MAGIC Set theory lecture 2 - University of East Anglia › ~bfe12ncu › MAGIC-Aspero14-set... ·...

MAGIC Set theory

lecture 2

David Aspero

University of East Anglia

18 October 2018

Axiomatic set theory: ZFC

Z is for Ernst Zermelo, F is for Abraham Fraenkel, C is for theAxiom of Choice.

The objects of set theory are sets. As in any axiomatic theory,they are not defined (they are feature–less objects; in thecontext of thetheory there is nothing to them apart from what the theory says).

ZFC expresses facts about sets expressible in the first orderlanguage of set theory. The same is true for any other first ordertheory in the language of set theory, like ZF, ZFC+“There is asupercompact cardinal”, ZFC+GCH, ZFC+V = L, ZFC+PFA, ...

Most ZFC axioms will be axioms saying that certain “classes”(built out of given sets) are actual sets (they are objects in theset–theoretic universe): Axiom 0, The Axiom of unorderedpairs, Union set Axiom, Power set Axiom, Axiom Scheme ofSeparation, Axiom Scheme of Replacement and Axiom ofInfinity will be of this kind.

Here, a class is any collection of objects, where this collectionis definable possibly with parameters. For example the class ofall sets. A proper class will be a class which is not a set.

ZFC will also have an axiom guaranteeing the existence of setswith a given property, even if these sets are not definable: TheAxiom of Choice

We will also have two “structural” axioms: Axiom ofExtensionality and Axiom of Foundation.

A classification of the ZFC axioms

1 Structural axioms: Axioms of Extensionality, Axiom ofFoundation.

2 Constructive set–existence axioms: Axiom 0, TheAxiom of unordered pairs, Union set Axiom, Power setAxiom, Axiom Scheme of Separation, Axiom Scheme ofReplacement and Axiom of Infinity.

3 Non–constructive set–existence axiom: Axiom ofChoice.

The axioms

Axiom of Extensionality: Two sets are equal if and only if theyhave the same elements:

8x8y(x = y $ 8z(z 2 x $ z 2 y))

In other words, the identity of a set is completely determined byits members:

The sets• ;• {(a, b, c, n) : a

n+b

n = c

n, a, b, c, n 2 N, a, b, c � 2, n � 3}are the same set.

Axiom 0: ; exists.

9x8y(y 2 x $ y 6= y)

(of course y 6= y abbreviates ¬(y = y)).

Strictly speaking this axiom is not needed: It follows from theother axioms.

It is convenient to postulate it at this point, though.

In the theory given by the Axiom of Extensionality together withAxiom 0 we can only prove the existence of one set:

;

Not so interesting yet.

The theory T = { Axiom 0, Axiom of Extensionality } surely isconsistent: For any set a,

({a}, ;) |= T

But ({a, b}, ;) 6|= T if a 6= b.

Axiom of unordered pairs: For any sets x , y there is a setwhose members are exactly x and y ; in other words, {x , y}exists.

8x8y9z8w(w 2 z $ (w = x _ w = y))

Of course: If x = y , then {x , y} = {x}.

[Prove this using the Axiom of Extensionality.]

Recall:Definition: (x , y) = {{x}, {x , y}}

The theory laid down so far gives us already the existence ofinfinitely many sets! :

;, {;}, {{;}}, {{{;}}}, {{{{;}}}}, {;, {;}}, {;, {;, {;}}},{{;}, {;, {;}}}, {;, {;, {;, {;}}}}, ...

With the definition of the natural numbers given in lecture 1,these sets are: 0, 1, {1} = (0, 0), {{1}} = {(0, 0)},((0, 0), (0, 0)), 2 = (0, 1), {0, 2}, {1, 2} = (0, 1), {0, {0, 2}}, ...

All sets whose existence is proved by the theory given so farhave at most two elements (!).

In fact this theory is consistent together with the sentence thatsays “Every set has at most two elements” (starting from ; andclosing under unordered pair gives rise to a model of it where“Every set has at most two elements” also holds).

This theory proves the existence of (a, b) for all a, b.

Union set Axiom: For every set x ,[

x = {y : (9w)(w 2 x ^ y 2 w)}

exists.

8x9v8y(y 2 v $ (9w)(w 2 x ^ y 2 w))S

x is the set consisting of all the members of members of x ,SSx is the set of all the members of members of members of

x , etc.

Notation: Given sets x , y ,x [ y = {a : a 2 x _ a 2 y} =

S{x , y}.

Note: Given sets x , y , x [ y exists (by the Axiom of unorderedpairs and the Union set Axiom).

With the theory given so far we can prove the existence of:

{0} [ {1, 2} = {0, 1, 2} = 3, {0, 1, 2} [ {3} = {0, 1, 2, 3} = 4,{0, 1, 2, 3} [ {4} = {0, 1, 2, 3, 4} = 5, ....

So we can prove the existence of every individual naturalnumber! Similarly, we can prove the existence of every finite setof natural numbers, every ordered pair of natural numbers,every tuple of natural numbers, every finite set of tuples ofnatural numbers, ...

However: All particular sets proved to exist by the theory givenso far are finite.

Notation: z ✓ x means: Every member of z is a memberof x .

Power set Axiom: For every x there is y whose elements areexactly those z which are a subset of x :

8x9y8z(z 2 y $ (8w)(w 2 z ! w 2 x))

Notation: For every a, P(a) = {z : z ✓ a}.

The Power set Axiom says that P(a) is a set whenever a is aset.

With the theory T laid down so far we can prove the existenceof P(n) for any particular n 2 N.

For example:• P(0) = {;} = 1• P(1) = {;, {;}} = 2• P(2) = {;, {;}, {{;}}, {;, {;}}} 6= 4• ...

T is consistent:1 Let (Xn

)n2N defined recursively by

•X0 = {;}

•X

n+1 = X

n

[ {{a, b} : a, b 2 X

n

} [ {S

a : a 2X

n

} [ {P(a) : a 2 X

n

}Then (

Sn2N X

n

,2) |= T .

Actually it would be enough to start with ; and takeX

n+1 = P(Xn

) at each stage n + 1.

Note: All particular sets proved to exist by T are still finite.

1Isn’t it?

Axiom Scheme of Separation: Given any set X and any firstorder property P,

{y 2 X : P(y)}

exists; in other words: any definable subclass of a set exists asa set.

8x8v0, . . . vn

9y8z(z 2 y $ (z 2 x ^ '(z, x , v0, . . . vn

)))

for every L–formula '(y , x , v0, . . . , vn

) such that none of x , y , z

occur as free variables in '.

In the theory laid down so far we can prove the existence, for allx , y , of

x ⇥ y = {(a, b) : a 2 x , b 2 y},

and much more.

x ⇥ y exists: Let z = x [ y , which we know exists in our theory.Note that x ⇥ y is a definable sub-collection of P(P(x [ y)).Hence x ⇥ y exists using Power set twice and Separation once.

For a formula '(v0, . . . vn

, u, v), ‘'(v0, . . . vn

, u, v) is functional ’is an abbreviation of the formula expressing

“for all u there is at most one v such that '(v0, . . . vn

, u, v)”

Axiom Scheme of Replacement: Given any set X and anydefinable (class)–function F , range(F � X ) is a set: [F � X isthe restriction of F to X , i.e. {(a, b) 2 F : a 2 X}]

“For all x , v0, . . . , vn

, if '(v0, . . . vn

, u, v) is functional, then thereis y such that for all v , v 2 y if and only if there is some u 2 x

such that '(v0, . . . vn

, u, v),”

for every formula '(v0, . . . vn

, u, v) such that none of x , y , z

occur as free variables in '.

Caution: The Axiom schemes of Separation and Replacementare not axioms but infinite sets of axioms (!)

However, it is obviously possible to write down a computerprogram which, given a sentence �, recognises whether or not� belongs to either of these schemes.

Given a set X such that a 6= ; for all a 2 X , a choice function

for X is a function f with dom(f ) = X and such that f (a) 2 a forall a 2 X .

Axiom of Choice (AC): Every set consisting of nonempty setshas a choice function.

Exercise: Write down a sentence expressing the Axiom ofChoice.

AC is needed in a lot of mathematics. For example, to provethat every vector space has a basis, that there are sets of realswhich are not Lebesgue measurable, etc.

Nevertheless, historically AC has often been seen withsuspicion:

Finite sets clearly have choice functions, but if X is infinite,where did the choice function for X come from?

Also: AC has “strange consequences”: It is possible todecompose a sphere S into finitely many pieces and rearrangethem, without changing their volumes—in fact by moving themaround and rotating them, and without running into oneanother—in such a way that we obtain two spheres with thesame volume as S! (Banach–Tarski paradox) The pieces arenot Lebesgue measurable.

The Banach–Tarski is not an actual paradox, in the sense thatRussel’s paradox is, but a counterintuitive fact.

AC has interesting equivalent formulations (modulo the rest ofZF). For example AC is equivalent to “For every two nonemptysets A, B, either |A| |B| or |B| < |A|”.

The Axiom of Foundation: If X 6= ; is a set, there is somea 2 X such that b /2 a for every b 2 X .

Inother word: Every nonempty sets has some 2–minimal element.

Modulo the other axioms (in particular AC), the following areequivalent:

• Foundation• There are no x0, x1, . . . , xn

, xn+1, . . . such that

. . . 2 x

n+1 2 x

n

2 . . . 2 x1 2 x0.

Idea behind Foundation: Sets are generated at different stages.If a set X is generated at stage ↵, then all members of X havebeen generated at some stage before ↵.

Foundation, together with Extensionality, of course, is perhapsthe most fundamental axiom in set theory (!).

As with AC, one could perhaps also complain: Where did the2–minimal element a of X come from? But wait. a was already

in X . If you remove a from X , X is no longer X !

In fact, most people like Foundation: It says that the universe isgenerated in an orderly fashion. And it provides a veryconvenient tool to use in proofs, which we will be using all thetime: Induction.

Let (Vn

)n2N be defined by recursion as follows.

•V0 = ;

•V

n+1 = P(Vn

)

The theory laid down so far, T = Ax0+ Extensionality +Unordered Pairs + Union + Power Set + Separation +Replacement + AC + Foundation, is consistent.2 In fact

([

n

V

n

,2) |= T

Still, all sets proved by T to exist are finite. In fact,

([

n

V

n

,2) |= “Every set is finite”

2Isn’t it?

Finite?

For the moment let us say:

X is finite if and only if for every a 2 X , |X \ {a}| < |X |.

X is infinite if and only if X is not finite.

The above is not the official definition of ‘finite’ but is equivalentto the official definition. But it makes things easier to deal withthe above ‘definition’ (which does not involve the notion ofordinal, which we haven’t defined yet).

Axiom of Infinity: There is an infinite set.

Definition: Given a set x , S(x) = x [ {x}

(the successor of x).

So, S(0) = 1, S(1) = 2, ... S(n) = n + 1.

The Axiom of Infinity is equivalent to:

(9x)(; 2 x ^ (8y)(y 2 x ! S(y) 2 x))

This is also phrased as: There is an inductive set.

Note: Every inductive set is infinite (in our present sense).

Proof: Try to prove it yourself. ⇤

One could also define “↵ is an ordinal” (which we’ll do soon).Then:

Definition: A natural number is an ordinal ↵ such that1 ↵ is either ; or of the form S(y) for some y 2 ↵ and2 for every x 2 ↵, x is either ; or of the form S(y) for some

y 2 ↵.The Axiom of Infinity is then equivalent to:

Axiom of Infinity’: The class of all natural numbers is a set.

In other words: There issome x such that for all y , y 2 x if and only y is a natural number.

Note that Axiom of Infinity’ is a constructive set–existenceaxiom, whereas Axiom of Infinity was not, strictly speaking.

Axiom of Infinity’ and Axiom of Infinity turn out to be equivalentmodulo the other axioms.

This, for one thing, shows that the previous classification of‘existence axioms’ into constructive and non-constructiveexistence axioms is perhaps not a very good one: Theconstructive aspect of an axiom may depend on the (remaining)background theory.

The Axiom of Infinity completes the list of ZFC axioms.

Notice the big leap when adding Infinity to the list of axioms.ZFC certainly proves the existence of infinite sets, by design!

Before adding Infinity we had a theory T which ‘surely’ wasconsistent (since (

Sn2N V

n

,2) |= T ).

Now, with the addition of Infinity, it’s not so obvious that ZFC isconsistent... .

Challenge: Construct a model of ZFC.

ZFC vs PAPA (Peano Arithmetic): A first order theory for (N,S,+, ·, 0),where S(n) = n + 1 (in the language of arithmetic, i.e., thelanguage with S, +, ·, 0):

• 8x(S(x) 6= 0)• 8x , y , (S(x) = S(y) $ x = y)

• 8x(x + 0 = x)

• 8x , y(x + S(y) = S(x + y))

• 8x(x · 0 = 0)• 8x , y(x · S(y) = x · y + x)

• 8y(('(0, y) ^ (8x('(x , y) ! '(S(x), y))) ! 8x'(x , y))

for every first order formula '(x , y) in the language ofarithmetic

(First order Induction Axiom Scheme)

First order arithmetical facts can be expressed in this language:“· is distributive with respect to +”, Fermat’s last theorem,Goldbach’s conjecture,...

PA does provemany facts about (N,S,+, ·, 0). But it does not prove everything!

Theorem (Godel, 1930’s, Incompleteness Theorem (specialcase)) If PA is consistent then there is a sentence � in thelanguage of arithmetic such that

• PA 0 � and• PA 0 ¬�

Godel’s Incompleteness theorem(s), in their generalformulation, are very profound facts that we will look back intoin a moment.

The sentence � in the Incompleteness Theorem does notexpress any fact that mathematicians would have looked intoprior to proving the incompleteness theorem. � is designed forthe purpose of the proof only.

Notation: Given a set X and n 2 N, let

[X ]n = {a ✓ X : |a| = |n|}

Consider the following statement HP:

“For all n, k , m there is some N such that for every colouring f

of [N]n into k colours there is some Y ✓ N such that Y has atleast m many members and at least min(Y ) many membersand such that all members of [Y ]n have the same colour underf .”

Here, n, k , m and N range over natural numbers.

HP can be easily expressed by a sentence, which I will call HP,in the language of arithmetic.

ZFC proves that (N,S,+, ·, 0) |= HP.

On the other hand:

Theorem (L. Harrington and J. Paris, 1977): If PA is consistent,then

PA 0 HP

ZFC vs PAConsider the theory T = (ZFC \{Infinity}) [ {¬Infinity}.

It turns out that T and PA are essentially the same theory:There are effective translation procedures

' �! �(')

between the sentences in the language of set theory and thesentences in the language of arithmetic and

�! �( )

between the sentences in the language of arithmetic and thesentences in the language of set theory such that for all ', ,

•T ` ' if and only if PA ` �(')

• PA ` if and only if T ` �( )

The Harrington–Paris theorem gives an example of a simple“natural” (purely combinatorial) statement � talking only aboutfinite sets which is true if there is an infinite set but need not betrue if there are no infinite sets (!)

Other examples have been found since then.

The consistency question

We pointed out that T = ZFC \{Infinity} was ‘surely’ consistent,based on the fact that (

Sn2N V

n

,2) |= T

(assuming, in our metatheory, that P(a) exists for every a, thatN exists, that the recursive construction of F = (V

n

)n2N is

well–defined class–function, and thatS

range(F ) exists, i.e.,assuming something like ZFC in our metatheory !)

Question: Can we prove, in T , that T is consistent? Can weprove, in ZFC, that ZFC is consistent?

The above questions do make sense: Both T and PA haveenough expressive power to make “T is consistent”, “PA isconsistent”, etc. expressible in the theory:

For example, we can code formulas, proofs, and othersyntactical notions as natural numbers and reduce a statementlike “PA is consistent” to an arithmetical statement (somespecific, but extremely complex, polynomial equation p(x) = 0in many variables does not have solutions). It then makessense to ask whether T proves that p(x) = 0 does not havesolutions.

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