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Unformatted text preview: on: We want to show that the set
Ý Hn
n1 cannot include any nonempty open set. Suppose that U is a nonemty open subset of X and, using the fact
that H 1 has no interior points we choose a point y 1 U H 1 and using the fact that the set U H 1 is a
neighborhood of y 1 we choose a positive number 1 1 such that
B y1, 1
U H1.
Using the fact that the open set B y 1 , 1 cannot be included in the set H 2 (because H 2 has no interior
points) we choose a point
y2 B y1, 1
H2 B y1, 1
X H1
1
and, using the fact that the latter set is a neighborhood of y 2 we choose a positive number 2
such
2
that
B y2, 2
B y1, 1
H2.
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
H n 1 .
We observe that
n B yn, n U Hj
j1 In order to show that the set
Ý U Hn
n1 is nonempty we shall show that the sequence y n is a Cauchy sequence and then we shall observe that the
limit of the sequence y n must belong to the set
Ý U Hn.
n1 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 181 For each positive integer N, since the sequence y n is eventually in the closed ball B y N , N we have
N y B yN, N U Hj.
j1 Thus
Ý y Hn U n1 and the proof is complete. Exercises on Compact Metric Spaces
1. Prove the following converse of the Cantor intersection theorem: If X is a metric space and if every
contracting sequence of nonempty closed subsets of X has a nonempty intersection then X is compact. Solution: To show that every sequence in X has a partial limit, suppose that x n is a sequence in X. For each positive integer n we define
Using the fact that the sequence H n
point x such that Hn xm m n .
is a contracting sequence of nonempty closed subsets of X, choose a
Ý Hn. x n1 It is easy to see that x is a partial limit of the sequence x n . Write out the details.
2. Given a metric space X, prove that the following conditions are equivalent:
a. The space X is compact.
b. For every family
covers X. of open balls that covers the space X there exists a finite subfamily of that still It is obvious that condition a implies condition b because every open ball is an open set. Now
suppose that condition b holds. Suppose that is a family of open sets that covers X. We define
to be the family of open balls B that are included in at least one member of the family . Given
any point x X, since covers X we know that x belongs to at least one member U of the family
and, since such a set U is open, there exists an open ball B such that x B U. Therefore the
family
covers the space X and, using condition b, we choose finitely many members, that we
shall call...
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 Fall '08
 STAFF
 Math, Calculus

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