at a set other than the empty set. We will howevercontinue to simply say that x is constructible if it is constructed from the emptyset at some ordinal. Note that a set which is constructible from some set is notnecessarily constructible from the empty set, and so we may have that a set x isconstructible from a set a, and yet x is not in L, the class of constructible sets.This distinction is important to keep in mind when Cohen's model is considered,because we will see that there are nonconstructible sets which are neverthelessconstructed from some other set. This allows Cohen's model to fail V = L whilestill maintaining a convenient structure.
6. The Minimal Model
Within Cohen's proof that there is a model for ZF āŖ {Ā¬CH}, we will assumethe following additional axiom.
Axiom 6.1. The Standard Model Axiom (SM): There is a set M with a relationR = {(x, y) | x ā y & x ā M & y ā M} such that M is a model for ZF under therelation R.
This axiom gives us not just the existence of a model for ZF, but a model wherethe relation ā in the axioms of ZF is interpreted simply as the usual set membershiprelation restricted to the elements of some set M . A model which interprets ā insuch a way is called standard. The above axiom allows us to infer the existenceof a particular model which will be useful.
Theorem 6.2. ZF āŖ {SM} implies that there exists a unique transitive countablemodelM such that if N is any standard model, then there is an isomorphism from
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M onto a subset of N which preserves the ā relation. Furthermore, M satisļæ½esV = L.
Notation. Throughout the rest of this paper, M will refer to the unique minimalmodel of ZF whose existence is given by the above theorem. We have that M isstandard, transitive, and countable, and it satisļæ½es V = L. Additionally, Ī±0 willdenote sup{Ī± | Ī± āM}, the supremum of the ordinals inM.
There is one more theorem concerning this model which will be useful later on.
Theorem 6.3. For each element x inM, there is a formula A(y) in ZF such thatx is the unique element ofM that satisļæ½es AM(x), the formula A restricted to theelements of the modelM.
This theorem asserts that there is a way to uniquely identify each of the x inMusing formulas in ZF .
7. Motivation for Cohen's Model
Gƶdel's proof of the consistency of the Generalized Continuum Hypothesis in-volved establishing that (V = L) ā (AC & GCH) holds within ZF set theory.This tells us that if we are to ļæ½nd a model where GCH fails, that model must failto satisfy V = L. So, we will need to build a model which contains sets that arenot constructible relative to that model.
Moreover, the following theorem tells us that we cannot hope to ļæ½nd such amodel by restricting the sets of ZF to those which satisfy some formula.
Theorem 7.1. Let A(x) be any formula in ZF set theory. It cannot be proven inZF that the axioms of ZF and the statement Ā¬(V = L) hold when restricted to theclass of all sets x which satisfy A(x).
Proof. Assume we have a formula A(x) for which it can be proven that the classof all x for which A(x) is true satisļæ½es ZF āŖ {Ā¬(V = L)}. We will show that thisleads to a contradiction.
Recall thatM is the unique minimal model, and letMā = {x | x āM&AM(x)}.Now, sinceM is a standard model of ZF , and using the assumption, we have thatMā is a standard model of ZF which also satisļæ½es Ā¬(V = L).
We have that Mā ā M by deļæ½nition, and by Theorem 6.2 there is an iso-morphism from M onto a subset of Mā, so we must have that M and Mā areisomorphic. Thus, since V = L holds in M, it must also hold in Mā. Hence Māsatisļæ½es both V = L and Ā¬(V = L), which is impossible. ļæ½
The above two results make it clear that if we wish to ļæ½nd a model for ZF āŖ{Ā¬CH}, we will need sets that appear non-constructible relative to the model, andwe cannot ļæ½nd this model by restricting some collection of sets or by looking insidea model likeM; we will have to add new sets. Cohen's forcing technique gives usa way to introduce the new sets we need.
8. Forcing
Plainly, forcing will allow us to introduce a new set by using a sequence of ļæ½nitesets of membership statements. The idea is similar to trying to specify the set of
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even natural numbers E = {0, 2, 4, ...} by the following sequence.
P0 = {0 ā E}P1 = {0 ā E, 2 ā E}P2 = {0 ā E, 2 ā E, 4 ā E}
...
Pn = {2 Ā·m ā E | m ā¤ n & m ā N}Note that for no single n does Pn contain enough information to uniquely determinethe set, and it is this vagueness in the determination of the set that we want totake advantage of. Forcing will allow us to add sets to the modelM for which anyquestion about membership will be eventually resolved, and yet a set determinedin this way provides a minimal enough amount of information so that relative tothe model, it will not appear constructible.
The following concept is necessary because as we are building our new model, wewill need a way to refer to its members before the model has been explicitly deļæ½ned.To facilitate this, we will essentially label formulas that will later be satisļæ½ed bythe elements of the new model.
Deļæ½nition 8.1. A labeling is a mapping deļæ½ned in ZF, which assigns to eachordinal 0 < Ī± < Ī±0
ā¢ a set SĪ± called a label space,ā¢ a function ĻĪ± deļæ½ned over SĪ±
We deļæ½ne S =āĪ± SĪ±, and we have a subset G ā S whose elements are called the
generic sets. The sets SĪ± are mutually disjoint, and if c ā SĪ±, then ĻĪ±(c) is aformula A(x) which may contain elements from SĪ² with Ī² < Ī± as constants. Wewill eventually denote by c the set of elements that that satisfy the formula givenby ĻĪ±(c).
In our usage of this idea, each c ā S will act as a label for a unique formuladeļæ½ned over the variables and constants that came before c. Since each SĪ± isrequired to be disjoint from the others, we have that each c ā S is in a unique SĪ±.The set G of generic sets is a collection of labels, some of which will refer to thesets that we will use forcing to identify and introduce to our model.
Notation. We will write āĪ±x or ā
Ī±x to indicate that a variable x bound by a
quantiļæ½er is restricted to the set XĪ±.
Deļæ½nition 8.2. A limited statement is a statement in ZF in which every quan-tiļæ½er is of the form āĪ± or āĪ± for some ordinal Ī± < Ī±0 and in which elements of Smay appear as constants.
Deļæ½nition 8.3. An unlimited statement is a statement in ZF in which elementsof S may appear as constants.
Deļæ½nition 8.4. Given a generic set G, a forcing condition P is a ļæ½nite set oflimited statements each of the form n ā a or Ā¬(n ā a) , where n ā N and a ā G,and where for any such n and a, at most one of (n ā a) and Ā¬(n ā a) is in P.
INDEPENDENCE OF THE CONTINUUM HYPOTHESIS 12
The last part of the above deļæ½nition ensures that no forcing condition containsconļæ½icting information as to whether something is an element of one of the sets a.
Deļæ½nition 8.5. A condition set is a set U of forcing conditions deļæ½ned in Mand a relation < on U , also deļæ½ned inM, such that < is reļæ½exive and transitive,along with a map Ļ inM such that for all forcing conditions P ā U , Ļ(P ) is someset of statements of the form c1 ā c2 with c2 in G, and if c2 ā SĪ±, then c1 ā SĪ²for some Ī² < Ī±. The relation < is required to satisfy the property that wheneverP < Q, we have Ļ(P ) ā Ļ(Q).
Now we will deļæ½ne a hierarchy on the limited statements which will be used forinduction in the deļæ½nitions and proofs that follow.
Deļæ½nition 8.6. Given a limited statement A, we deļæ½ne rank(A) = (Ī±, i, r) whereā¢ Ī± is the least ordinal such that whenever āĪ² or āĪ² appear in A, we haveĪ² ā¤ Ī±, and whenever c ā SĪ² appears in A, we have Ī² < Ī±.
ā¢ i = 0 if Ī± = Ī² + 1 for some ordinal Ī² and neither āĪ± nor āĪ± appear in A,and no term of the form c ā x, c = x, or x = c appears in A with c ā SĪ² .Otherwise, i = 1.
ā¢ r is the length of the statement A; i.e., it is the number of symbols in A.
The way Ī± is chosen in the above deļæ½nition is directly tied to the deļæ½nition ofXĪ± and ensures that if rank(A) = (Ī±, i, r), then A only references variables andconstants from XĪ±.
Now, with rank(A) deļæ½ned, we can deļæ½ne the forcing of statements.
Deļæ½nition 8.7. Given a labeling with S =āĪ± SĪ±, a generic set G ā S, and a
condition set U , we deļæ½ne P forces A, where P is a forcing condition in U and Ais a limited statement, by induction on rank(A) as follows.
(1) P forces āĪ±x B(x) if for some c ā SĪ² with Ī² < Ī±, P forces B(c).(2) P forces āĪ±x B(x) if for all forcing conditions Q ā P and all c ā SĪ² with
Ī² < Ī±, we have that Q does not force Ā¬B(c).(3) P forces Ā¬B if for all Q ā P we have that Q does not force B.(4) P forces B&C if P forces B and P forces C.(5) P forces B āØ C if P forces B or P forces C.(6) P forces Aā B if P forces Ā¬A or P forces B.(7) P forces Aā B if P forces Aā B and P forces B ā A.(8) P forces c1 = c2, where c1 ā SĪ±, c2 ā SĪ² , and Ī³ = max(Ī±, Ī²), if either
Ī³ = 0 and c1 = c2 as elements in S0, or if Ī³ > 0 and P forces āĪ³x (x āc1 ā x ā c2).
(9) P forces c1 ā c2, where c1 ā SĪ±, c2 ā SĪ² with Ī± < Ī², if(a) Ā¬c2 ā G and P forces A(c1) where A(x) = ĻĪ²(c2), the unique formula
corresponding to c2 in the labeling.(b) c2 ā G and for some c3 ā SĪ³ with Ī³ < Ī², we have {c3 ā c2} ā Ļ(P )
and P forces c1 = c3.(10) P forces c1 ā c2 where c1 ā SĪ± and c2 ā SĪ² with Ī± ā„ Ī², if for some c3 ā SĪ³
with Ī³ < Ī², we have P forces āĪ±x (x ā c1 ā x ā c3) & (c3 ā c2).
We will soon see an analogous list of deļæ½nitions for unlimited statements, butļæ½rst we will consider two examples that explore the consequences of the deļæ½nitionof forcing.
INDEPENDENCE OF THE CONTINUUM HYPOTHESIS 13
Example 8.8. Suppose we have the following sequence that is intended to describea set a.
P0 = {1 ā a}P1 = {1 ā a, 2 ā a}
...
Pn = {m ā a | 0 < m ā¤ n}The sequence of conditions ļæ½rst asserts 1 ā a and then at each step asserts that
the next natural number is also in a, along with all of the assertions from theprevious condition. We can see that for no n does Pn ever contain the statement0 ā a, and so for every n we have that Pn does not force 0 ā a. Yet, we donot have that any Pn forces Ā¬(0 ā a) either, because each Pn could be containedin a forcing condition Qn which forces 0 ā a. In particular, we can simply letQn = Pn āŖ {0 ā a}. Interestingly, this means that every Pn in this sequence failsto force (0 ā a) āØ Ā¬(0 ā a).
On the other hand, every Pn in the sequence forces 1 ā a, and also forcesĀ¬Ā¬(1 ā a) because for any Q ā Pn, we have (1 ā a) ā Q, and so there exists anR ā Q which forces (1 ā a) since R contains this statement; Q itself can serve asthis R. Thus Q does not force Ā¬(1 ā a) for any Q ā Pn, and we have that Pn forcesĀ¬Ā¬(1 ā a).
We will later only consider certain sequences which eventually, for any statementA, force either A or Ā¬A at some step in the sequence.
Example 8.9. Consider the statement ļæ½a is inļæ½niteļæ½ written as ā1x ā1y (y >x & y ā a). Cohen notes that while > is not an admissible symbol in our statements,we can informally see how every forcing condition P would force that ļæ½a is inļæ½niteļæ½.This is because for any natural number n, no P forces the statement ļæ½there is noelement of a greater than nļæ½, Ā¬ā1y (y > n & y ā a), since P can always be containedin a larger forcing condition Q which has a statement y ā a where y is larger thann.
We now proceed with the deļæ½nition of forcing for unlimited statements.
Deļæ½nition 8.10. P forces A, where P is a forcing condition and A is an unlimitedstatement, is deļæ½ned by induction on the length of A by the following.
(1) P forces āx B(x) if for some c ā S, P forces B(c).(2) P forces āx B(x) if for all c ā S, if Q ā P , then Q does not force Ā¬B(c).
The deļæ½nitions for P forces Ā¬B, B&C, BāØC, B ā C, B ā C, c1 ā c2, and c1 = c2are exactly as in the case for limited statements.
9. Properties of Forcing Conditions
In this section we will prove a few results regarding when a condition forcesa statement, and we will prove the existence of a sequence which, given somestatement A, will force either A or Ā¬A after a ļæ½nite number of steps. It is this sortof sequence with which we are primarily concerned.
Throughout this section, A denotes a statement which is either limited or un-limited, and P,Q denote forcing conditions.
INDEPENDENCE OF THE CONTINUUM HYPOTHESIS 14
Lemma 9.1. For any P and any A, we do not have both P forces A and P forcesĀ¬A.
Proof. If P forces Ā¬A, then by Deļæ½nition 8.7.3, since P ā P , we have that P doesnot force A. ļæ½
Lemma 9.2. If P forces A and Q ā P , then Q forces A.
Proof. We prove this for limited statements A by induction on rank(A).If P forces āĪ±x B(x), then P forces B(c) for some c ā SĪ² with Ī² < Ī±. By
induction, Q forces B(c) and so also forces āĪ±x B(x).If P forces āĪ±x B(x), then if R ā Q, we have that R ā P , so by the deļæ½nition
of forcing, R does not force Ā¬B(c) for any c ā SĪ² with Ī² < Ī±. So, Q also forcesāĪ±x B(x).
If P forces Ā¬B, then if R ā Q, we have R ā P and so by the deļæ½nition of forcingR does not force B. Thus, Q also forces Ā¬B.
The remaining cases of P forces A where A is of some other form reduce to Pforces B where B has a lower rank than A, and so they hold by induction. Thebase cases, where the rank is lowest, are simple membership statements. The lemmaholds here since if P forces (c1 ā c2), then we have that (c1 ā c2) is a statementin P , but P ā Q, so (c1 ā c2) is a statement in Q and so Q forces (c1 ā c2). Theproof for unlimited statements is similar, but inducts on the length of A. ļæ½
Lemma 9.3. For all P and A, there is a Q ā P such that either Q forces A or Qforces Ā¬A.
Proof. If P does not force Ā¬A, then by Deļæ½nition 8.7.3, there is some Q ā P suchthat Q forces A. Thus, either P ā P forces Ā¬A, or some Q ā P forces A. ļæ½
Deļæ½nition 9.4. A complete sequence is a sequence {Pn} of forcing conditionssuch that for every n, we have Pn ā Pn+1, and for each A, there is an n such thateither Pn forces A or Pn forces Ā¬A.
Lemma 9.5. A complete sequence exists.
Proof. Since M is countable, we can enumerate all statements An deļæ½ned in M.We can construct a complete sequence inductively by ļæ½rst setting P0 as a forcingcondition that forces either A0 or Ā¬A0, and then selecting Pn+1 as any forcingcondition Q ā Pn such that Q forces An+1 or Q forces Ā¬An+1.
Intuitively, we have created the sequence by picking each successive forcing con-dition so that it decides one way or the other on the next membership statement.Thus we have a sequence where each condition is a subset of the next condition,and for any statement An, we know that either An or Ā¬An is forced by Pn. ļæ½
10. Cohen's Model N
Recall that the cardinality of the continuum is the cardinality of the set of subsetsof N. So, if we are to build a model where the continuum hypothesis fails, then weneed to ensure that in our model, there are more than āµ1 subsets of N. We willaccomplish this by selecting an arbitrary cardinal āµĻ with Ļ ā„ 2 and introducingāµĻ many subsets toM which are identiļæ½ed by a sequence of forcing conditions.
First, ļæ½x some cardinal āµĻ in M with Ļ ā„ 2. We now select our labeling andgeneric sets. Our goal with this labeling is to eventually be able to use a label
INDEPENDENCE OF THE CONTINUUM HYPOTHESIS 15
W ā S such that W is the set of ordered pairs consisting of elements Ī“ < āµĻ pairedwith the new subsets aĪ“ that we will introduce to the model. Ordered pairs (Ī“, aĪ“)in set theory are sets of the form {{Ī“}, {Ī“, aĪ“}}, so getting to the point where wecan refer to the ordered pairs will require a few steps of building up.
For each ordinal Ī± < āµĻ , we set SĪ± = {cĪ±} where cĪ± will later be sent by Ļ to aformula A which deļæ½nes the set cĪ± = Ī±. We include each cĪ± in our set G of genericsets. For Ī± = āµĻ , we include āµĻ many elements in SĪ± and denote these elements byaĪ“ with Ī“ < āµĻ . Each aĪ“ is included in G, and each is such that aĪ“ will be a subset ofthe natural numbers. Still with Ī± = āµĻ , we set SĪ±+1 as the union of two sets, eachcontaining āµĻ many elements. The ļæ½rst set contains elements denoted by c{Ī²} withĪ² < āµĻ , which will be eventually be such that ĀÆc{Ī²} = {Ī²}. The second containselements denoted c{Ī“,aĪ“} with Ī“ < āµĻ , which will be such that ĀÆc{Ī“,aĪ“} = {Ī“, aĪ“}. Allof these elements are also included in G. The label space SĪ±+2 is where the labelsfor the ordered pairs appear; it has āµĻ many elements c(Ī“,aĪ“) with Ī“ < āµĻ , each suchthat ĀÆc(Ī“,aĪ“) = (Ī“, aĪ“). These are again included in G. The next space, SĪ±+3 consists
of a single element denoted by W , which is such that W = {(Ī“, aĪ“) | Ī“ < āµĻ}, andonce more, we include this element W in G. For Ī± > āµĻ + 3, we have that SĪ±contains no elements of G and that SĪ± is in one-to-one correspondence with the setof formulas that range over
ā(SĪ³ | Ī³ < Ī±).
We will use only use forcing conditions which contain statements of the formn ā aĪ“ or Ā¬n ā aĪ“ where n is a natural number and Ī“ < āµĻ . So, the forcingconditions will be ļæ½nite sets of statements saying that a natural number is or isnot a member of one of our aĪ“, which are each a generic set in G. As a reminder,there are āµĻ many of these elements aĪ“, and we are introducing them to the modelM as new subsets of the natural numbers. This is what will cause the ContinuumHypothesis to fail: there will be at least āµĻ many subsets of the natural numbersin our model, and so the cardinality of the continuum, which is the cardinality ofP(N), will be at least āµĻ , which was selected to be greater than āµ1.
Our condition set will consist of the above forcing conditions, and our ordering< will be deļæ½ned by P < Q whenever P ā Q. We deļæ½ne our function Ļ by settingĻ(P ) equal to the union of the following sets
{cn ā aĪ“ | (n ā aĪ“) ā P}{cĪ± ā cĪ² | Ī± < Ī² < āµĻ}{cĪ± ā c{Ī±} | Ī± < āµĻ}{cĪ“ ā c{Ī“,aĪ“} | Ī“ < āµĻ}{aĪ“ ā c{Ī“,aĪ“} | Ī“ < āµĻ}{c{Ī“} ā c(Ī“,aĪ“) | Ī“ < āµĻ}{c{Ī“,aĪ“} ā c(Ī“,aĪ“) | Ī“ < āµĻ}{c(Ī“,aĪ“) āW | Ī“ < āµĻ}
Let {Pn} be any complete sequence of forcing conditions. We will now deļæ½ne allof the c corresponding to the elements of our label spaces. We proceed by inductionon the ordinal index Ī± of the label spaces SĪ± as follows. For c ā S0, let c = ā . Forc ā SĪ± when all previous c ā SĪ² with Ī² < Ī± have had their c deļæ½ned, there are twocases.
INDEPENDENCE OF THE CONTINUUM HYPOTHESIS 16
(1) If c is not also in G, then c is the set of x such that x satiļæ½es the formulaA = ĻĪ±(c) and such that x ā SĪ² for some Ī² < Ī±. That is, c is the set ofalready deļæ½ned elements which satisfy the formula that corresponds to c.
(2) If c is in both SĪ± and G, then c = {x | ān ā N s.t. (x ā c) ā Ļ(Pn)}.Essentially, this means that if c is a generic set, then c contains exactly theelements that the forcing conditions in our complete sequence tell us belongto c.
Note that since each aĪ“ is one of the elements of SĪ± ā©G, they fall into the secondcase above, and the deļæ½nition of Ļ ensures that each aĪ“ will be exactly the subsetof the natural numbers which our forcing sequence says belong to aĪ“.
We can now ļæ½nally build our new model, which we call N . We will use the usualset membership relation in our structure, since our intention is to extend the modelM, which is a standard model. Let Gā = {aĪ“ | Ī“ < āµĻ} and deļæ½ne the domain ofthe new structure N to be ā
{MĪ±(Gā) | Ī± < Ī±0}or equivalentlyā{XĪ±0(Gā)}
Recall that Ī±0 is the supremum of the ordinals in the unique minimal modelM.The above set is thus the set of elements that can be constructed before stage Ī±0
from our new sets, each of which is deļæ½ned by our complete sequence of forcingconditions.
Lemma 10.1. A statement A is true in the model N if and only if for some n, Pnforces A.
Proof. We prove the lemma for limited statements A by induction on rank(A).If A is of the form āĪ±x B(x) and Pn forces A, then we have that Pn forces B(c)
where c ā SĪ² with Ī² < Ī±. By induction, since Ī² < Ī±, we have B(c) is true in Nand so A is true in N . Conversely, if A is true in N , then we have that for somec ā SĪ² with Ī² < Ī±, B(c) is true. By induction, there is a Pn which forces B(c) andthus also forces A.
If A is of the form āĪ±x B(x) and Pn forces A, then for some c ā SĪ² with Ī² < Ī±,some Pm with m > n forces B(c), because for āĪ±x B(x) to be forced by Pn, no Pmcan force Ā¬B(c), but our conditions are in a complete sequence. Induction givesthat B(c) is true in N and so A is also true in N . Conversely, if A is true in N ,then no Pn can force āĪ±x Ā¬B(x) since that would imply that for some c ā SĪ² withĪ² < Ī±, Pn forces Ā¬B(c). Then Ā¬B(c) would hold in N and A would be false in N .So, since no Pn can force āĪ±x Ā¬B(x), but our sequence is complete, we have thatsome Pn forces Ā¬āĪ±x Ā¬B(x). By the deļæ½nition of forcing, this gives that for everyQ ā Pn, Q does not force āĪ±x Ā¬B(x). This means that every Q ā Pn does notforce Ā¬B(c) for any c ā SĪ² with Ī² < Ī±, which again by the deļæ½nition of forcingmeans that Pn forces āĪ±x B(x).
If A is of the form Ā¬B and Pn forces A, then if Ā¬B were false in N , B is true inN and so by induction, some Pm forces B. However, by Lemma 9.2, we would havethat Pmax(n,m) forces B and forces Ā¬B, which Lemma 9.1 tells us is impossible.Thus, Ā¬B must be true in N . Conversely, if Ā¬B is true in N , then supposing Ā¬Bis not forced by any Pn, we have that B is forced by some Pn since our sequenceis complete. But B is a lower rank statement than Ā¬B, so by induction this means
INDEPENDENCE OF THE CONTINUUM HYPOTHESIS 17
that B is true in N , which is impossible. So, we must have that some Pn forcesĀ¬B.
The other cases just reduce the rank of A and are covered by induction, and thebase cases are membership statements, for which the lemma holds because our cand aĪ“ have been deļæ½ned so that the membership statements that are true aboutthem are exactly the statements contained in some Ļ(Pn). The proof for unlimitedstatements is similar but inducts on the length of A. ļæ½
We can now see that the sets we are introducing to the model will be distinctwithin the model.
Theorem 10.2. Given aĪ“1 , Ī±Ī“2 with Ī“1 6= Ī“2, we have that aĪ“1 and aĪ“2 will givedistinct sets in N .
Proof. Any forcing condition P is ļæ½nite and can be contained in a forcing conditionQ which has (m ā aĪ“1) and Ā¬(m ā aĪ“2) for some m ā N, and thus no P forces theelements of the two sets to be the same. The deļæ½nition of forcing tells us that thismeans each of the Pn in our complete sequence actually forces Ā¬(aĪ“1 = aĪ“2), andso by the above lemma, the sets are not equal in N . ļæ½
Theorem 10.3. N is a model of ZF .
Proof. We have to verify that each of the axioms of ZF set theory is true in N . Wedo so by checking that the axioms hold when they are restricted to N . That is,for each axiom A in ZF , we show that AN holds. Throughout the proof, we arepermitted to use the usual axioms of ZF set theory.
(1) The Axiom of the Null SetWe want to show that ā ā N , or in other words that ā is constructible
from Gā at some stage before Ī±0. We can easily obtain ā as the set of xin M0(Gā) satisfying the property x 6= x. This property ranges only overM0(Gā) and so ā appears in M1(Gā). Hence, ā is in N .
(2) The Axiom of ExtensionalityWe must show that if x, y are two sets in N which share the same
elements in N , then x = y. If x and y are in N , then x ā MĪ±1(Gā)and y ā MĪ±2(Gā) for some Ī±1, a2 < Ī±0. Let Ī± be the larger of Ī±1, Ī±2
and note that by deļæ½nition MĪ±(Gā) = (āĪ²<Ī±MĪ²(Gā))ā² and thus MĪ±(Gā)
contains both x and y. Since every MĪ²(Gā) is transitive, all elements of xand all elements of y are also inside MĪ±(Gā). Thus, if x and y share thesame elements inside N , then they truly share all of their elements, and soby the usual Extensionality of ZF, we have x = y.
(3) The Axiom of Unordered PairsGiven x, y in N , we have that x āMĪ±1(Gā) and y āMĪ±2(Gā) for some
Ī±1, a2 < Ī±0. Both x and y are thus in MĪ±(Gā) where Ī± is the larger ofĪ±1, Ī±2. So, we can obtain the set {x, y} as the set
{z | z āMĪ±(Gā) & (z = x or z = y)}This is a set deļæ½ned by a property restricted to MĪ±(Gā), and so {x, y} willbe an element of MĪ±+1(Gā). Thus, {x, y} is in N .
(4) The Axiom of UnionGiven x in N , we have that x ā MĪ±(Gā) with Ī± < Ī±0. Since MĪ±(Gā)
is transitive, we have that all elements y ā x are also in x ā MĪ±(Gā). We
INDEPENDENCE OF THE CONTINUUM HYPOTHESIS 18
can obtain the union over x as the set
{z | z āMĪ±(Gā) & āy(y āMĪ±(Gā) & z ā y & y ā x}which is deļæ½ned by a property containing variables restricted to MĪ±(Gā)and thus is an element of MĪ±+1(Gā).
(5) The Axiom of Inļæ½nityWe will show this by showing the stronger condition that every ordinal
Ī± is in the set MĪ±+1(Gā), and hence the least inļæ½nite ordinal Ļ is in N .If we have a set x inside a transitive set X, then x has the same elementswhen membership is restricted to X as it does when membership is unre-stricted. The deļæ½nition of an ordinal refers only to the elements of the setin question, and so, x will be an ordinal when the deļæ½nition is restrictedto X if and only if it is actually an ordinal. Written with the restrictedformula notation, this means that for any transitive set X and any elementx ā X, we have On(x) if and only if OnX(x).By our argument for (1), we know that 0 = ā is in M1(Gā). Assume Ī± isthe least ordinal for which Ī± 6ā MĪ±+1(Gā). We will ļæ½nd a contradiction.Since Ī± is assumed to be least, we have that Ī² < Ī± implies Ī² āMĪ²+1(Gā)where Ī² + 1 ā¤ Ī±. If we let X =
āĪ³ā¤Ī±MĪ³(Gā) then this means that for
Ī² < Ī±, we have Ī² ā X. We now consider the set Ī³ = {x ā X | OnX(x)}.We can see that X is transitive, and so by the above paragraph, we havethat Ī³ is a set of ordinals and is actually itself an ordinal itself. The or-dinal Ī³ is larger than any ordinal it contains, and it contains all Ī² < Ī±,since each Ī² < Ī± is an ordinal in X. So we have that Ī³ ā„ Ī± and henceeither Ī³ = Ī± or Ī³ contains Ī±. Now we observe that the set X is XĪ±+1 andthus X ā² is transitive and is a subset of MĪ±+1(Gā). Since Ī³ is deļæ½ned by aproperty ranging over XĪ±+1, we have that Ī³ is in MĪ±+1(Gā). So, if Ī³ = Ī±,we are done, and if Ī³ contains Ī±, then we have that Ī± is in MĪ±+1(Gā) bytransitivity.We have established that Ī± āMĪ±+1(Gā) for every ordinal Ī±. We now sim-ply note that the least inļæ½nite ordinal Ļ is an element of MĻ+1(Gā) andhence is an element in N , so the Axiom of Inļæ½nity holds.
(6) The Axiom of RegularityWe are required to show that any x in N has a member y, also in N ,
which contains no elements from x. So, suppose we have some x āMĪ±(Gā).Since we are able to use the usual Axiom of Regularity from ZF, we knowthat x has an element y containing no elements from x, and we need onlyshow that this y is in N . But, we know that MĪ±(Gā) is transitive, and soy ā x and x āMĪ±(Gā) implies that y āMĪ±(Gā). Thus, the y that we wantis in N .
(7) The Power Set AxiomThe proof for this axiom requires using some properties of M, but we
have intentionally made our new sets very vague from the perspective ofM,so we have to try to infer enough information from the forcing conditionsto ļæ½nd the power set of our set.Let x be a set in N . We have that x ā SĪ± with Ī± < Ī±0. We want to showthat the power set of x is also in N . Note that all of the labels from ourlabel spaces and all of the forcing conditions are statements in ZF and sothey are statements inM since it is a model of ZF . Our ļæ½rst goal will be
INDEPENDENCE OF THE CONTINUUM HYPOTHESIS 19
to ļæ½nd some bound for the label spaces that could contain elements thatrefer to the elements of our power set. For each c ā S, deļæ½ne the followingsets inM.
R(c) = {P | P forces c ā x}T (c) = {(P, cā) | P forces (cā ā c) where cā ā SĪ² with Ī² < Ī±}U(c) = {(R(c), T (c))}
Each R(c) is a set of forcing conditions, so it is an element of the power setof the set of all forcing conditions relative toM. Each T (c) is similarly anelement of the power set of the set of pairs of elements from the set of allforcing conditions and the set
āĪ²<Ī± SĪ±. These sets of forcing conditions
and labels are all elements of M, so we have that each U(c) will be anelement in M. So, we can deļæ½ne the set Uā = {U(c) | c ā S} in M. Wedeļæ½ne a function f such that for every u ā Uā, f(u) is the least ordinal Ī²such that there exists a c ā SĪ² for which U(c) = u, or f(u) = 0 is there isno such ordinal. We then let Ī²0 = sup{f(u) | u ā Uā}. We have that Ī²0 isinM.Now, if c ā x, then some Pn forces c ā x. The function f will send U(c) tothe least Ī² for which some c1 ā Ī² is such that U(c1) = U(c). Such a Ī² willexist, and so by the way we have deļæ½ned Ī²0, we have a c1 ā Ī² with Ī² < Ī²0
for which U(c) = U(c1). If c 6= c1 were true, then either some element ofc is not in c1 or some element of c1 is not in c. We will assume the ļæ½rstwithout loss of generality, so there is some c2 ā SĪ² with Ī² < Ī± such thatc2 ā c and not c2 ā c1. Then, by Lemma 10.1, some Pn forces c2 ā c. ButU(c) = U(c1) implies T (c) = T (c1), so Pn must also force c2 ā c1. Againby Lemma 10.1, this means that c2 ā c1,which contradicts our assumption.Thus, we have that c = c1. So, we now know that any subset of x will be cfor some c ā SĪ² with Ī² < Ī²0.Any subset of x will thus be an element of XĪ²0(Gā) in N . So, we can obtainthe powerset of x as the set {y | y ā XĪ²0 & y ā x} which is deļæ½ned overXĪ²0 and thus will be an element in (XĪ²0(Gā))ā² = MĪ²0(Gā), and as such isan element in N .
(8) The Axiom of ReplacementWe need to show that given some function f deļæ½ned in N , the image of
f on any set is also a set in N . So, suppose we have a formula A(x, y) whichdeļæ½nes a function f in N that gives y as a function of x. Let c0 ā SĪ± beļæ½xed. In a way similar to the above proof, we will ļæ½nd a bound for whichlabel spaces SĪ² the range of f will hit.Deļæ½ne in M the function g which sends any pair of a forcing conditionand an element c ā S to the least Ī² for which there is some cā² ā Ī² suchthat P forces f(c) = cā², or if no such Ī² exists, then g(P, c) = 0. Let Ī²0 besup{g(P, c) | P is a forcing condition & c ā SĪ² , Ī² < Ī±}. So, in a similarway to the above proof, we know that the range of f on c0 will be containedin MĪ²0(Gā). This tells us where the elements of the range are, but we stillneed to know that we can obtain the range of f on c0 as a set in N .What follows will allow us to conclude that the formula A can be restrictedto some set c and still deļæ½ne the range of f in N . The formula A will be
INDEPENDENCE OF THE CONTINUUM HYPOTHESIS 20
of the form
Q1y1 Ā· Ā· Ā·Qmym B(x1, ..., xm, y1, ..., ym)
where the Qi are quantiļæ½ers, and the formula B contains no quantiļæ½ers.For every r with 1 ā¤ r ā¤ m, there is a function fr(P, c1, ..., cn, cā²1, ..., c
ā²rā1),
deļæ½ned on ci, cā²j in
āĪ“<Ī± SĪ±, such that
(a) if Qr is ā, then fr maps each tuple to the least Ī³ such that P forces
Qr+1yr+1 Ā· Ā· Ā·Qmym B(c1, ..., cn, cā²1, ..., cā²rā1, yr+1, ..., ym)
for some cĪ³ ā S, or to 0 if there is no such Ī³.(b) if Qr is ā, then fr maps each tuple to the least Ī³ such that P forces
Ā¬Qr+1yr+1 Ā· Ā· Ā·Qmym B(c1, ..., cn, cā²1, ..., cā²rā1, yr+1, ..., ym)
for some cĪ³ ā S, or to 0 if there is no such Ī³.Now, for each r, let gr be the function such that gr(c1, ..., cn, cā²1, ..., c
ā²n)
is the supremum of {fr(P, c1, ..., cn, cā²1, ..., cā²rā1) | P is a forcing condition}.Deļæ½ne another function, h, such that h(Ī³) is the supremum of the range ofgr across all r with 1 ā¤ r ā¤ m and all ci, c
ā²j in
āĪ“<Ī³ SĪ³ . The function h is
still deļæ½nable inM.Let Ī²1 = Ī² and Ī²n+1 = h(Ī²n), and take Ī²ā² = sup{Bn}. If we take c1 asthe set deļæ½ned by the formula that says c1 is
āĪ³<Ī²ā² MĪ³(Gā), then c1 will
contain the range of f on c0. So, we can obtain the range of f on c0 as theset {c | āx ā c0 Ac1(x, c)}.
ļæ½
The following lemma is useful because constructibility provides a natural well-ordering on the elements that are constructed. We needed to design a model inwhich V = L failed, but in order to have the Axiom of Choice in N , we need tohave that all sets can be well-ordered. The Axiom of Choice is desirable because itmakes some of the remaining theorems easier to prove, and because satisfying ACwill mean that our model demonstrates the consistency of Ā¬CH with ZFC, whichis a stronger condition than consistency with ZF .
Lemma 10.4. Every element in N is constructible from W .
Proof. We want to verify that every x in N is a member of MĪ±(W ) for someordinal Ī±. In other words, we want to show the underlying set of N is a subset ofā{
MĪ±(W ) | Ī± āM}.
Remember that we deļæ½ned N as the set of elements constructible from thecollection of our new subsets aĪ“ before stage Ī±0, and recall that W is the set ofordered pairs of the form (Ī“, aĪ“) = {{Ī“}, {Ī“, aĪ“}}. It is clear that all of the aĪ“ are inthe transitive closure of W , and so the set of elements that are constructible fromW contains the set of elements that are constructible from the set of aĪ“, which wasthe set of elements in N . ļæ½
We can see also see now why the aĪ“ are not constructible in N . Fix one of theaĪ“ and some ordinal Ī±. Let cĪ± ā S be the label such that cĪ± = Ī±. Let x be theconstructible set constructed at step Ī± in some well-ordering of the constructionprocess (for example, ordered by MĪ² , then by formulas). The construction processwill be the same in N as it is in ZF set theory, so within N , x is the Ī±-th elementconstructed. Any forcing condition P is ļæ½nite and so can be contained in a condition
INDEPENDENCE OF THE CONTINUUM HYPOTHESIS 21
Q such that for some n ā N, either n ā x is true and Ā¬(n ā aĪ“) is in Q or Ā¬(n ā x)is true and (n ā aĪ“) is in Q. That is, P can be contained in a Q that makes aĪ“diļæ½erent from x. If aĪ“ were the Ī±-th element constructed in N , then some Pn wouldhave to force this to be true. Any of the Q containing Pn which cause aĪ“ to diļæ½erfrom x would then also have to force that statement, but if we consider a completesequence {Qn} starting with Q0 = Q, we see that in a model deļæ½ned by it in thesame way that we have made our N , we would have that both x and aĪ“ are theĪ±-th element constructed, of which there can only be one, and yet x 6= aĪ“.
Theorem 10.5. The Axiom of Choice is true in N . (So N is also a model ofZFC.)
Informally, the above theorem holds since every element being constructible fromW provides us with a well-ordering on the elements of N . The elements can beordered based on the stage at which they are constructed, an ordering on theformulas used to deļæ½ne them, and the well-ordering of W itself. This in turn makesevery set well-ordered in our model, which is a condition equivalent to the Axiomof Choice.
Since we know that we have a model of ZFC in which we have at least āµĻ manydistinct subsets of N, our only concern is whether by introducing new sets toM wehave made some bijective functions available which were not already present, andso perhaps may have made it so that some ordinals that appeared to have diļæ½erentcardinalities in M due to the lack of a one-to-one correspondence now have thesame cardinality in N . This would be a problem because it could potentially resultin the cardinalities between āµ1 and āµĻ collapsing together. We need to know thatāµĻ > āµ1 still actually holds in N . We will see that, in fact, every pair of ordinalswith diļæ½erent cardinalities inM still have diļæ½erent cardinalities in N , but ļæ½rst weneed two more lemmas.
Deļæ½nition 10.6. Let P and Q be forcing conditions. We call P and Q incom-
patible if there is no forcing condition R such that both P < R and Q < R.
In our case, since we have chosen ā as the relation for our condition set, we havethat P andQ are incompatible when they cannot both be a subset of another forcingcondition. This happens simply when P and Q contain contrasting membershipstatements, for example if 1 ā a were in P and Ā¬(1 ā a) were in Q.
Lemma 10.7. If B is a set inM of mutually incompatible P , then B is countableinM.
Proof. Assume B is uncountable. Deļæ½ne Bn to be
{P ā B | P contains fewer than n statements}for each n ā N. SinceM satisļæ½es V = L, it also satisļæ½es the Axiom of Choice, andso we know that the union of countably many countable sets is a countable set inM. We have that B =
ānāN Bn, and so, since B is uncountable, but
ānāN Bn is the
union of countably many sets, we must have that some Bn is uncountable. Now,let k be the largest integer such that for some P0, not in B, containing exactlyk statements, there are uncountably many P ā Bn where P0 < P . Let BP0 bethe set of those P ā B such that P0 < P . Take any condition P1 ā B and letA1, A2, ..., Am be the statements in P1 ā P0, of which there are ļæ½nitely many sinceall forcing conditions are ļæ½nite, and of which there is at least one since P0 6= P1
INDEPENDENCE OF THE CONTINUUM HYPOTHESIS 22
and P0 < P1. Since BP0 is still a set of mutually incompatible conditions, P1 isincompatible with every other P ā BP0 . Also, every P ā BP0 contains P0 bydeļæ½nition of BP0 . So, we must have that one of the Ai is such that uncountablymany P ā BP0 contain Ā¬Ai, because if there were no such Ai, then only countablymany sets would be incompatible with P1. However, Ā¬Ai cannot be in P0 sincewe have P0 < P1, and so P0 āŖ {Ā¬Ai} is a condition with k + 1 elements which iscontained in uncountably many P ā Bn. This contradicts our selection of k, andthus we have that B must be countable. ļæ½
The next lemma will allow us to conclude that if a function deļæ½ned in N hasa domain with cardinality Īŗ in N , then the cardinality of its range cannot exceedāµ0 Ā· Īŗ in M. This is what prevents any new functions in N from bridging a gapbetween cardinals.
Lemma 10.8. Let f be a function deļæ½ned in N . There is a function g, deļæ½ned inM, which sends each c ā S to a countable subset g(c) ā S and such that for all c,f(c) = cā² for some cā² ā g(c).
Proof. In N , set cā² as the earliest element of S for which f(c) = cā². The deļæ½nitionof each SĪ± can be carried out in N , and a well-ordering on S can be found in N ,so there is an unlimited statement T which says that cā² is the ļæ½rst element of S forwhich f(c) = cā². Since this statement is true in N due to our selection of cā², andsince f is a function in N , both the statement T and that f is a function must beforced by some Pn according to Lemma 10.1.
InM, for every c, cā² in S, deļæ½ne A(c, cā²) to be the set of all forcing conditions Psuch that P forces f to be a function and such that cā² is the earliest element of Sunder some well-ordering so that for all Q > P , we have that Q forces f(c) = cā². Wecan see that if cā², cā²ā² are distinct elements of S, then each of the elements in A(c, cā²)must be incompatible with each of the elements in A(c, cā²ā²), since if P ā A(c, cā²)and Q ā A(c, cā²ā²), then any R such that P < R and Q < R would have to forcef(c) = cā² = cā²ā². Since the Axiom of Choice holds in M, and by Lemma 10.7,there can be at most countably many cā² such that A(c, cā²) 6= Ļ, the function fromour labeling. Let g(c) be the countable set of these cā². We have that A(c, cā²) isnonempty because of the condition Pn we found above, and we have that the cā²
such that f(c) = cā² holds is in g(c). ļæ½
Theorem 10.9. Given inļæ½nite ordinals Ī±, Ī² such that inM, the cardinality of Ī±,denoted |Ī±|, is less than the cardinality of Ī², denoted by |Ī²|, we have that |Ī±| < |Ī²|in N as well.
Proof. Suppose that f is a function in N from |Ī±| onto |Ī²|. Deļæ½ne
ZĪ³ =ā{SĪ“ | Ī“ < Ī³}
for every ordinal Ī³, and note that Ī³ ā ZĪ³+3 for all Ī³. In particular, we have thatĪ± ā ZĪ±+3. Extend the function f to a function f ā² on XĪ±+3 such that f ā²(x) = 0 forall x 6ā Ī± and f ā²(x) = f(x) for all x ā Ī±. Lemma 10.8 tells us that the range of f ā²
is contained in T where T is a subset of ZĪ² =ā{SĪ“ | Ī“ < Ī²} and the cardinality of
T is inM is at most āµ0 Ā· |Ī±| and so is at most |Ī±| since Ī± is inļæ½nite. Now, because|T | ā¤ |Ī±| < |Ī²| in M, there is an ordinal Ī³ < Ī² such that T ā {SĪ“ | Ī“ < Ī³}.However, this means that T is constructed earlier than Ī², and so cannot contain
INDEPENDENCE OF THE CONTINUUM HYPOTHESIS 23
Ī², and thus f cannot be onto. So, there are no onto functions from Ī± to Ī² in themodel N , which means that |Ī±| < |Ī²| in N . ļæ½
So, we have introduced āµĻ many subsets of N into our model, which views eachof these subsets as distinct sets, and we know that the cardinals are still distinct, soāµĻ ā„ āµ2 inN as we intended. Thus, the cardinality of the continuum inN is at leastāµĻ , which is strictly greater than āµ1, and so the Continuum Hypothesis is false.This establishes the consistency of ZF āŖ {Ā¬CH & AC}. Together with Gƶdel'swork, this demonstrates that both the Continuum Hypothesis and GeneralizedContinuum Hypothesis are independent of the axioms of ZF and ZFC set theory.
INDEPENDENCE OF THE CONTINUUM HYPOTHESIS 24
References
[1] Paul J. Cohen. Set Theory and the Continuum Hypothesis. Dover Publications: Mineola, NewYork, 2008.
[2] Timothy Gowers, ed. The Princeton Companion to Mathematics. Princeton University Press:Princeton, New Jersey, 2008.
[3] Richard Hodel. An Introduction to Mathematical Logic. PWS Publishing: Boston, Mas-sachusetts, 1995.
[4] Karel Hrbacek and Thomas Jech. Introduction to Set Theory, 3rd ed. CRC Press: Boca Raton,Florida, 1999.