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 <!is a transitive set of ordinals and therefore also an ordinal. It is called ! + !. Then we can build (! + !)+1, and so on. 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. 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’. 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 β = ↵ : ↵ = 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 β is an ordinal. β is not countable: If f : ! β were a bijection, −! 2n+13m+1 : f (n) f (m) would be a subset of ! encoding a { 2 } well–order of order type β. But then β β, which is impossible 2 for ordinals. It is easy to see that the ordinal β 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 β such that there is an injective function f : β 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 β such that there is an injective function f : β 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, + = β : β = 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 <! is the least @n {@n } cardinal bigger than for all n <!. Hence @n S = : n <! . In general, prove that if X is a set of @! {@n } cardinals, then X is a cardinal and is the least cardinal λ S such that λ for every X.] ≥ S 2 Theorem (Cantor–Bernstein–Schroder¨ Theorem) (ZF) For all sets X and Y , the following are equivalent. (1) X Y and Y X . | || | | || | (2) X = Y | | | | 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 σ be the –maximal sequence with 2 [ c ✓ domain included in Z such that σc(0)=c, σc(z + 1)=f (σc(z)) or σ (z + 1)=g(σ (z)) depending on whether σ (z) X or c c c 2 σ (z) Y , such that σ (z 1)=c¯ if c¯ X is such that c 2 c − 2 f (c¯)=σ (z) (if σ (z) Y and if there is such a c¯), and such c c 2 that σ (z 1)=c¯ if c¯ Y is such that g(c¯)=σ (z) (if c − 2 c σ (z) X and if there is such a c¯). c 2 We call σc the orbit of c. We say that σc starts in X if there is some z Z in the domain 2 of σ such that σ (z) X and such that there is no c¯ Y with c c 2 2 g(c¯)=σc(z) (so σc(z) is the first member of σc). Similarly we define ‘σc starts in Y ’. And we say that σc does not start in the remaining case (i.e., if and only if dom(σc)=Z). We also say that a set σ is an orbit if σ 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 σ is an orbit, then f σ is a bijection between σ X and σ Y if σ starts in X, • \ \ g σ is a bijection between σ Y and σ X if σ starts in • \ \ Y , and f σ is a bijection between σ X and σ Y and g σ is a • \ \ bijection between σ Y and σ X if σ 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.