# In order to show this we will suppose that every

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Chapter 2 / Exercise 61
Calculus
Stewart Expert Verified
In order to show this, we will suppose that every element x A has an immediate successor; this assumption will lead to a contradiction. If every element of A has an immediate successor, then we can define a function f : A A such that for each x A, f (x ) is an immediate successor of x . Indeed, let T x be the set of all the immediate successors of x ; by the Axiom of Choice, there exists a choice function g such that g(T x ) T x . We define f by letting f(x ) = g(T x ); clearly, f(x ) is an immediate successor of x . 5.5 Definition A subset B A is called a p-sequence if the following conditions are satisfied. α ) p B , β ) if x B , then f(x ) B , γ ) if C is a chain of B , then sup C B . There are p -sequences; for example, A is a p -sequence. 5.6 Lemma Any intersection of p -sequences is a p -sequence. The proof is left as an exercise for the reader.
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Chapter 2 / Exercise 61
Calculus
Stewart Expert Verified
Let P be the intersection of all the p -sequences. (Note that P ≠ Ø because p P ). By 5.6 , P is a p -sequence. 5.7 Definition An element x P is called select if it is comparable with every element y P . 5.8 Lemma Suppose x is select, y P , and y < x . Then f(y ) x . Proof . y P , P is a p -sequence, hence by (β), f (y ) P . Now, x is select, so either f(y ) x or x < f(y ). By hypothesis y < x ;so if x < f(y ),we have y < x < f(y ), which contradicts the assertion that f(y ) is the immediate successor of y . Hence f(y ) x . 5.9 Lemma Suppose x is select. Let Then B x is a p -sequence. Proof. We will show that B x satisfies the three conditions which define a p - sequence. α ) Since p is the least element of A , p x , hence p B x . β ) Suppose y B x ; then y x or y f(x ). Consider three cases: 1) y < x . Then f(y ) x by 5.8 , hence f(y ) B x . 2) y = x . Then f(y ) = f(x ), thus f(y ) ≥ f(x ); hence f(y ) B x . 3) y f(x ). But f(y)>y ,so f(y)>f(x ); hence f(y ) B x . In each case we conclude that f(y ) B x . γ ) If C is a chain of B x , let m = sup C . For each y B x ,y x or y f(x ). If y C y f(x ), then (since m y) m f(x ),so m B x . Otherwise, y C, y x ; thus x is an upper bound of C ,so m x . Thus again m B x . 5.10 Corollary If x is select, then y P, y x or y f(x ).
Proof . B x is a p -sequence; P is the intersection of all p -sequences; hence P B x . But B x P by definition, hence P = B x . So y P, y x or y f(x ). 5.11 Lemma The set of all select elements is a p -sequence. Proof α) p is select because it is less than (hence comparable to) each y P . β ) Suppose x is select, by 5.10 . y P , either y x (in which case y f(x ) because x < f(x) ) or y f(x ). Thus f(x ) is select. γ ) Let C be a chain of select elements and let m = sup C ; let y P . If x C y x , then y m (because x m ). Otherwise, x C, x y , hence y is an upper bound of C ,so m y . Thus m is select. 5.12 Corollary P is fully ordered. Proof. The set S of all the select elements is a p -sequence; P is the intersection of all the p -sequences; hence P S . But S P (by definition a select element is in P ), so P = S . Thus each element of P is select, that is, is comparable to each element of P .
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