Appendix to Basic cardinal arithmetic

Transcription

1 Appendix to Basic cardinal arithmetic Notations 1.1. (a) For sets a and b define a+ set b:={0} a {1} b. (b) Let a (set b) :={f f : b a}. (c) Let α,β On. Then α+ ord β, α ord β and α (ord β) denote respectively their ordinal sum, their ordinal product and their ordinal exponentiaton. Note that if a b =, then a+ set b a b. The following theorem requires only ZF. The axiom of choice is not needed. Theorem 1.2. (Assume ZF) (a) ω + set ω ω ω ω. (b) For every a, 2 (set a) P(a). (c) For every a,b,c, (a b) (set c) a (set c) b (set c). (d) For every a,b,c, a (set b+setc) a (set b) a (set c). (e) For every a,b,c, (a (set b) ) (set c) a (set b c). (e) For every m,n ω, (i) m+ ord n m+ set n. (ii) m ord n m n. (iii) m (ord n) m (set n). Proof The proof is simple and is left to the students. The next proposition too, requires only ZF, and the Axiom of Choice is not needed. Proposition 1.3. (a) For every a the following conditions are equivalent. (1) ω < a; (2) a+ set 1 a; (3) for every n ω, a+ set n a; (4) a+ set ω a. (b) For every a the following conditions are equivalent. (1) a+ set a a; (2) a 2 a; (3) for every n ω \{0}, a n a; (4) a ω a. 1

2 Proof The proof is simple and is left to the students. Relying on AC, we wish to prove the following theorem. Theorem 1.4. (Assume ZFC) Let a be an infinite set and b < a. Then (a) a+ set b a. (b) If b, then a b a. Theorem 1.4 does not follow from ZF. In fact, there is a model of ZF and in this model an infinite set a such that a a + set 1. Since a, it follows that 1 < a. So in this model, Part (a) of Theorem 1.4 does not hold. The fact a a+ set 1 implies a a 2. So Part (b) is also refuted. If AC is not assumed, then still, the claims of Theorem 1.4 hold for sets a,b which can be well ordered. So at first we shall prove in ZF the version of Theorem 1.4 for well orderable sets. Since AC implies that every set is well orderable, Theorem 1.4 follows from ZFC. Theorem 1.5. (Assume ZF) Let α be an infinite ordinal. Then Proof α+ set α α+ ord α α. Every β On has a unique representation in the form β = β lim + ord β nat, where β lim is a limit ordinal or 0 and β nat ω. Let α be an infinite ordinal. We define f : α α + set α. Let β α. If β nat is an even number, then f(β) = 0,β lim ord βnat +. If β nat is an odd 2 number, then f(β) = 1,β lim + ord βnat 1. 2 It is easy to see that if α is a limit ordinal, then f is a bijection between α and α+ set α. (A bijection is a 1 1 onto function.) Assume now that α is a successor. Then α α lim α lim + set α lim α+ set α. Now we prove that α α+ ord α. Suppose that α is a limit ordinal. Define g : α α + ord α as follows: Let β α. If β nat is even define 2

3 g(β) = β lim ord βnat +. If 2 βnat is odd define g(β) = α+ ord β lim + ord βnat 1. It 2 is easy to see that g is a bijection between α and α+ ord α. Supposet now that α is a successor ordinal. Then α+ ord α = α lim + ord α lim + ord α nat (α lim + ord α lim )+ set α nat α lim + set α nat α lim + ord α nat = α. We prove an analogous result for multiplication. Theorem 1.6. (Assume ZF) Let α be an infinite ordinal. Then α α α ord α α. Proof Let us recall some facts from preceding exercises. Let a,< a and b,< b be linearly ordered sets. We define their product. It is linearly ordered set too, and it is denoted by a,< a lin b,< b. a,< a lin b,< b := a b,< [<a,< b], where < := < [<a,< b] is defined as follows: ( ) a 1,b 1 < a 2,b 2 if b 1 < b b 2 or (b 1 = b 2 and a 1 < a a 2 ). It was mentioned in an earlier exercise that for every α,β On, α, α lin β, β = α ord β, α ordβ. So α β α ord β, and in particular, α α α ord α. Next we show that α α α. We define a relation on On On. Let 3

4 α 1,β 1, α 2,β 2 On On. Then α 1,β 1 α 2,β 2 if ( ) (i) max(α 1,β 1 ) < max(α 2,β 2 ) or ( ) (ii) max(α 1,β 1 ) = max(α 1,β 1 ) and min(α 1,β 1 ) < min(α 2,β 2 ) or ) (iii) (max(α 1,β 1 )=max(α 1,β 1 ) and min(α 1,β 1 )=min(α 2,β 2 ) and α 1 <α 2 We state the following fact. On On, is a well ordered class structure. To check this is completely straight-forward, but let us check this anyway. Claim 1 On On, is a well ordered class structure. Proof Clearly is irreflexive and antisymmetric. Note the following fact. Let α,β On and denote max(α,β) by γ. Then { δ,ε δ,ε α,β } { δ,ε δ,ε γ,γ } = (γ+1) (γ+1)\{ γ,γ }. So is left-narrow. Next we check that is connected on On On. Let α,β, γ,δ On On be distinct. Suppose first that {α,β} {γ,δ} =. Then max(α,β) max(γ,δ). So α,β and γ,δ are -comparable. Next assume that {α,β} {γ,δ} has cardinality 2. Then {α,β} = {γ,δ}, α β and the second pair is equal to β,α. So the maxima and the minima of the two sets {α,β} and {γ,δ} are equal, and the first elements of α,β and γ,δ are unequal. So again, α,β and γ,δ are -comparable. Finally, we address the case that {α,β} and {γ,δ} have exactly one elementincommon. Callthiscommonelementε. Ifε = max(α,β) = max(γ,δ), then min(α,β) min(γ,δ). If, vice versa, ε = min(α,β) = min(γ,δ), then max(α,β) max(γ,δ). In both cases α,β and γ,δ are -comparable. Lastly assume without loss of generality that ε = max(α,β) and ε = min(γ,δ). Then α,β γ,δ. 4

5 We have shown that in all the different cases α,β and γ,δ are -comparable. So is connected on On On. We check that is transitive. Suppose that α,β γ,δ ε,ι. If among the maxima of the three sets {α,β}, {γ,δ} and {ε,ι} there are two which are different, then it may not happen that max(α,β) = max(ε,ι). Neithermayithappenthatmax(α,β) > max(ε,ι). Somax(α,β) < max(ε,ι) and hance α,β ε,ι. Assume now that the three maxima are equal. If two of the three minima are unequal, then an argument similar to the above shows that α,β ε,ι. It cannot happen that the three maxima are equal and the three minima are equal. Because this implies that {α,β} = {γ,δ} = {ε,ι}. And in that case either (i) α,β = γ,δ or (ii) γ,δ = ε,ι or (iii) α,β = ε,ι. (i) and (ii) cannot happen since is irreflexive, and (iii) cannot happen because is antisymmetric. We have shown that in the two possible cases that we dealt with α,β ε,ι. So is transitive. It remains to show that if a On On, then a has a -minimal element. Let γ = min({max(α,β) α,β a}) and b = { α,β a max(α,β) = γ}. Let δ = min({min(α,β) α,β b}) and c = { α,β b min(α,β) = δ}. Let ε = min({α there is β such that α,β c}) and d = { α,β c α = ε}. Then c has one or two elements, d is the singleton x = { γ,δ } or d is the the singleton x = { δ,γ }, and x is the -minimum of a. If r is a relation and a is a set, then r a denotes r (a a), namely, the two sided restriction of r to a. We prove the following claim by induction on cardinals. 5

6 Claim 2 If µ is an infinite cardinal, then µ µ, (µ µ) = µ,< µ. Let µ be an infinite cardinal and suppose that for every infinite cardinal λ < µ, λ λ, (λ λ) = λ,< λ. We abbreviate (λ λ) by. Suppose first that µ = ω. The set { n,n n ω} is unbounded in ω,, and for every n ω, the set { x,y x,y n,n } is equal to the set (n+1) (n+1)\{ n,n }, so it is finite. So every initial segment of ω ω, is finite, and ω ω is infinite. This implies that ω ω, = ω, ω. We prove this implication. The structures m:= ω ω, and n:= ω, ω are well ordered sets. So there are 3 possibilities: (i) m is isomorphic to a proper initial segment of n; (ii) n is isomorphic to a proper initial segment of m; (iii) m = n. Possibility (i) does not happen because every initial segment of n is finite, whereas the universe of m is infinite. Possibility (ii) does not happen because every initial segment of m is finite, but the universe of n is infinite. So possibility (iii) happens, and this is what we wished to prove. Now assume that µ > ω. Denote On On by Ond, and for every x Ond, let Ond x denote the set {y Ond y x}. We have not assumed AC. So there may be sets which are not well orderable. Hence a cannot be defined for all sets. Let WO denote the class of well orderable sets. For a WO we define a to be the first ordinal α such that there is a relation r on a such that a,r = α, α. The set D µ :={ α,α ω α < µ} is unbounded in µ µ,. Let α D µ and set λ = α. Then λ is infinite and λ < µ. Obviously, α α λ λ. By the induction hypothesis, λ λ, = λ, λ. In particular, λ λ λ. So α α λ. The following fact has been noted earlier in the proof. Ond α α = (α+1) (α+1)\{ α,α }. 6

7 Recall that α is an infinite ordinal. So ω < α. Hence α+1 α. Hence α α α (α+1) (α+1)\{ α,α } (α+1) (α+1) α α α. By Cantor Bernstein Theorem, λ α (α+1) (α+1)\{ α,α } = Ond α α. That is, Ond α α λ. It follows that for every α D µ, Ond α α < µ. And since D µ is unbounded in µ µ,, we also have that for every α < µ, Ond α α < µ. That is, every proper intial segment of µ µ, has cardinality less than µ. We have two well ordered sets: µ µ, and µ, µ. Either they are isomorphic, or one of them is isomorphic to a proper initial segment of the other. Clearly, µ, µ is not isomorphic to a proper initial segment of µ µ,. This is so, since proper initial segments of µ µ, have cardinality < µ, whereas the cardinality of µ is µ, and it is not less than µ. Now, the other direction. Since µ is a cardinal every proper intitial segment of µ, µ has cardinality less than µ. But the cardinality of µ µ is not less than µ. So µ µ, cannot be isomorphic to a proper intitial segment of µ, µ. The remaining possibility is µ µ, = µ, µ. We have proved Claim 1. Claim 1 implies in particular that for every infinite cardinal µ, (1) µ µ µ. Let α be an infinite ordinal. Then α is an infinite cardinal. So by (1), α α α α α α. 7

8 Corollary 1.7. (Assume ZF) Let a,b be well orderable sets. Suppose that a is infinite and that b < a. Then (a) a+ set b a. (b) If b, then a b a. Proof (a) By Theorem 1.5, a < a+ set b < a+ set a a. So by Cantor Bernstein Theorem, a+ set b < a. (b) By Theorem 1.6, a < a b < a a a. So by Cantor Bernstein Theorem, a b < a. Proof of Theorem 1.4 Theorem 1.4 follows from Theorem 1.6 and the fact that under the assumption of ZFC, every set is well orderable. Exercise 1.8. (Assume ZF) Let α,β On. Suppose that α 2 and α or β are infinite. Then α (ord β) = max( α, β ). 8