MAGIC Set Theory Lecture 4

MAGIC Set theory

lecture 4

David Aspero´

University of East Anglia

1 November 2018 Last week we ﬁnished with:

Theorem Every well–order (L, ) is order–isomorphic to a unique ordinal.  Many sets can be well–ordered in different ways (so that the corresponding well–orders have different order types).

Example: ! can be well–ordered by in order type !. And it 2 can be well–ordered by putting 0 on top of every n > 0 and well-ordering ! 0 according to . This well–order has order \{ } 2 type ! + 1.

[Exercise: Characterize the sets that can be well–ordered in different ways.] We have seen the ordinals ! + 1, (! + 1)+1, ((! + 1)+1)+1, etc., aka !, ! + 1, ! + 2, etc.

The set consisting of all natural numbers and ! + n for every n

All these ordinals are countable (i.e., they are bijective with !).

Is there an inﬁnite ordinal which is not bijective with !? A cardinal is an ordinal  such that  is not bijective with any ordinal ↵<.

So, each natural number is a cardinal, ! is a cardinal, but no ! + n is a cardinal for n 1. Similarly, none of ! + !, (! + !)+1, etc., are cardinals.

When regarded as a cardinal, ! is also denoted (so @0 N = ! = 0). @ Notation: If X is bijective with a cardinal , we say that  is the cardinality of X and write X = . | | We will see that ZFC proves that every set is bijective with an ordinal.

Caveat: In a context without AC one can extend the notion of cardinals to things that are not ordinals in a perfectly meaningful way. We don’t need to do that for the moment. So, for us, at least for the moment, cardinals are ordinals.

Cardinals, in our sense, i.e., cardinals that are ordinals, are sometimes also called ‘alephs’. We will see that ZFC proves that every set is bijective with an ordinal.

Cardinals, in our sense, i.e., cardinals that are ordinals, are sometimes also called ‘alephs’. Definition: ! , also denoted , is the first uncountable cardinal 1 @1 (in other words, the first infinite cardinal not bijective with !).

Does !1 exist? Proposition

(ZF) !1 exists. Proof. Say that X ! encodes a well–order if ✓ (n, m) ! ! : 2n+13m+1 X { 2 ⇥ 2 } is a well–order. Every inﬁnite initial segment of a well–order encoded by a subset of ! can be encoded by a subset of !. Hence = ↵ : ↵ = ot( ) for some encoded by some X ! is {   ✓ } transitive and is a set since it is range(F), where F : (!) Ord is the function sending X to ot( ) if X P !  encodes and to 0 otherwise. Hence is an ordinal.  is not countable: If f : ! were a bijection, ! 2n+13m+1 : f (n) f (m) would be a subset of ! encoding a { 2 } well–order of order type . But then , which is impossible 2 for ordinals. It is easy to see that the ordinal in the above proof is in fact !1; that is,

! = ↵ : ↵ = ot( ) for some encoded by some X ! 1 {   ✓ } [Exercise] Similarly, one can prove in ZF that there is a least cardinal strictly bigger than ! . It is called ! , or . In general, we 1 2 @2 deﬁne:

Deﬁnition: Given an ordinal ↵, , also denoted ! , is the ↵–th @↵ ↵ inﬁnite cardinal.

Notation: The !↵ notation is usually used in contexts where the focus is the ordinal structure. The notation is instead @↵ used when the focus is the cardinality of this set (i.e., the property of being bijective to ). @↵ Deﬁnition: Given a cardinal , +, the successor of , is the least cardinal strictly bigger than .

Hence, ( )+ = , ( )+ = , and in general, ( )+ = . @0 @1 @1 @2 @↵ @↵+1 Proposition (ZF) For every inﬁnite cardinal  Ord, + exists. 2 This follows immediately from: Theorem (Hartogs) Given any set X there is an ordinal ↵ for which there is no injective function f : ↵ X. ! Proof (of Hartogs’ Theorem): Let ↵ be the collection of ordinals such that there is an injective function f : X. ↵ is the ! collection of of ordinals of the form ot(Y , ), where Y X and  ✓ is an order type of Y . Hence ↵ is a set by Replacement  (being the image of the function with domain (X X) sending P ⇥ W to ot(W ) if W is a well–order of some Y X, and to ✓ ; otherwise), and it is trivial to see that it is transitive. Hence ↵ is an ordinal. Finally, there is no injective f : ↵ X; otherwise ! ↵ ↵, which is impossible for ordinals. 2 ⇤ Proposition (ZF) For every inﬁnite cardinal  Ord, + exists. 2 This follows immediately from: Theorem (Hartogs) Given any set X there is an ordinal ↵ for which there is no injective function f : ↵ X. ! Proof (of Hartogs’ Theorem): Let ↵ be the collection of ordinals such that there is an injective function f : X. ↵ is the ! collection of of ordinals of the form ot(Y , ), where Y X and  ✓ is an order type of Y . Hence ↵ is a set by Replacement  (being the image of the function with domain (X X) sending P ⇥ W to ot(W ) if W is a well–order of some Y X, and to ✓ ; otherwise), and it is trivial to see that it is transitive. Hence ↵ is an ordinal. Finally, there is no injective f : ↵ X; otherwise ! ↵ ↵, which is impossible for ordinals. 2 ⇤ It is easy to see that if X =  is a cardinal, then the ordinal ↵ in the above proof is in fact +; that is,

+ = : = ot(R) for some well–order R   { ✓ ⇥ } [Exercise]

[Exercise: We have seen that exists, and so does and, in @0 @1 general, for every n. Prove that : n

We’ll see that if f : X Y and g : Y X ire injective ! ! functions, then there is a bijection h : X Y than can be ! effectively constructed from f and g.

Example: f : 2n : n ! ! and g : ! 2n : n ! { 2 } ! !{ 2 } are injective non–surjective functions, where f is the identity and g(n)=4n for all n. We’ll see that there is a bijection h : ! 2n : n ! that can be effectively constructed from !{ 2 } f and g. Proof: We only need to prove that (1) implies (2). For this, let f : X Y and g : Y X be injective functions. By replacing ! ! if necessary X and Y by, for example, X 0 and Y 1 , ⇥{ } ⇥{ } respectively, we may assume that X and Y are disjoint in the ﬁrst place.

Given c X Y , let be the –maximal sequence with 2 [ c ✓ domain included in Z such that c(0)=c, c(z + 1)=f (c(z)) or (z + 1)=g( (z)) depending on whether (z) X or c c c 2 (z) Y , such that (z 1)=c¯ if c¯ X is such that c 2 c 2 f (c¯)= (z) (if (z) Y and if there is such a c¯), and such c c 2 that (z 1)=c¯ if c¯ Y is such that g(c¯)= (z) (if c 2 c (z) X and if there is such a c¯). c 2

We call c the orbit of c. We say that c starts in X if there is some z Z in the domain 2 of such that (z) X and such that there is no c¯ Y with c c 2 2 g(c¯)=c(z) (so c(z) is the ﬁrst member of c). Similarly we deﬁne ‘c starts in Y ’. And we say that c does not start in the remaining case (i.e., if and only if dom(c)=Z). We also say that a set is an orbit if is (the range of) the orbit of some c X Y in the above sense. 2 [

Remark 1: Every two distinct orbits are disjoint and the orbits partition X Y . [ Remark 2: If is an orbit, then f is a bijection between X and Y if starts in X, • \ \ g is a bijection between Y and X if starts in • \ \ Y , and f is a bijection between X and Y and g is a • \ \ bijection between Y and X if does not start. \ \ Using these two remarks we can now deﬁne a bijection

h : X Y ! by ‘gluing together’ suitable restrictions of f and/or of the inverse of g:

Given a X: 2 If the unique orbit to which a belongs starts in X or does • not start, then h(a)=f (a). If this orbit starts in Y , let h(a) be the unique b Y such • 2 that g(b)=a. ⇤ Dual Cantor–Bernstein: For all sets X, Y , the following are equivalent: (1) X = Y | | | | (2) There is a surjection f : X Y and there is a surjection ! g : Y X. ! Proposition (ZFC) Dual Cantor–Bernstein is true. Proof. Suppose (2) holds. Using AC we find functions ¯f : Y X and ! g¯ : X Y as follows: ! For every b Y , ¯f (b) is some a X such that f (a)=b. • 2 2 For every a X, g¯(a) is some b Y such that g(b)=a. • 2 2 Then ¯f and g¯ are injective functions, so by C–B–S, X = Y . | | | | Open question: It is not known whether or not, modulo ZF, Dual Cantor–Bernstein is equivalent to the Axiom of Choice. Definition A set X is countable if and only if X = . A set is uncountable | | @0 if it is not finite or countable.

Examples of countable sets: ! !; in general, n! := s : s : n ! for any n !, • ⇥ { ! } 2 n 1.(n! is often denoted !n.)

An injective f : ! ! ! is given for example by n+1 m⇥+1 ! f (n, m)=2 3 (we used this coding in the proof that !1 exists).

The existence of a bijection between n! and ! can then be proved by induction on n since n+1! = (n!) ! . | | | ⇥ | n This gives a deﬁnable sequence (fn)1 n

n To see that there is a bijection hn :[!] ! for every n 1, n 1 ! send x [!] to fn ((x0,...,xn 1)), where (x0,...,xn 1) is the 2 strictly increasing enumeration of x.

We can also encode all (the inverses of) these bijections together into a bijection h : ! [!]

Using also the above bijections, we can well–order all polynomials with coefﬁcients in Q in length !. Once this is done, we can easily ﬁnd a bijection between a subset of ! ! ⇥ and the set of algebraic numbers (which gives what we want by C–B–S):

Given (n, k), if p(x) is the n–th polynomial with rational coefﬁcients and p(x) has at least k distinct roots, then we send (n, k) to the k–th root of p(x) in (say) the linear order union of every countable collection of countable sets is countable: If (Xn)n ! is such that each Xn is countable, then 2 n ! Xn is countable. 2 S

Proof: X is clear: There is a bijection f : ! X , @0 | n ! n| ! 0 and f : ! 2X is an injection. !S n n S X : For every n Axiom of Choice: If ZF is consistent, then there are models of ZF in which !1 is a countable union of countable sets (!) Let F : X ! ! be the function sending x to n ! n ! ⇥ (n, f (x)) if2n is ﬁrst k

! , and in fact all xs ↵ ! . • 1 1 (!) (by Cantor’s Theorem that we saw in the ﬁrst lecture); • P let’s recall the proof: Suppose f : ! (!). Let !P X = n ! : n / f (n) (!) { 2 2 }2P Then, if n ! is such that f (n)=X, then 2 n f (n)=X 2 iff n / f (n)=X 2 This is a contradiction, so there is no such n !. 2 Proposition R = (!) . In particular, R is uncountable. | | |P | Proof. Let I be the closed–open interval [0, 1) R. Since of course ✓ [0, 1) R , by C–B–S it sufﬁces to show (!) [0, 1) and | || | |P || | R (!) . | ||P |

✏n Let f : (!) [0, 1) send X ! to n ! 2n+1 , where ✏n = 0 if P ! ✓ 2 n / X and ✏n = 1 if n X (i.e., (✏n)n ! is the characteristic 2 2 2 P function of X).

Let h : Q ! be a bijection and let g : R (!) send x R ! !P 2 to h(q):q < x (this < is of course the natural order on R). { } f and g are injective functions, so by C–B–S, [0, 1) = R = (!) . | | | | |P | Remark: Even if Q = 0 < (!) = R , the rationals are | | @ |P | | | dense in the reals, i.e., between every two reals there is some (in fact, inﬁnitely many) rationals (!)

Exercise: C = (!) . | | |P | Hence, since the set of algebraic numbers is countable, most complex numbers are transcendental (i.e., non–algebraic). In fact

x C : x transcendental = C = R = (!) |{ 2 }| | | | | |P | Remark: Even if Q = 0 < (!) = R , the rationals are | | @ |P | | | dense in the reals, i.e., between every two reals there is some (in fact, inﬁnitely many) rationals (!)

Exercise: C = (!) . | | |P | Hence, since the set of algebraic numbers is countable, most complex numbers are transcendental (i.e., non–algebraic). In fact

x C : x transcendental = C = R = (!) |{ 2 }| | | | | |P | Almost disjoint families

Note: Every collection of pairwise disjoint subsets of ! has to be ﬁnite or countable.

Similarly: Suppose (!) and n Continuum Hypothesis Deﬁnition (ZFC) If  is a cardinal, () is denoted by 2. |P |

In particular, R = 2@0 . We have seen that 0 < 2@0 (Cantor’s | | @ Theorem), and therefore 2 0 by deﬁnition of as the @1  @ @1 least uncountable cardinal (we need the Axiom of Choice to conclude that there is an injection from !1 into R; without AC this is not true in general!).

Question Is = 2 0 true in ZFC? In other words, does ZFC prove that if @1 @ X R is uncountable, then X = R ? ✓ | | | | This is perhaps the most famous question in set theory.

Deﬁnition: Cantor’s Continuum Hypothesis (CH): 2 0 = . @ @1 Cantor’s Continuum Hypothesis Deﬁnition (ZFC) If  is a cardinal, () is denoted by 2. |P |

Question Is = 2 0 true in ZFC? In other words, does ZFC prove that if @1 @ X R is uncountable, then X = R ? ✓ | | | | This is perhaps the most famous question in set theory.

Deﬁnition: Cantor’s Continuum Hypothesis (CH): 2 0 = . @ @1