1873_solutions

A i for each i and then define f i to be the least

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: asses tells us that the number f i f j is irrational. 10. Prove that if ß is an equivalence relation in a set X then there exists a subset S of X such that the following two conditions hold: a. Whenever x and y belong to S and x y, we have xßy. b. For every x X there exists a unique member y of S such that x ß y. We use the axiom of choice to choose a set S that contains precisely one member of every equivalence class of the relation ß. 11. Given any real numbers x and y we define the statement x ß y to mean that there exists an integer n such that y  n 2. x a. Prove that ß is an equivalence relation in R. The fact that ß is an equivalence relation should be clear by now. There are many exercises just like this in the exercises on equivalance relations. b. Without using the axiom of choice, find a subset S of R the properties described in the above exercise. The interval 0, 2 has the desired properties. 12. This exercise introduces the concept of the product of a family of sets. Suppose that I is a given set and that to each member i of I there is associated a given nonempty set A i . The set of choice functions relative to this association is called the product of the indexed family of set A i and is written as Ü Ai. iI In other words, the set Ü i I A i is the set of all functions h with domain I that satisfy the condition h i Ai for every member i of the set I. The axiom of choice tells us that if I is a given set and that if to each member i of I there is associated a given nonempty set A i then the product Ü i I A i must be nonempty. The purpose of this exercise is to provide an even stronger statement about the size of the product Ü i I A i . This stronger statement is known as König’s inequality. a. (König’s inequality) Suppose that I is a given set and that to each member i of the set I are associated two sets A i and B i and suppose that for each i I the set A i is strictly subequivalent to the set B i . Prove that the set  i I A i is strictly subequivalent to the set Ü i I B i . Solution: We need to show that there is no function from the set  i I A i onto the set Ü i I B i . Suppose that f:  Ai Ü Bi. iI iI We recall that if g is any member of the product then g is a function whose domain is I and for each i I we have g i B i . Given any member j of the set I we define the function  j from Ü i I B i to B j by the equation j g  g j for every member g of the product Ü i I B i . We observe that whenever j I the function  j f is a function from  i I A i into B j . In fact, if x  i I A i and j I then 67 j f x  j f x  f x j Bj. For each j, since the set A j is strictly subequivalent to B j we know that the set B j j f Aj and, using the axiom of choice, we choose a choice function  whose domain is I and that satisfies the condition j f Aj j Bj for every member j of I. This function  belongs to the product Ü i I B i . However, whenever x  i I A i there must exist a member j of the set I such that x A j and for s...
View Full Document

Ask a homework question - tutors are online