This preview shows page 1. Sign up to view the full content.
Unformatted text preview: r f of the set
A B C such that g f . Given any member x of the set C we define f x to be the function from B to A 63 whose value at each member y of the set B is g y, x . The function f from C to A B that we have defined
clearly satisfies the condition g f .
Z 9. Show that R ß 0, 1 and then use this exercise to show that R Z ß R. Solution: We already know that
R ß p Z ß 0, 1 Z . Therefore
RZ ß
10. Prove that R Z 0, 1 Z ß 0, 1 Z Z Z ß 0, 1 ß R, R ß R. Solution: We know that R ß
whenever x, y
theorem that R R R. On the other hand, if we define f x, y x, y, 0, 0, 0,
R R then we can see that R R ß R Z ß R and so it follows from the equivalence
R ß R. 11. Prove that R 3 ß 0, 1 .
We know that R ß R 3 . On the other hand, if we define
f x, y, z x, y, z, 0, 0,
3
whenever x, y, z
R then we can see that R 3 ß R Z ß R and so it follows from the equivalence
theorem that R 3 ß R.
12. Prove that p R Z ß pR R ßpR. Solution: On the one hand we know that p R ß Z pR and on the other other hand we know that
pR Z ß R pR ß 0, 1 R R ß 0, 1 RR ß 0, 1 R ßpR and so the result follows from the equivalence theorem.
13. Prove that R R ß p R .
On the one hand, p R ß 0, 1
RR ß 0, 1 R ß Z R R and on the other hand
R ß 0, 1 Z R ß 0, 1 RR ß 0, 1 R ßpR and so the result follows from the equivalence theorem.
14. Prove that if A ß R and A has more than one member then A R ß p R .
Since A has more than one member we have
p R ß 0, 1 R ß A R .
On the other hand,
AR ß RR ß p R
and so the result follows from the equivalence theorem.
15. Given that S is a set with more than one member and that S S ß S, prove that S S ß p S
shall show in a later theorem that the condition S S ß S is satisfied by every infinite set.
Since S has more than one member we have
p S ß 0, 1 S ß S S ß p S S .
On the other hand,
pS S ß 0, 1 S S ß 0, 1 SS ß 0, 1 S S ß p S . We ßpS and so the result follows from the equivalence theorem.
16. Given that S S ß S, that A ß S and that A has more than one member, prove that A S ß p S .
Since A has more than one member we have 64 p S ß 0, 1 S ß AS. On the other hand,
AS ß SS ß p S
by Exercise 15. So the result follows from the equivalence theorem.
17. The preceding exercises show that when A ß B and the set A has more than one member and the set B
satisfies the condition B B ß B, then we have A B ß p B . Now what happens when B is the smaller set? In
other words, what happens when B is strictly subequivalent to A? In this case, Exercises this one, this one,
this one and this one seem to suggest that we should have A B ß A. In this exercise we see that this
statement is false even if A is uncountable and B is countable:
a. Suppose that S n is a sequence of sets, that
S Sn n Z and that, for each n, the set S n is strictly subequivalent to S. Prove that the sets S and S Z are not
equivalent to each other. Show, in fact, that there is no function from S onto S Z .
Solutio...
View Full
Document
 Fall '08
 STAFF
 Math, Calculus

Click to edit the document details