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Unformatted text preview: m members and B has n members, then the set A B has m n
members. (See also this exercise.)
This exercise is identical to the earlier one and the solution was given to the earlier one in the
instructor’s manual.
4. Given A ß U and any set B, prove that A B ß Solution: Using the fact that A ß function from AB to UB UB. U we choose a oneone function h from A to U. We now define a as follows: B
hf
f h
U
A
For each member f of the set A B we define f to be the function h f : B U. In other words,
f x hfx
whenever x B. To see that is oneone, suppose that f 1 and f 2 are different members of the st A B .
f 2 x . Since h is oneone we have
Choose x B such that f 1 x
h f1 x
h f2 x , 62 in other words, f 1 x f 2 x and therefore the functions f 1 and f 2 are unequal to each other.
V, prove that A B ß 5. Given that A and B are nonempty sets and that B ß Solution: Using the fact that B is nonempty and B ß
now define a function from A B into A V as follows: AV. V we choose a function h from V onto B. We B
h
f fh
V
A For each member f of the set A B we define f to be the function f h. In other words
f x fhx
whenever x V. To see that the function is oneone, suppose that f 1 and f 2 are different members of the
f 2 y . Using the fact that the function h is onto the
set A B and choose a member y of B such that f 1 y
set B, choose a member x of V such that h x y. Thus
f1 x f1 h x f1 y
f2 y f2 h x f2 x
and so the functions f 1 and f 2 are different from each other.
,Aß 6. Combine this exercise and this exercise to show that if A U and B ß 7. Use this exercise and the equivalence theorem to give a quick proof that if A
AB ß UV. V, then A B ß UV. , A ß U and B ß V, then 8. Given any sets A, B and C, prove that
AB C ßA BC . Solution: We begin by observing that a typical member of the set A B C is a function f from C to A B .
For such a function f we know that, whenever x C, the function f x is a function from B to A. Thus if
f
A B C then for every member x of the set C and every member y of the set B we have
fx y
A.
Now we take a similar look at the set A
. A typical member of this set is a function g from B
Thus if g A B C then whenever x C and y B we have
g y, x
A.
BC C into A. With these thoughts in mind we define a function from A B C to A B C as follows: Given any member f
of the set A B C , the function f is the function from B C to A defined by
f y, x f x y
whenever x C and y B.
We need to show that the function is oneone and that the range of is the entire set A B C .
To see that is oneone, suppose that f 1 and f 2 are different members of the set A B C . Choose a member
x of C such that f 1 x
f 2 x . Using the fact that f 1 x and f 2 x are different functions from B to A we
now choose a member y of B such that
f1 x y
f2 x y
and we observe that
f 1 y, x
f 2 y, x
which tells us that f 1
f2 .
Finally, to see that is onto the set A B C , suppose that g A B C . We need to find a membe...
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 Fall '08
 STAFF
 Math, Calculus

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