Iain T. Adamson

Answers to Chapter 7

76. Let R be a well-ordering on a class X. Let a and b be elements of X. Then the subset {a,b} has an R-least member. If the R-least member is a then (a,b) ∈ R; if the R-least member is b then (b,a) ∈ R.

Answers to Chapter 5

53. (1) Let (a,b) ∈ R∪R-1. If (a,b) ∈ R then (b,a) ∈ R-1 and so (b,a) ∈ R∪R-1. Similarly if (a,b) ∈ R-1 then (b,a) ∈ R∪R-1. So R ∪R-1 is symmetric. Of course R ∪R-1 ⊇ R.

Answers to Chapter 4

40. (1) Let x ∈ ∪k∈KX f (k). Then there is an index k1 in K such that x ∈ X f (k1). Sincefk1) ∈ I it follows that x ∈ ∪i∈IX. So ∪k∈KX f (k) + ∪i∈IX i .

Answers to Chapter 3

23. Let R be functional and let Y be any class. Let b be any element of R→(R←(Y)). Then there is an element a of R←(Y) such that (a,b) ∈ R. Since a ∈ R←(Y) there is an element y of Y such that (a,y) ∈ R. Since R is functional and both (a,b) and (a,y) belong to R it follows that b = y. So b ∈ Y.

Answers to Chapter 13

135. Let f and g be bijections from a to A and b to B respectively. Define the mapping h from (a × 0) ∪ (b × 1) to A ∪ B by setting n n

Answers to Chapter 9

hleft( x right) = left{ {begin{array}{*{20}c}{fleft( a right) if x = left( {a,0} right)} {gleft( b right) if x = left( {b,1} right).} end{array} } right.

Answers to Chapter 1

n nThen h is a bijection and hence it follows that a+b = Card ((a × 0)∪ (b × 1)) = Card (A ∪ B).

Answers to Chapter 6

121. As suggested, let x = n: (n ∈ ω)∧( every subset of n is finite). Then O ∈ x since the only subset of ∈ is ∈ itself which is finite, being equipotent to the natural number O.