1873_solutions

X y z r then we can see that r 3 r z r and so it

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: n: Suppose that f is any function from S to the set S Z . Given any member x of S we know that f x is a function from Z  to S which means that f x is a sequence of members of S. For each positive integer n and each x S we define f n x to be the nth member of the sequence f x . In other words, if x S then f x is the sequence f1 x , f2 x , f3 x ,  .  We need to show that the range of f must be a proper subset of S Z and for this purpose we shall find a sequence x 1 , x 2 , x 3 ,  of members of S such that x 1 , x 2 , x 3 ,  is not in the range of f. For each positive integer n we use the fact that S fn Sn to choose a member x n of S such that xn S fn Sn Now given any member x of the set S it follows from the fact that S  Sn n Z S n . For any such n we have f n x fx x1, x2, x3,  and so the sequence x 1 , x 2 , x 3 ,  has the desired properties. that for some n we have x x n and therefore  b. Prove that if S is the gigantic set defined earlier then the sets S and S Z are not equivalent. 18. Suppose that to each member i of a given set I there is associated a set S i that is strictly subequivalent to a given set U. Prove that  i I S i is strictly subequivalent to the set U I . Solution: Suppose that f is a function from  i I S i to U I and for each i fi : i I Si I we define a function U by the equation fi x  f x i for every x  i I S i . For each i we use the fact that f i S i U to choose a member that we shall call g i of the set U such that g i U f S i . In this way we have defined a member g of U I . To see that g does not lie in the range of f we observe that if x  i I S i then for some i we have x S i and for any such i we have fx i gi. Exercises on The Axiom of Choice 65 1. One of the assertions of an earlier theorem was that if A and B are given sets and if there exists a function g from B onto A then there must exist a one-one function from A into B. Rewrite the proof of this part of the theorem and show how and where the axiom of choice is used. Solution: We assume that g is a function from B onto A. For every member x of the set A we define Ex  y B g y  x . Since the function g is onto A we know that all of the sets E x are nonempty. The axiom of choice therefore guarantees the existence of a function f defined on A such that f x E x for every x A. We observe that for every member x of the set A we have g f x  x. To see that the function f is one-one, suppose that x 1 and x 2 belong to A and that f x 1  f x 2 . We see that x1  g f x1  g f x2  x2. 2. Suppose that I is a given set and that to each member i I there is associated a nonempty set A i of natural numbers. Explain why the axiom of choice does not have to be used to produce a choice function relative to this association. Solution: We can provide a specific definition of a choice function in this example by defining f i be the least member of the set A i for each i to I. 3. Suppose that to each member i of a given set I there is associated a nonempty finite set A...
View Full Document

This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

Ask a homework question - tutors are online