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Unformatted text preview: um values, respectively, of the
function f. 84 4. Prove that if f and g are bounded above on a nonempty set S then
sup f g
sup f sup g. Solution: Given any member x of the set S, since f x sup f and g x sup g, we have fg x f x g x
sup f sup g.
Thus the number sup f sup g is an upper bound of the function f g and so
sup f g
sup f sup g.
5. Give an example of two bounded functions f and g on the interval 0, 1 such that
sup f g sup f sup g.
6. Given that f is a bounded function on a nonempty set S and that c is a real number prove that
c sup f if c 0 sup cf c inf f if c 0 Solution: We give the solution here for the case c 0 and leave it to you to handle the case c 0. Suppose that c 0.
Given any member x of the set S, it follows from the fact that inf f f x that
cf x
c inf f
and we conclude that c inf f is an upper bound of the function cf. To show that c inf f is the least upper
bound of cf we need to show that no number less than c inf f can be an upper bound of cf. Suppose that
p c inf f.
Since p/c inf f we know that p/c fails to be a lower bound of f. Using this fact, we choose a member x of S
such that
p
c fx.
Since p cf x we have shown, as promised, that p fails to be an upper bound of the function cf.
7. Prove that if f is a bounded function on a nonempty set S then
sup f  supf . Solution: For every x S we have
fx f x  supf  and so
sup f
Futhermore, for every x supf . S we have fx
supf .
f x 
Using the fact that the set S is nonempty, choose a member x of the set S. We see that
supf  f x
sup f
and so
supf  sup f supf 
and we have shown that
sup f  supf . Exercises on Sequences of Sets
1. Evaluate 85 Ý 1 ,1 .
n n1 0, 1 . Now given any number x
We observe first that if n is a positive integer then 1 , 1
n
1
1
there exist positive integers n for which n x and for such n we have x
n , 1 . Therefore
Ý
n1 1 ,1
n 0, 1 0, 1 . 2. Evaluate
Ý
n1 1 1 ,5
n 2.
n Using an argument similar to the one used in Exercise 1 we can see that
Ý
n1 1 1 ,5
n 2
n 1, 5 . 3. Evaluate
Ý 1 ,2 1 .
n
n 1 n1 Using an argument similar to the one used in Exercise 1 we can see that
Ý
n1 1 1 ,2 1
n
n 1, 2 . 4. Explain why if A n is an expanding sequence of subsets of R then the sequence R A n is a contracting
sequence.
Suppose that A n is an expanding sequence of subsets of R. For each n it follows at once from the
fact that A n
A n1 that
R A n 1
R An.
5. Suppose that A n is a sequence of sets.
a. Prove that if we define
Bn Ý Aj
jn for each n then the sequence B n is a contracting sequence of sets.
Given any positive integer n, the inequality
Ý Ý Aj jn1 Aj jn
Ý
A
jn1 j follows at once from the observation that if x
then there is an integer j
which x A j and so there must certainly exist an integer j n for which x A j .
b. Prove that if we define
Bn Ý Aj
jn for each n then the sequence B n is an expanding sequence of sets.
Given any positive integer...
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 Fall '08
 STAFF
 Math, Calculus

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