aximatica de conjuntos suppes

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Axiomatic Set Theory

January 14, 2013

1 IntroductionOne of our main aims in this course is to prove the following: 1 2 3

Theorem 1.1 (G odel 1938) If set theory without the Axiom of Choice (ZF)is consistent (i.e. does not lead to a contradiction), then set theory with the axiom of choice (ZFC) is consistent.

Importance of this result: Set theory is the axiomatization of mathematics,and without AC no-one seriously doubts its truth, or at least consistency.However, much of mathematics requires AC (eg. every vector space has abasis, every ideal can be extended to a maximal ideal). Probably most math-ematicians dont doubt the truth, or at least consistency, of set theory withAC, but it does lead to some bizarre, seemingly paradoxical resultseg. theBanach-Tarski paradox Hence it is comforting to have G odels theorem.

I formulate the axioms of set theory below. For the moment we have:

(AC.) Axiom of Choice (Zermelo) If X is a set of non-empty pairwisedisjoint sets, then there is a set Y which has exactly one element in commonwith each element of X .

To complement Godels theorem, there is also the following result whichis beyond this course:

1 These lecture were originally written by Peter Koepke many years ago and subse-quently modied and tought at Oxford for a number of years by Alex Wilkie. In 2006 thecourse was given by R.Knight, who edited and typed them up. I have introduced a fewmore editorial changes.

2 Prerequisite for this course is B1.3 See Andreas Blass, On the inadequacy of inner models, JSL 37 no. 3 (Sept 72)

569571.

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Proposition 1.2 (Cohen 1963) If ZF is consistent, so is ZF with AC.

We shall also discuss Cantors continuum problem which is the following.Cantor dened the cardinality, or size, of an arbitrary set. The cardinality

of A is denoted card A. He showed that card R > card N, but could not ndany set S such that card R > card s > card N, so conjectured:

(CH.) Cantors Continuum Hypothesis For any set S , either card S card N, or card S card R .

Again Godel (1938) showed:

Theorem 1.3 If ZF is consistent, so is ZF+AC+CH,

and Cohen (1963) showed:

Proposition 1.4 If ZF is consistent, so is ZF+AC+ CH.

We shall prove G odels theorem but not Cohens.Of course Godels theorem on CH was perhaps not so mathematically

pressing as his theorem on AC since mathematicians rarely want to assumeCH, and if they do, then they say so.

We rst make G odels theorem precise, by dening set theory and its

language.

2 BasicsSee D. Goldrei Classic Set Theory , Chapman and Hall 1996, or H.B. EndertonElements of Set Theory , Academic Press, 1977.

The language of set theory , LST, is rst-order predicate calculus withequality having the membership relation (which is binary) as its only non-logical symbol.

Thus the basic symbols of LST are: =, , , , , ( and ), and an

innite list v0, v1, . . . , vn , . . . of variables (although for clarity we shall oftenuse x ,y,z , t , . . . ,u ,v, . . . etc. as variables).

The well-formed formulas , or just formulas , of LST are those expressionsthat can be built up from the atomic formulas : vi = v j , vi v j , using therules: (1) if is a formula, so is , (2) if and are formulas, so is (),and (3) if is a formula, so is vi .

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2.1 Some standard abbreviationsWe write ( ) for ( ); ( ) for ( ); ( ) for(( ) ( )); x for x; !x for x ( y((y) x = y));x y for x(x y ); x y for x(x y ); x,y (etc.) forxy; x / y for x y.

We shall also often write as (x) to indicate free occurrences of a variablex in . The formula (z ) (say) then denotes the result of substituting everyfree occurrence of x in by z . Similarly for (x, y), (x,y,z ),..., etc.

2.2 The Axioms

(A1.) Extensionality

x, y(x = y t(t x t y))

Two sets are equal iff they have the same members.

(A2.) Empty set xy y / x

There is a set with no members, the empty set , denoted .

(A3.) Pairing x, yz t(t z (t = x t = y))

For any sets x, y there is a set, denoted {x, y}, whose only elements are xand y.

(A4.) Union xyt(t y w(w x t w))

For any set x, there is a set, denoted x, whose members are the membersof the members of x.(A5.) Separation Scheme If (x , y) is a formula of LST, the following is anaxiom:

xuz y(y z (y u (x , y))

For given sets x, u there is a set, denoted {y u : (x , y)}, whose elementsare those elements y of u which satisfy the formula (x , y).

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(A6.) Replacement Scheme If (x, y) is a formula of LST (possibly with

other free variables u , say) then the following is an axiom:

x,y,y (((x, y) (x, y )) y = y ) sz y(y z x s (x, y))

The set z is denoted {y : x(x, y) x s}. The image of a set under a functionis a set.

(A7.) Power Set

xyt(t y z (z t z x))

For any set x there is a set, denoted P(x), whose members are exactly thesubsets of x.

(A8.) Innity

x[y(y xz (z / y)y(y x z (z xt(t z (t yt = y))))]

There is a set x such that x and whenever y x, then y {y} x.(Such a set is called a successor set . The set of natural numbers is a successor set.

(A9.) Foundation

x(z z x z (z x y(y z y / x))

If the set x is non-empty, then for some z x, z has no members in commonwith x.

(A10.) Axiom of Choice

u[[x uy y x x, y((x u y u x = y) z (z / x / y))]

vx u!y(y x y v)]We write ZF for the collection of axioms A1A8; ZF for A1A9; ZFC

for A1A10.

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2.3 Proofs in principle and proofs in practiceSuppose that T is one of the above collections of axioms. If is a sentence of LST (ie. a formula without free variables), we say that is a theorem of T ,or that can be proved from T , and write T , if there is a nite sequence1, . . . , n of LST formulas such that n is , and each i is either in T or elsefollows from earlier formulas in the sequence by a rule of logic. Clearly everytheorem of ZF is a theorem of ZFC and every theorem of ZF is a theoremof ZF. To say that T is consistent means that for no sentence of LST is( ) a theorem of T (which is in fact equivalent to saying that there issome sentence which is not provable from T ). This now makes theorem 1.1precise: we must show that if ZF is consistent, then so is ZFC.

Now in proving this theorem we shall need to build up a large stock of theorems of ZF (and we shall discuss some theorems of ZFC as well) butto give formal proofs of these would not only be tedious but also infeasible.We shall therefore employ the standard short-cut of adopting a Platonic viewpoint . That is, we shall think of the collection of all sets as being a clearlydened notion and whenever we want to show a sentence, , say, of LST hasa formal proof (from ZF say) we shall simply give an informal argument thatthe proposition asserted by about this collection is true. Indeed, we shalloften not bother to write out as a formula of LST at all; we shall simplywrite down (in English plus a few logical and mathematical symbols) what

it is saying. Of course we shall take care that, in our informal argument,we only use those propositions about the collection of all sets asserted by theaxioms of ZF. Thus, for example, if I write:

Theorem 2.1 (ZF ) There is no set containing every set.

then I mean that from the axioms of ZF there is a formal proof of the LSTsentence

xy y / x.

Actually, it probably wouldnt be too difficult to give a formal proof of

this, but we shall supply the following as a proof:Proof. Suppose A were a set containing every set. By A5 {x A : x / x} isa set, call it B . Then B B iff B A and B / B . But B A is true (as Acontains every set), so B B iff B / Ba contradiction.

Of course in all such cases, the reader should convince him- or herself that (a) the informal statement we are proving can be written as a sentence

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of LST, and (b) the given proof can be converted, at least in principle, to a

formal proof from the specied collection of axioms.

2.4 InterpretationsThe Completeness Theorem for rst-order predicate calculus (also due toGodel) states that a sentence (of any rst-order language) is provable froma collection of sentences S (in the same language) if and only if every modelof S is a model of . Equivalently, S is consistent if and only if S has amodel. Let us examine this in our present context. Firstly, a structure for LST is specied by a domain of discourse M over which the quantiers x . . .

and x . . . range, and a binary relation E on M to interpret the membershiprelation . If is a sentence of LST which is true under this interpretation wesay that is true in M, E or M, E is a model of , and write M, E .If T is a collection of sentences of LST we also write M, E T iff M, E for each sentence in T . (If (x1, . . . , x n ) is a formula of LST with freevariables among x1, . . . , x n and a1, . . . , a n are in the domain M , we alsowrite M, E (a1, . . . , a n ) to mean (x1, . . . , x n ) is true of a1, . . . , a n in theinterpretation M, E .)

For example, suppose M contains just the two distinct elements a and b,and E is specied by a b, ie. E (a, b), not E (b, a), not E (a, a ), not E (b, b).Then M, E A2, ie. M xyy / x, since it is true that there is an x inM (namely a) such that for all y M , not E (y, x). It is also easy to see thatM, E A1 and M, E A3. Notice that, by the completeness theorem,

this implies that A3 is not provable from the axioms A1, A2 since we havefound a model of the latter two axioms which is not a model of the former.

Exercise 2.2 Let Q be the set of rational numbers and the usual ordering of Q. Which axioms of ZF are true in Q , ?

Note that the Platonic viewpoint adopted here amounts to regarding asentence, , say, of LST as true, if and only if V , , where V is the

collection of all sets, and is the usual membership relation.The completeness theorem provides a method for establishing theorem1.1. For we can rephrase that theorem as: If ZF has a model then so doesZFC. Indeed we shall construct a subcollection L of V such that if we as-sume V , ZF, then L, ZFC. (Actually our proof will yield some-what more which ought to be enough to satisfy any purist. Namely, it will

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produce an effective procedure for converting any proof of a contradiction

(ie. a sentence of the form ( )) from ZFC to a proof of a contradictionfrom ZF.)

We now turn to the development of some basic set theory from the axiomsZF.

2.5 New sets from oldThe axioms of ZF are of three types: (a) those that assert that all sets havea certain property (A1, A9), (b) those that sets with certain properties exist(A2, A8), and (c) those that tell us how we may construct new sets out of

given sets (A3A7). Our aim here is to combine the operations implicit in theaxioms of type (c) to obtain more ways of constructing sets and to introducenotations for these constructions (just as, for example, we introduced thenotation x for the set y given by A4). It will be convenient to use theclass notation {x : (x)} for the collection (or class ) of sets x satisfyingthe LST formula (x).4 As we have seen, such a class need not be a set.However, in the following denitions it can be shown (from the axioms ZF )that we always do get a set. This amounts to showing that for some seta, if b is any set such that (b) holds (ie. V (b)) then b a, so that{x : (x)} = {x a : (x)} which is a set by A5. I leave all the requiredproofs as exercisesthey can also be found in the books.

In the following, A,B , . . . , a, b, c, . . . , f , g, a 1, a2, . . . , a n , . . . etc. all denotesets.

1. {a1, . . . , a n } : = {x : x = a1 . . . x = an }.

2. a b : = {a, b} = {x : x a x b}.3. a b : = {x : x a x b}.

4. a \ b : = {x : x a x / b}.

5. a : ={x :yax y}

if a =

undened if a = .4 Actually, (x ) will be allowed to have parameters (ie. names for given sets), so is not

strictly a formula of LST. Notice, however, that parameters are allowed in A5 and A6 (thex and u ).

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6. a, b : = {{a}, {a, b}}.

(Lemma. a, b = c, d (a = c b = d).)

7. a b : = {x : c ad bx = c, d }.

8. a b c : = a (b c),. . . , etc.

9. a2 : = a a, a3 : = a a a,..., etc.

10. We write a b for x a(x b).

11. c is a binary relation on a we take to mean c a2. (Similarly forternary,. . . , n-ary, . . . relations.)

12. If A is a binary relation on a we usually write xAy for x, y A.

A is called a (strict) partial order on a iff

(a) x, y a(xAy yAx),

(b) x,y,z a((xAy yAx) xAz ).

If in addition we have (3) x, y a(x = y xAy yAx), then A iscalled a (strict) total ( or linear) order of a.

13. Write f : a b (f is a function with domain a and codomain b, orsimply f is a function from a to b) if f a b and c a!d b c, d f . Write f (c) for this unique d.

14. If f : a b, f is called injective (or one-to-one ) if c, d a(c = d f (c) = f (d)), surjective (or onto) if d bc af (c) = d, and bijective if it is both injective and surjective.

15. We write a b if f (f : a b f bijective).

16. a b : = {f : f : a b}.

17. A set a is called a successor set if

(a) a and

(b) b(b a b {b} a).

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Axiom A8 implies a successor set exists and it can be further shown

that a unique such set, denoted , exists with the property that afor every successor set a. The set is called the set of natural numbers.If n, m we often write n + 1 for n {n} and n < m for n m and0 for (in this context). The relation (ie. < ) is a total order of (more precisely { x, y : x , y x y} is a total order of ).

18. The set satises the principle of mathematical induction , ie. if (x)is any formula of LST such that (0)n ((n) (n +1)) holds,then n (n) holds.

19. The set also satises the well-ordering principle , ie. for any set a, if a and a = then b ac a(c > b c = b).

20. Denition by recursion

Suppose that f : A A is a function and a A. Then there is aunique function g : A such that:

(a) g(0) = a, and

(b) n g(n + 1) = f (g(n)).

(Thus, g(n) = f (f (f

n timesa)) )).)

More generally, if f : B A A and h : B A are functions,then there is a unique function g : B A such that

(a) b Bg(b,0) = h(b), and

(b) b Bn g(b, n + 1) = f (b,n,g(b, n)).

Using this result one can dene the addition, multiplication and expo-nentiation functions.

(Remark I have adopted here the usual convention of writing g(b, n+1)

for g( b, n + 1 ). Similarly for f .)21. A set a is called nite iff n a n.

22. A set a is called countably innite iff a .

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23. A set a is called countable iff a is nite or countably innite. (Equiva-

lently: iff f (f : a f injective).)(Theorem P is not countable. In fact, for no set A do we haveA PA. (Cantor))

3 Classes, class terms and recursionV =the collection of all sets (assuming only ZF ).

We call collections of the form {x : (x)}, where is a formula of LST,classes .

Every set is a class, a = {x : x a}. (so (x) is x a here).We must be careful in their usewe cannot quantify over them but someoperations will still apply, eg. if U 1 = {x : (x)} and U 2 = {x : (x)}, then

U 1 U 2 = {x : (x) (x)}U 1 U 2 = {x : (x) (x)}U 1 U 2 = {x : y(y = s, t (s) (t))}

(1)

and so on. x U 1 means (x) and U 1 U 2 means x((x) (x)).

Classes are only a notationwe can always eliminate their use.Note that V is a classV = {x : x = x}.If F, U 1, U 2 are classes with the properties that F U 1 U 2 and x

U 1!y U 2 x, y F , then F is called a class term , or just a term, and wewrite F (x) = y instead of x, y F . We also write F : U 1 U 2, althoughF may not be a function, as U 1 may not be a set. So if F = {x : y1, y2(x =y1, y2 y2 = y1)}, so for all sets F (x) = x, then F is a class term. We

need class terms for higher recursion.

3.1 The recursion theorem for (We assume only ZF throughout.)

Theorem 3.1 Suppose G : U U is a class term and a U . Then there is a term F : U (which is therefore a function) such that

1. F (0) = a and

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2. n F (n + 1) = G(F (n)) .

Proof.

Lemma 3.2 Suppose that n . Then there is a unique function f , with domain n + 1 , such that

1. f (0) = a and

2. m nf (m + 1) = G(f (m)).

(Recall n = {m : m < n }.)Proof. Existence: By induction on n.

For n = 0: Let f = { 0, a }. Then f is a function with domain {0} =1 = 0 + 1, such that f (0) = a and m 0 f (m + 1) = G(f (m)). (trivially)

Suppose true for n. Let f have domain n and satisfy (1) and (2). Letb = f (n). Let f = f { n + 1, G(b) }. Then f is a function with domainn+1{n+1 } = ( n+1)+1. Further f (0) = f (0) = a (since 0 n+1 = dom f )(using (1)) and if m n + 1 = n {n}, then either m n, in which casem + 1 n + 1 = dom f , so f (m + 1) = f (m + 1) = G(f (m)) (using (2)) (byproperties of f ), or m = n, so f (m + 1) = f (n + 1) = G(b) = G(f (n)) =G(f (m)), as required. So the proposition is true for n + 1.

The uniqueness is also by induction.

We now dene F by

F = {z : x y U z = x, y f (f is a function with domain x + 1

such that f (0) = a w xf (w + 1) = G(f (w)) f (x) = y)}

the stuff after the colon is a formula of LST.It is easy to show that x !y U x, y F , and that F satises (1)

and (2) of Theorem 3.1.

Some applications:

Denition 3.3 A set a is called transitive if x ay x y a. (ie. x a x a, or a a.)Lemma 3.4 is transitive; and if n , then n is transitive.

Proof. See the books.

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Theorem 3.5 For any set a, there is a unique set b, denoted T C (a), and

called the transitive closure of a, such that 1. a b,

2. b is transitive,

3. whenever a c and c is transitive, then b c.

Proof. Uniqueness is clear since if a b1 and a b2, b1 and b2 transitive andboth satisfying (3), then b1 b2 and b2 b1, so b1 = b2.

For existence (idea: b = a a a . . .).Let G be the class term given by G(x) = x (for x V ). Apply 3.1, to

get a term F such that

1. F (0) = a, and

2. n F (n + 1) = G(F (n)) = F (n).

By replacement, there is a set B such that B = {y : x F (x) = y}.Let b = B = {F (n) : n }. Then

1. Since a = F (0) and F (0) B, we have a B , so a B = b.

2. Suppose x b and y x. We must show y b. But x b impliesx B implies x F (n) for some n implies x F (n), soy F (n), so y F (n + 1), so y B, so y b.

3. Suppose a c, c transitive.We prove by induction on n that F (n) c.F (0) = a c.Suppose F (n) c.We want to show that F (n + 1) c, so suppose x F (n + 1), iex

F (n). Then for some y F (n), x y. Thus x y F (n) c,

so x y c, so x c, since c is transitive, as required.Thus, by induction, n F (n) c, so {F (n) : n } c, ie. b c,as required.

Recursion on .

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Theorem 3.6 (Requires Foundationie. assume ZF) For (x) any formula

of LST (with parameters) if x(y x (y) (x)) , then x(x). (The hypothesis trivially implies ().)

Proof. Suppose x(y x (y) (x)), but that there is some set a suchthat (a). Then a = . Let b = T C (a), so a b, and hence b = .Let C = {x b : (x)}. Then C = , since otherwise we would havex b (x), hence x a (x) (since a b), and hence (a), contradiction.

By foundation there is some d C such that d C = , ie. d b,(d), but x d x b (since b is transitive) and x / C . But this meansx d (x), so (d)contradiction.

Our present aim is to prove that if ZF is consistent then so is ZFso wewont use 3.6. Instead we nd another generalization of induction.

Denition 3.7 Suppose that a is a set and R is a binary relation on a.Then R is called a well-ordering of a if

1. R is a total ordering of a.

2. If b is a non-empty subset of a, then b contains an R-least element.ie. x by b(y = x xRy ).

Remark. AC iff every set is well-orderable.

Denition 3.8 Suppose that R1 is a total order of a, and R2 is a total order of b. Then we say that a, R 1 is order-isomorphic to b, R2 , written a, R 1 b, R2 , if there is a bijective function f : a b such that x, y

a(x < y f (x) < f (y)) .

Denition 3.9 We say x is an ordinal, On(x), or x On, if

1. x is transitive, and

2. is a well-ordering of x.We usually use , , etc., for ordinals.On is a class.

Theorem 3.10 (Enderton)

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3. Show < () ( ), for limit

Theorem 3.14 (Denition by recursion on On) Suppose F : V V is a class term, and a V . Then there is a unique class term G : On V such that

1. G(0) = a

2. G( + 1) = F (G())

3. G( ) = for a limit.

Proof. Let (g, ) be the formula of LST expressing:g is a function with domain + 1 such that < g ( + 1) = F (g( ))

and if is a limit g( ) = {g() : < } and g(0) = a.((*) Note that if (g, ) and , then (g ( + 1) , ).)

Lemma 3.15 On !g (g, ).

Proof. Induction on . = 0: Clearly g = { 0, a } is the only set satisfying (g, 0).Suppose true for . Let g be the unique set satisfying (g, ). (Note

g : + 1 V .) Certainly g= g { + 1 , F (g()) } satises (g, +1).If g also satised (g , + 1), then (g + 1 , ) holds, so by the inductivehypothesis g = g + 1. But (g , + 1) implies g ( + 1) = F (g ()) =F (g()). So g = g { + 1 , F (g()) } = g, as required.

Suppose is a limit and < !g (g, ). For given < let theunique g be g . Notice that S = {g : < } is a set by Replacement. But 1 < 2 implies g 1 = g 2 1 + 1. Let g= S . Then g is a function withdomain { : < } = , and < g( + 1) = F (g()) and if is alimit < , then g( ) = {g() : < } and g(0) = a. (Since for any < , g coincides with g on + 1, and the g s satisfy the condition bythe inductive hypothesis.) Further g is the only such function by (*).

Now dene g = g { , {g() : < } }. Then g is unique such that

(g, ).Now set G = { x, : g((g, ) g() = x)).Then G satises the required conditions since by the lemma for each

On, G + 1 is the unique g such that (g, ).We get uniqueness of G by induction.

The following can be easily proved using the above.

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Theorem 3.16 Suppose F : V On V and H : V V are class

terms. Then there is a unique class term G : V On V such that 1. G(x, 0) = H (x)

2. G(x, + 1) = F (x, G (x, ))

3. G(x, ) =


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4 The Cumulative Hierarchy and the consis-tency of the Axiom of Foundation

We apply Theorem 3.14 with a = and F (x) = Px, to get V : On V dened by

1. V (0) =

2. V ( + 1) = PV (), and

3. V ( ) =


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Corollary 4.2 (1) V is a transitive class (ie. x y V x V ) contain-

ing all the ordinals.(2) < V V .

Theorem 4.3 (V,) ZF.

Proof. (Note that ( V,) is a substructure of ( V ,), so for a, b V , (V,) a b iff a b, and (V,) a = b iff a = b.)

A1. Suppose x, y V , and V, t(t x t y) (*). We must showV, x = y, ie x = y. Suppose x = y. Say a x, a / y. Since a x V

we have a V (by Corollary 4.2). But by (*), t V , t x t y. Inparticular a x a ycontradiction.

So x = y.A2. We must show V, xy y / x. Since V , we have V ,

and clearly y V, / .A3. Suppose a, b V . We must show V, z t(t z (t = a t =

b)). Let c = {a, b}. Now by 4.2 (ii), there is some such that a, b V . Soc V , so c V +1 , so c V . It remains to show t V (t c (t =a t = b)), which is clear since this is true t V .

A4. V, Unionsexercise.A7. Power Set Suppose a V . We must show V, yt(t y

z (z t z a)).Now suppose a V .Exercise: On, if b a V , then b V .It follows that b P(a), b V . Thus P(a) V , so P(a) V +1 . So

P(a) V . Let c = P(a).We show V, t(t c z (z t z a)).So suppose t V . ): If V, t c, then t c, so t a, ie. z V (z t z a),

thus z V (z t z a). ): Suppose V, z (z t z a) (*) (ie. V, t a). We show

that really, t a. Suppose d t. Since t V , we have d V (by 4.2 (i)).

Hence, by (*), d a. Thus t a, so t c, so V, t c as required.A8. Innity Exercise (Note: V +1 , so V ).A9. Foundation Suppose a V , a = . We must nd b a such that

b a = .[Since then b V , by transitivity, and V, y by / a.]

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Let x a. Then x V , so x V for some . This shows

On,a V = . Choose minimal such that a V = . Then is asuccessor ordinal since, for a limit, a V = a


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A6. Replacement Suppose (x, y) is a formula of LST (possibly involving

parameters from V ).Suppose V, x,y,y (((x, y) (x, y )) y = y ).

Let (x, y) beV (x )

x V V (y)

y V V (x, y). [Note V (x) has no parameters.]Then we have (in V ) x,y,y (((x, y) (x, y )) y = y ), by lemma4.5.

Let s V .Hence there is a set z such that

y(y z x s(x, y)) (*)

(by replacement in V ). We want to show z V .Now by (*), if y z , then x s(x, y), so x s(x V y

V V (x, y), so y V .Thus for each y z , On, y V .Let (u, v) be u z v is the least ordinal such that u V v.Then by replacement in V , there is a set S such that

v(u z ( (u, v)) v S ).

Clearly S is a set of ordinals, so

S is an ordinal, say.

Clearly y z, y V . Hence z V , so z V +1 , so z V .We must show V, y(y z x s(x, y)). ): So suppose y V and y z .By (*), x s(x, y), ie. x s(x V y V V (x, y)), so V,x s(x, y). ): Conversely, if y V , and V, x s(x, y), then x S (x

V V (x, y)), so x s(x V y V V (x, y)), ie x s(x, y), so by(*), y z .

Corollary 4.6 If ZF is consistent, then so is ZF.

Proof. If is an axiom of ZF, we have shown that ZF V . Henceif 1, 2, . . . , k were a proof of a contradiction from ZF, then (roughly)V 1 , . . . , V k could be converted into one from ZF.

From now on we assume Foundation, and hence may assume (exercise)that V = V .

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5 Levys Reection Principle

5.1Theorem 5.1 (LRP) (ZFfor each individual )

Suppose W : On V is a class term, and write W for W (). Suppose W satises:

1. < W W ( , On)

2. W = W for all limit ordinals .

Let W =

On W ( = {x : On,x W }, so W is a class; each

W is a set.)Suppose (v1, . . . , vn ) is a formula of LST (without parameters). Then,

for any On, there is On such that , and such that a1, . . . a n W , W, (a1, . . . , a n ) iff W , (a1, . . . , a n ); ie. for all a1, . . . , a n W , W (a1, . . . , a n ) W (a1, . . . , a n ).

Proof. For any formula of LST, by the collection of subformulas of ,SF (), we mean all those formulas that go into the building up of . Formally

1. SF () = {} if is of the form x = y or x y;

2. SF () = {}SF ();

3. SF ( ) = { } SF () SF ();

4. SF (x) = {x}SF ().

Clearly SF () is a nite colleciton for any formula , and SF ().Suppose now that S is any nite collection of formulas, which is closed

under taking subformulasie. if S , then SF () S .Dene T S = { On : S a W ( W (a ) W (a )}. (Abuse of

notation here.) ( T S is a class since S is nite.)We must show that T S is unbounded in the ordinals. (LRP follows by

taking S = SF ( ).)We rst show.

Lemma 5.2 For any S as above, T S is a closed class of ordinals, ie. it contains all its limits, ie. when X is a subset of T S , then sup X T S .

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Proof. We prove this by induction on the total number n of occurrences of

connectives in formulas of S . We write this n as # S .If n = 0, then all formulas of S are of the form x = y or x y (for

variables x and y), so T S = On, so T S is denitely closed.Now suppose that # S = n + 1. Let be a formula in S with maximal

number of connectives.Let S = S \ { }. Clearly S is also closed under taking subformulas and

# S n. Also since S S , we have T S T S .Let X T S , a subset, and suppose X has no greatest element. Note that

X T S , so sup X T S by the inductive hypothesis.We want to show that sup X T S .

Case 1. is . Note S , so T S = T S . So sup X T S .Case 2. is 1 2. Then again 1, 2 S , so we can easily checkT S = T S , and the result follows by the inductive hypothesis.

Case 3. is vn +1 (v1, . . . , vn , vn +1 ).Then (v1, . . . , vn , vn +1 ) S . Let = sup X . Now since X has no

greatest element is a limit ordinal, so W =


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Suppose a W and W (a). Since W =

X W we have a W

for some X . Now vn +1 W W (a , vn +1 ). Since W W , we havevn +1 W W (a , vn +1 ). Now let an +1 W . Then W (a , a n +1 ). Henceby (*), W (a , an +1 ). But X T S (and S ), so W (a , a n +1 ). Sincean +1 W was arbitrary, we have vn +1 W W (a , vn +1 ), ie. W (a ).Hence by (**), W (a ) as required.

To complete the proof of the theorem we now show that On On ( > T S ).The proof is again by induction on # S , and the only difficult case is when

is vn +1 (v , vn +1 ) and S = S \ { }, S closed under taking subformulas.By our inductive hypothesis we have

> T S . (***)

It remains to show that given any On, > T S , such thata W ( W (a ) W (a ). (For then such a will be in T S .)

Let On be given.Now (v) is vn +1 (v1, . . . , vn , vn +1 ).Dene the term f : On V n On so that Ona1, . . . , a n V

f (, a 1, . . . , a n ) is the least On such that > and an +1 W suchthat W (a1, . . . , a n , an +1 ), if such a exists.

Now dene the term F : On On so that On F ( ) is the least T S such that > sup{f (, a 1, . . . , a n ) : a1, . . . , a n W n }. (This lastthing is a set by replacement since W n is. exists using (***).)

Notice that for all , F ( ) > , F ( ) T S , and if a1, . . . , a n W ,vn +1 W F ( ) W (a1, . . . , a n , vn +1 ) vn +1 W W (a1, . . . , a n , vn +1 ) ()(For otherwise, an +1 W W (a1, . . . , a n , a n +1 ), so for some minimal

, an +1 W W (a1, . . . , a n , a n +1 ) (since W = On W ), so F ( ) f (, a 1, . . . , a n ) , so an +1 W F ( )W (a1, . . . , a n , a n +1 ) since W F ( ) W contradiction.)

Now by the recursion theorem on dene the function g : On by

1. g(0) = F (),2. g(n + 1) = F (g(n));

let X = ran g. Clearly X has no greatest element and X T S . Let =sup X . Since T S is closed (Lemma above), we have T S . We also have > , and:

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For all a1, . . . , a n W ,

if vn +1 W W (a1, . . . , a n , vn +1 ), then vn +1 W W (a1, . . . , a n , vn +1 ).(****)Proof. Suppose a1, . . . , a n W . Since W = X W , we have a1, . . . , a n W , for some X . Suppose vn +1 W W (a1, . . . , a n , vn +1 ).

Since F ( ) X , and hence W F ( ) W , we havevn +1 W F ( )W (a1, . . . , a n , vn +1 ).Hence by () we have vn +1 W W (a1, . . . , a n , vn +1 ), as required.

Now show that (****) implies T S as required (exercise, Problem sheet4).

6 G odels Constructible Universe6.1For any set a and n we dene n a to be {f : f : n a}, and


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Denition 6.1 (The constructible hierarchy)

We dene the class term L : On V (writing L for L()) by recursion on On as follows:

1. L0 = ;

2. L +1 = Def (L );

3. L =


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(4) By IH L V .

Also x L +1 x L x V x PV = V +1 .Thus L +1 V +1 .(5) By IH On L = .Suppose x On L +1 . Then x On and x L .But every member of x is an ordinal, so x L On , so x . Thus

either x or x = . In either case x { } = +1. Thus On L +1 + 1.

Suppose x + 1. Then either x , in which case x On L On L +1 (by (1)), or x = . So it remains to show L +1 .

Let s = .

Then A = G( On (v0) , L , s) = {b L : L , On(b)}, and A Def (L ) = L +1 . We show A = .But On(v0) is an absolute formula, that is has the same meaning in any

transitive class (exercise).Thus b L , L , On( ) iff b On.Thus A = L On = by IH, as required.The Limit Step Suppose > 0 is a limit ordinal and (1)(5) hold for all

< . Since L =


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2.

a L.

3. Pa L L.

4. L.

Proof. (1) Suppose a, c L . Dene s = { 0, a , 1, c }, so s

for all ran f .Clearly Pa L L (using 6.2 (1)).We may also suppose that a L .Let s = { 0, a }.Then L +1 G( v2 v1(v2 v0) , L , s) = {b L : L , v2

b(v2 a)} = A, say.Suffices to show A = Pa L.Suppose b A. Then b L (so b L) and L , v2 b(v2 a).Now suppose d b. Then d L since L is transitive. Hence L ,

d b d a, so d a.Hence b a, so b Pa L. Thus A Pa L.Conversely suppose b Pa L. Then b L .Also v2 b(v2 a). Hence v2 L (v2 b v2 a), so L ,v2 b(v2 a).

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So b A.

Hence Pa L = A. It is now easy to check that

Corollary 6.5 Extensionality, empty-set, pairs, unions, power-set are all true in L.

Lemma 6.6 L, separation.

Proof. Suppose u L, and a0, . . . , a n L. Say u, a 0, . . . , a n L . Let(v0, . . . , vn +1 ) be a formula of LST. By Levys Reection Principle, there is

some such that c, c1, . . . , cn +1 LL , (c cn +1(c0, . . . , cn , c)) L, (c cn +1 (c0, . . . , cn , c)) . ()

Let (v0, . . . , vn +2 ) = ( vn +2 vn +1 (v0, . . . , vn , vn +2 ).Let s = { 0, a0 , . . . , n, a n , n + 1 , u }.Then L +1 G( (v0, . . . , vn +2 ) , L , s) = {b L : L , (a0, . . . , a n ,u ,b)} =

{b L : L , (b u (a0, . . . , a n , b)} = A, say. (So A L.)Sufficient to show L, x(x A (x u (a0, . . . , a n , x))). ): Suppose x L and x A. Then x L , and L , x

u (a0, . . . , a n , x).

By (*), L, x u (a0, . . . , a n , x), as required. ): Suppose x L, and x u(a0, . . . , a n , x). Then x L , since x

L and L is transitive. Hence, using (*), ( L , x u (a0, . . . , a n , x),so x A, as required.

Lemma 6.7 L, replacement.

Proof. Suppose a0, . . . , a n L, a = a0, . . . , a n , u L, (x , y, z ) a formulaof LST, and L, z,y,y (((a , z , y) (a , z , y )) y = y )

.

Now choose so large that a0, a

1, . . . , a

n, u L

, and such that (by LRP)

for all z L L, y((a , z , y)z u) L , y((a , z , y)z u), and for all c, d L , L, (a , c ,d) iff L , (a , c ,d).

Now let A = {b L : L , z u((a , z , b)}, so A L +1 .Then, as in the proof of separation, L, z u(y(a , z , y) y

A((a , z , y)), as required.

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Lemma 6.8 L, Foundation.

Proof. Suppose a L. Choose b V such that b a b a = . Since L istransitive, b L and clearly L, b a b a = .

Theorem 6.9 L, ZF.

7 Absoluteness

7.1

Denition 7.1 The 0-formulas of LST are dened as follows:1. x y, x = y, x y, x = y are 0-formulas for any variables x and

y.

2. If , are 0-formulas, so are , , x y and x y (where x and y are distinct variables).

3. Nothing else is a 0 formula.

Lemma 7.2 If is a 0 formula, then is logically equivalent to a 0 formula.

Proof. Easy induction on . Note that x y is logically equivalent tox y .

Lemma 7.3 If (x1, . . . , x n ) is a 0-formula and U 1 and U 2 are transitiveclasses such that U 1 U 2, then for all a1, . . . , a n U 1,

U, (a1, . . . , a n ) U 2, (a1, . . . , a n ).

We say is absolute between U 1 and U 2.

Proof. Exerciseinduction on .

Denition 7.4 The 1-formulas of LST are dened as follows:

1. x y, x = y, x y, x = y are 1-formulas for any variables x and y.

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2. If , are 1-formulas, so are , , x y and x y

(where x and y are distinct variables), and x .3. Nothing else is a 1 formula.

Remark 7.5 Note that every 0 formula is 1.

Lemma 7.6 If (x1, . . . , x n ) is a 1-formula, and U 1 and U 2 are transitive classes with U 1 U 2, then for all a1, . . . , a n U 1

U 1, (a1, . . . , a n ) U 2, (a1, . . . , a n ).

(ie. is preserved up or is upward absolute between U 1 and U 2.)

Denition 7.7 (1) A formula (x) is called ZF 0 (respectively ZF 1 ) if there is a 0 (or 1) formula (x) such that ZF x((x) (x)) .

(2) A formula is called ZF 1 if and are ZF 1 .(3) Suppose n and F : V n V is a class term. Then F is called

ZF 1 if the formula (x1, . . . , x n , xn +1 ) dening F (x1, . . . , x n ) = xn +1 is ZF 1 ,and if ZF proves that F is a class term.

Remark 7.8 We need only verify that in part (3) is ZF 1 , since is ZF 1thus:

ZF x1, . . . , x n , xn +1 ((x1, . . . , x n , xn +1 ) y((x1, . . . , x n , y)y = xn +1 ))

and the bit on the right is ZF 1 if is.

Remark 7.9 Every ZF 0 formula is ZF 1 by 7.2 and 7.5.

Theorem 7.10 Suppose (x1, . . . , x n ) is ZF 1 and U 1 and U 2 are transi-tive classes such that U 1 U 2 and U i , ZF ( i = 1, 2). Then for all a1, . . . , a n U 1,

U 1, (a1, . . . , a n ) U 2, (a1, . . . , a n ).

(ie. is ZF-absolute .)

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Proof. Let (x1, . . . , x n ) be 1 such that ZF x((x) (x) (*).

Then

U 1, (a ) U 1, (a) (*) and U 1, ZF U 2, (a) by 7.6 U 2, (a ) (*) and U 1, ZF

(2)

Now let (x1, . . . , x n ) be 1 such that ZF x((x) (x) (*).Then as above,

U 1, (a ) U 1, (a ) (*) and U 1, ZF U 2, (a ) by 7.6 U 2, (a) (*) and U 1, ZF

(3)

Theorem 7.11 The following formulas and class terms are all ZF 0 (and hence ZF 0 ):

1. x = y

2. x y

3. x y

4. F (x1, . . . , x n ) = {x1, . . . , x n } (for each n)

5. F (x1, . . . , x n ) = x1, . . . , x n (for each n)

6. (where n 1 and 0 i n 1) F (x) = xi if x is an n-tuple x0, . . . , x n 1 , otherwise.

7. F (x, y) = x y.

8. F (x, y) = x y.

9. F (x) = x.

10. F (x) = x if x = , F (x) = otherwise.

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11. F (x, y) = x \ y.

12. x is an n-tuple.

13. x is an n-ary relation on y.

14. x is a function.

15. F (x) = dom x if x is a function, otherwise.

16. F (x) = ran x if x is a function, otherwise.

17. F (x, y) = x[y] ( = {x(t) : t y}) if x is a function, otherwise.

18. F (x, y) = x y if x is a function, otherwise.

19. F (x) = x 1 if x is a function, otherwise.

20. F (x) = x {x}.

21. x is transitive.

22. x is an ordinal.

23. x is a successor ordinal.

24. x is a limit ordinal.

25. x : y z .

26. x : y z .

27. x is a natural number.

28. x = .

29. x is a nite sequence of elements of y.

Proof. (Selections) (3) x y z x(z y) which is 0.Note that all the class terms F above are in ZF provably class terms, so

we only have to show that the statement F (x) = y can be put in 0 form.(4) F (x1, . . . , x n ) = y x1 y x2 y . . . xn y z y(z =

x1 . . . z = xn ).

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(5) F (x1, x2) = y z 1 yz 2 y(z 1 = {x1} z 2 = {x1, x2} t

y(t = z 1 t = z 2)), which is 0 by (4).(12) x is a 2-tuple iff z 1 x x1 z 1 x2 z 1 (x = x1, x2 ), which is

0 by (5).(13) x is a 2-ary relation on y iff z x y1 y y2 y (z = y1, y2 ),

which is 0 by (5).(29) x is a natural number iff ( x is an ordinal)(x is not a limit ordinal) (y

x y is not a limit ordinal), which is 0 by (24), (26) and the fact that ZF 0formulas are closed under .

Lemma 7.12 Suppose F and G are ZF 1 class terms. Then F (x) = G(y)

is ZF 1 .

Proof. Let (x , z ) adn (y , t ) be 1 formulas dening (in ZF) F (x) = y andG(y) = t respectively. Then

F (x) = G(y)

ZF z ((x , z ) (y , z )) ,

which is 1, and

F (x) = G(y)

ZF z t((x , z ) (y , t ) z = t),

which is 1.Hence F (x) = G(y) is ZF 1 .

Theorem 7.13 Suppose F : V V V is a ZF 1 class term. Then the class term G dened from F by recursion on On, ie:

1. G(0, x) = x

2. G( + 1 , x) = F (G(, x ), x) for all On

3. G(, x) =


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Proof. As in the proof of 3.14 dene (g,,x ) by

On () 1 g is a function 2 dom g = {} 3 g(0) = x 4 y1y2 (y1 = { } y2 = g( ) g(y1) = F (y2)) 5 ( is a limit ordinal g( ) = {g() : }). 6

(4)

1 is ZF 0

by 7.11 (24); 2 is ZF 0

by (14); 3 is by (15), (22) and 7.12; 4 can be rewritten as y((z y (z = z ) g(y) = x) so is ZF 1 by (17); 5 is ZF 1 by (22), (17) and the fact that F is ZF 1 , and using 7.12; 6 is ZF 1 by (26) and the fact that g( ) = {g() : } is equivalent toyz (y = g[ ] z = y g( ) = z ), which is ZF 1 by (18), (9) and (17).

Hence (g,,x ) is ZF 1 .Now recall from the proof of 3.14 that G can be dened by:

G(, x ) = y g((g,,x ) g() = y) (On () y = ).

This shows G is ZF 1 , and hence ZF 1 by 7.8.

Corollary 7.14 Assuming the class term G (from the beginning of subsec-tion 6) is ZF 1 , then so is the class term L : On V . (Strictly L : V V ,where L(x) = if x / On.)

Proof. By 7.13 it is sufficient to show Def is ZF 1 . Recall that Def : V V is dened by

Def (a) = {G(m,a,s ) : m , s


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Theorem 7.16 L, V=L.

Proof. Suppose a L. We must show L, (On () a L()). Nowchoose such that a L , ie. V, a L().

Let X be the set L() (ie. L ). Then X L +1 by 6.2 (2). Hence X L.Since V, a X we have L, a X . Now V, On() X =L(). But the formula x = L(y) is ZF 1 , and On() is ZF 1 , so by 7.10(since , X L),

L, On() X = L() a X.

Hence L, x(On ()x = L()a x), so L, (On ()a L()), as required.

Corollary 7.17 If ZF is consistent, so is ZF+V=L.

(Same argument as for Foundation.)Later well show ZF+ V=L AC, GCH.

8 G odel numbering and the construction of

Def 8.1(Throughout, if we say F : U 1 U n V is a ZF 1 term we meanthat the classes U 1, . . . , U n are ZF 1 (ie. dened by ZF 1 formulas) and thatF (x1, . . . , x n ) = y can be expressed by a 1 formula. By earlier sec-tions this guarantees that the extension F : V n V of F dened byF (x1, . . . , x n ) = F (x1, . . . , x n ) if x1 U 1, . . . ,xn U n and = otherwise, is ZF 1 in the sense given.)

To give numbers to formulas we rst dene F : 3 by F (n,m,l ) =

2n

3n

5l

. Then F is injective and easily seen to be ZF 1 . Write [n,m,l ] forF (n,m,l ). We now dene by induction on :

vi = v j = [0, i , j ];vi v j = [1, i , j ]; = [2, , ];

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= [3, , ];

vi = [4, i, ].(5)

Of course this denition does not take place in ZF and is not actually usedin the following denition of Def . However it should be borne in mind inorder to see whats going on.

Now dened the class term Sub : V 4 V by Sub(a,f , i ,c ) = f (c/i ) if f


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1. F (0, a) = 0

2. and n \ { 0}

F (n, a ) = H (F (1(n), (a)) , F (2(n), (a)) , F (3(n), (a)) ,a ,n ).

(Thus instead of dening F (n, a ) in terms of F (n 1, a), we are deningF (n, a ) in terms of three specied previous values.)Proof. Similar to the proof of the usual recursion theorem on .

Thus the denition of Sat in 8.1 is an application of 8.2 with 1(n) = iif for some j, k < n , [i,j,k ] = n, = 0 otherwise; and 2 and 3 are dened

similarly, picking out j and k respectively from [i , j ,k ], and with H : V 4

V dened so that

H (x,y,z,a,n ) =

{f


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Clearly one can prove in ZF that F r (x) is a nite set of natural numbers

for any set x, and we dened

(x) = max( F r (x)) .

is ZF 1 .It is easy to show that if is any formula of LST and m = , then

(m) is the largest n such that vn occurs as a free variable in , and that if f Sat (m, a ), for any a V , then dom f 1 + (m) (ie. 0, 1, . . . , (m) dom f ). This is proved by induction on and it is for this reason that wedened Sat ([3, j ,k ], a) as we did (rather than just as


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9 ZF + V=L AC

9.1We rst construct a class term H : V V such that if a, R V and R isa well-ordering of the set a, then H ( a, R ) =


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Corollary 9.2 (ZF) The formula (x, y) : = ( On x, y J ())

is a well-ordering of L. (ie. satises the axioms for a total ordering of L,and every a L has a -least element. In particular a L, { x, y a2 :(x, y)} is a well-ordering of a.)

Theorem 9.3 ZF + V=L every set can be well-ordered, so ZF + V=L AC.

Proof. Immediate from 9.2.

10 Cardinal Arithmetic10.1Recall A B means there is a bijection between A and B.

Denition 10.1 An ordinal is called a cardinal if for no < is .

Cardinals are usually denoted , , . Card denotes the class of allcardinals. Now every well-ordered set is bijective with an ordinal (using anorder-preserving bijection). (Provable in ZF.) Hence if we assume ZFC, as

we do throughout this subsection , then every set is bijective with an ordinal.Denition 10.2 (ZFC) The class term card : V On is dened so that card x is the least ordinal such that x.

Lemma 10.3 (ZFC) (1) The range of card is precisely the class of cardinals.(2) For all cardinals there is a cardinal such that > . ( + is the

least such .)(3) If X is a set of cardinals with no greatest element then sup X is a

cardinal.(4) card = for all cardinals .

Proof. (1) Exercise(2) Consider card P (though this result is provable in ZFC)(3) Let = sup X . Suppose < ( ). Choose X , > .

Then id is an injection from to . However X , so < , so by

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the Schr oder-Bernstein Theorem contradicting the fact that is a

cardinal.(4) Exercise.

(2) and (3) allow us to make the following

Denition 10.4 (ZFC) The class term : On Card is dened by (writ-ing for )

1. 0 = (ie. card N)

2. +1 = +

3. = .

5. (ZFC) If or is innite, + = . = max {, }.

6. + , . and exp are (weakly) order-preserving.

Proof. See the books.

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Denition 10.8 The Generalized Continuum Hypothesis (GCH) is the state-

ment of LST: for all innite cardinals , 2 = + (ie. On(2 = +1 )).

Denition 10.9 Suppose > 0 is an ordinal and = : < is a -sequence of cardinals (ie. is a function with domain and () = for all < ). Then we dene

1.


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Proof. Dene f :


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11 The Mostowski-Shepherdson Collapsing LemmaLemma 11.1 Suppose X is a set and M 1, M 2 are transitive sets. Suppose i : X M i are -isomorphisms (ie. x, y X (x y i (x) i (y))).Then 1 = 2 (and hence M 1 = M 2).

Proof. Dene (x) x / X 1(x) = 2(x).We prove x(x) by -induction (see 3.6).Suppose x is any set, and (y) holds for all y x. If x / X we are

done. Hence suppose x X , and 1(x) = 2(x). Then there is z such that(say) z 1(x) and z / 2(x). Since M 1 is transitive and pi1(x) M 1, wehave z M 1. Hence (since 1 is onto), y X such that 1(y) = z . Since1(y) 1(x), we have y x, and hence (by IH), z = 1(y) = 2(y) and2(y) 2(x). So z 2(x)a contradiction.

Thus (x) holds, hence result by 3.6.

Theorem 11.2 Suppose X is any set such that X, Extensionality(ie. if a, b X and a = b, then x X such that x a x / b or vice versa.) Then there is a unique transitive set M and a unique function such that is an -isomorphism from X to M .

Proof. Uniqueness is by 11.1. For existence, we prove by induction on On , that : X V M for some transitive set M .

Note that On, X V , Extensionality (since V is transitive).Now suppose , M exist for all < . Its easy to show (by 11.1) thatthey are unique and < < M M , and = M . Hence if is a limit ordinal, then take M =


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Thus (c) / +1 (b), so +1 (a) = +1 (b) and so +1 is injective.

We now show that if x X V ( X V +1 ), then (x) = +1 (x) (*)For, y (x) implies y (x) M implies y M (since M is

transitive), say (t) = y (t X V ).Then (t) (x), so t x, hence t x X .Thus +1 (x) = { (z ) : z x X } (t) = y.This shows (x) +1 (x).Conversely, suppose y +1 (x). Then y = (t) for some t x X .

Since t x X V , we have (t) (x) (since is an -isomorphism).Ie. y (x). So +1 (x) (x), and we have (*).

Now suppose a, b X V +1 , and a b (so a X V ).

Then +1 (b) = { (y) : y b X }. But a b X , so (a) +1 (b).Hence by (*) +1 (a) +1 (b).Finally, M +1 is transitive, since if a b M +1 , then b = +1 (x)

for some x X V +1 , and hence a = (y) for some y x X . Sincey X V , we have, by (*), (y) = +1 (y), so a ran +1 = M +1 , asrequired.

12 The Condensation Lemma and GCHTheorem 12.1 (The Condensation Lemma) Let be a limit ordinal and suppose X L (ie. a1, . . . , a n X , and formulas (v1, . . . , vn ) of LST,X, (a1, . . . , a n ) iff L , (a1, . . . , a n ), although we only need this

when is a 1 formula). Then there is unique and such that and : X L is an -isomorphism. Further if Y X and Y is transitive,then (y) = y for all y Y .

We prove this in stages.

Lemma 12.2 m , Lm X .

Proof. Clear for m = 0. Suppose Lm X and let a Lm +1 , so a ={a1, . . . , a n } Lm . Then L x((a1 x . . . an x) y x(y =a1 . . . y = an )). Hence X x((a1 x . . . an x) y x(y =a1. . .y = an )). Clearly such an x must be a, so a X . Hence Lm +1 X .Hence the result follows by induction.

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Lemma 12.3 X Extensionality.

Proof. For suppose a, b X and a = b. Then c, c a c / b (say), andc L since L is transitive. Thus L x(x a x / b), so X x(x a x / b), as required.

By 11.2 there is transitive M and : X M . Now since M is transitive,M On is a transitive set of ordinals so is an ordinal, , say. Then (exercisesuppose > , so 1() X . Show 1() = to getcontradiction). We show M = L .

An admission! For this proof we need the fact that most of the formu-las that we have proven ZF 1 are in fact absolute between transitive classes

satisfying much weaker axioms than ZFin fact BSbasic Set Theory (seeDevlin). BS is such that L BS for any limit ordinal > . In particu-lar, the formula On(x), and (x, y) : = On(x) y = Lx , is ZF 1 and henceabsolute between V and L and between V and M . (Since M is transi-tive.) As an application, suppose = { }. Since / M , and M ,and M On( ) (since On( ) really is 0 and M is transitive), we haveM x(On (x) yy = x {x}). Now X M , so X x(On (x) y y =x{x}), hence L x(On (x)y y = x{x}), which is a contradiction,since is a limit ordinal. Hence, we have shown:

Lemma 12.4 is a limit ordinal.Lemma 12.5 L M .

Proof. Since is a limit, L =


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Hence X (since X L ), hence M (since X M .

Let a M . Then for some c, d M ,

M On(c) (c, d) a d.

By absoluteness, c On, and hence c < , and d = Lc and a Lc. Hencea . Now x for some < , sox L . Then L {x} is transitive.

Using V=L, let be a limit ordinal such that , and L {x} L .By 12.8, with A = L and Y = L {x}, let X be such that L {x} X and X L , with card X card L {x} = card . Let : X L beas in 12.1. Then card = card L = card X card < , so < . ButL {x} is transitive so, in particular, (x) = x, so x L L , as required.

Corollary 12.10 ZF+V=L GCH. Hence if ZF is consistent, so is ZFC+GCH.Proof. By 12.9. ZF+V=L for all innite , P L + . But ZF for allinnite , card L + = + , hence ZF+V=L for all innite , card P + .So 2 + , and is obvious.

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