Exercise 8.1.13

(Principle of Dependent Choices) If R is a binary relation on M such that for each x M there is a y M for which 𝑥𝑅𝑦 , then there is a sequence x n n ω such that x n R x n + 1 holds for all n ω .

Answers

Proof. Suppose such a relation R on M . Define the set X x = { y M 𝑥𝑅𝑦 } for each x M . Then, by the given property of R , clearly X x for any x M . Then, by the Axiom of Choice, the system of sets { X x x M } has a choice function g . We also have that there is an a M since M . We then define a sequence recursively as follows:

x 0 = a x n + 1 = g ( X x n ) ,

noting that this is well defined since each X n is nonempty.

To show that the sequence x n n ω has the desired property, consider any n ω . Then by the recursive definition we have that x n + 1 = g ( X x n ) so that x n + 1 X x n since g is a choice function and X x n is nonempty. Then, from the definition of X x n , it follows that x n R x n + 1 . This shows the desired result since n was arbitrary. □

User profile picture
2024-07-15 11:42
Comments