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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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 Fall '08
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 Math, Calculus

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