1873_solutions

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

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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...
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