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Unformatted text preview: 2 n
m 1d x1, x2
.
1 179 m To show that x n is a Cauchy sequence, suppose that 0 and choose an integer N such that
N 1d x1, x2
.
1
Then whenever m N and n N, regardless of whether m n or m n or m n, we have
d xm, xn .
b. Suppose that f is a contraction on a complete metric space X. Prove that there exists a point x X such
that f x x.
Choose a point t X. We define the sequence x n as in part a. Since X is complete and x n is
a cauchy sequence in X we know that x n is convergent. We define
x n Ý xn.
lim
In order to show that f x x
c. Suppose that f is a contraction on a complete metric space X. Prove that there is one and only one point
x X such that f x x.
7. Prove that if X is a nonempty complete metric space and if every point of X is a limit point of X then X is
uncountable. Solution: Suppose that X is a complete metric space and that every point of X is a limit point of X. To
obtain a contradiction, suppose that x n is a sequence in X and that X is the range of the sequence x n .
1. Since x 1 is a limit point of X the set X
x 1 must be nonempty. Choose a point y 1 X
x 1 and, using
the fact that X
x 1 is a neighborhood of y 1 , choose a positive number 1 1 such that the closed ball
x 1 , in other words,
B y 1 , 1 is included in X
B y1, 1
X
x1
Since y 1 is a limit point of X and B y 1 , 1 is a neighborhood of y 1 , the ball B y 1 , 1 must contain
x 2 must be nonempty. Choose a point
infinitely many points and so the set B y 1 , 1
y2 B y1, 1
x2
1
and, using the fact that B y 1 , 1
x 2 is a neighborhood of y 2 , choose a positive number 2
such
2
that
B y2, 2
B y1, 1
x2 .
Continuing this process we obtain a sequence y n in the space X and a sequence n of positive numbers
such that, for each n,
1
n
n
and
B y n 1 , n 1
B yn, n
x n 1 .
We observe that
B yn, n
X
x 1 , x 2 , x 3 , , x n
for each n.
We shall obtain our contradiction by showing that the sequence y n is a Cauchy sequence and then we
shall observe that the limit of the sequence y n cannot belong to the range of the sequence x n .
To show that y n is a Cauchy sequence, suppose that 0. Choose an integer N 2/ . Whenever m and
n are integers and m N and n N we know that both y m and y n belong to the ball B y N , N and so
d ym, yN d yN, yn 1 1 .
d ym, yn
N
N
Thus y n is a Cauchy sequence and, since the space X is complete, the sequence y n is convergent. We
define
y n Ý yn.
lim
For each positive integer N, since the sequence y n is eventually in the closed ball B y N , N we have 180 x 1 , x 2 , x 3 , , x N
y B yN, N
and so y cannot belong to the range of the sequence x n and we have reached the desired contradiction.
8. This exercise is an improvement on Exercise 7 and is known as Baire’s theorem.
1. Suppose that H n is a sequence of closed subsets of a complete metric space X and that none of the sets H n
have any interior points. prove that the set
Ý Hn
n1 also fails to have any interior points. Soluti...
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 Fall '08
 STAFF
 Math, Calculus

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