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Unformatted text preview: any set then the metric space Ý S is complete. We saw the definition of this space earlier. Solution: Suppose that S is a set and that f n is a Cauchy sequence in the metric space
every member x of the set S we see from the fact that
f gÝ
f m x f n x  Ý S . For for all m and n that the sequence f n x is a Cauchy sequence in R. Since R is complete the sequence
f n x converges for every x S. For every x S we define
f x n Ý fn x
lim
and, in this way, we have defined a function f : S
Suppose that 0. R. Now we need to show that f n f Ý 0 as n Ý. 1. Using the fact that f n is a Cauchy sequence in Ý S , choose an integer N such that the inequality
fn fm Ý
holds whenever m and n are integers and m N and n N. Given any n
whenever m N and since
S we have f n x f m x 
lim
f n x f x  m Ýf n x f m x 
we have f n x fx  . Therefore if n N we have
f n f sup f n x f x 
and we have shown that f n f Ý 0 as n Ý. 178 x S N and any member x of the set Finally we need to observe that the function f belongs to Ý S , in other words, that f is bounded. Using the
fact that f n f Ý 0 as n Ý, we choose an integer n such that f n f Ý 1. We see that
f f Ý fn fn f Ý fn Ý fn Ý 1 fn Ý Ý. 3. Prove that if x n is a Cauchy sequence in a metric space X and if x X then the sequence of numbers
d x n , x converges in the metric space R.
Given any positive integers m and n we have
d x m , x d x n , x  d x m , x n
and it is therefore clear that the sequence d x n , x is a Cauchy sequence in R.
4. Give an example of a complete metric space X and a bounded sequence x n in X such that the sequence x n
has no partial limit. Hint: Look for an example in a space of the form Ý S.
We take S to be the set Z of positive integers. For each n
the equation
fn Ý R by 1 if x n fn x
From the fact that f m
ÝS. S we define the function f n : S 0 if x 1 whenever m S n n we see that f n has no partial limit in the space 5. Give an example of a complete metric space X and a contracting sequence H n of nonempty closed bounded
sets such that
Ý Hn . n1 We return to the functions f n that were defined in the solution to Exercise 4. For each n we define
Hn fm m n .
6. If X is a metric space then a function f : X X is said to be a contraction on X if there exists a number 1
such that whenever x and t belong to X we have
d f t ,f x
d t, x . a. Suppose that f is a contraction on a metric space X and that t X. Suppose that we have x 1 t and that
for every positive integer n we have
x n 1 f x n .
Prove that the sequence x n is a Cauchy sequence.
We observe first that
d x 1 , x 2
d x2, x3 d f x1 , f x2
d x3, x4 d f x2, f x3 d x 2 , x 3 2d x1, x2 and, in general,
n 1d x1, x2 .
d x n , x n 1
Now suppose that m and n are positive integers and that n m. We see that
d x m , x m 1 d x m 1 , x m 2 d x n 1 , x n
d xm, xn
m 1 d x 1 , x 2 m d x 1 , x 2 m 1 d x 1 , x 2 n 2 d x 1 , x 2
m 1d x1, x2 1 ...
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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.
 Fall '08
 STAFF
 Math, Calculus

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