Set theory - Chapter 6: The Axiom of Regularity

6.1

6.2

By induction on n < ω , we show that V n is finite (using 3.1 (iii)). Hence V ω n < ω V n = 0 sup n < ω V n 0 2 = 0 . (Note that we can apply Lemma 5.8, since only the first term is equal to zero). Moreover, V ω 0 since ω V ω . Finally V ω = 0 .

Now we shall show by induction on α that V ω + α = α . We see below that it is true for α = 0 . If it is the case for an ordinal α , then V ω + α + 1 = ( V ω + α ) = 2 α = α + 1 . Finally, consider a limit ordinal λ such that the property holds below it. Then V ω + λ = α < λ V ω + α α < λ V ω + α = λ sup α < λ α = λ λ . But Beth is a normal function, so λ λ and V ω + λ = λ .

6.3

For all κ we have κ V κ so κ V κ .

Suppose κ is inaccessible. Then by induction on α < κ , we have α < κ . For α = 0 , it is true because κ is uncountable. If the property holds for an α then it also holds for α + 1 since κ is strong limit. The regularity of κ allows to prove the property at limit step. Finally we have :

V κ α < κ V ω + α = κ sup n < κ α κ 2 = κ

6.4

Let x , y be two sets of rank less than α . Then by 6.1, rank ( { x ; y } ) = sup { rank ( x ) + 1 ; rank ( y ) + 1 } α + 1 . rank < x , y > = rank ( { { x } ; { x ; y } } ) . But we've just seen that the rank { x } and { x ; y } is α + 1 . So rank < x , y > α + 2 . If t x y then rank ( t ) < inf { rank ( x ) ; rank ( y ) } α , so rank ( t ) + 1 α and rank ( { x ; y } ) α . If t x then there is z x such that t x . So rank ( t ) < rank ( z ) , rank ( t ) + 1 rank ( z ) < rank ( x ) α . So rank ( x ) < α . If z ( x ) then for all t z , t x , rank ( t ) < rank ( x ) , rank ( t ) + 1 α . Hence rank ( z ) α . Finally  rank ( ( z ) ) α + 1 . x y y × x = < a , b > a y a x . For < a , b > x y , rank ( a ) + 1 α and rank ( b ) + 1 α so rank ( < a , b > ) = rank ( { { a } ; { a ; b } } ) sup { rank ( a ) + 2 ; rank ( b ) + 2 } α + 1 . Finally, rank ( x y ) α + 2 .

6.5

We can write = 2 R 1 , = × * R 2 and = I where R 1 , R 2 are relations of equivalence on 2 and × * and I the ideal u ( u n n + 0 ) . Hence we have :

6.6

Let A be a class and B = x A x B the class of sets that are hereditarily in the class A .

  1. ⇒ Let x B y TC ( x ) y n S n ( x ) .

  2. xxx
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