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Unformatted text preview: ve that the point z 1 t x ty belongs to the set B x,
S
x . Therefore x is a limit
point of S.
19. For the purpose of this exercise we shall call a subset S of a metric space X compressed if for every number
0 there exist two different points x and y in S such that d x, y . 131 a. Prove that whenever a subset S of a metric space has a limit point, it must be compressed.
This is just Exercise 17 again.
b. Give an example of a compressed subset of the metric space R that has no limit point.
The set
S ZÞ n 1
n Z
n
is compressed but, since no interval of length 1 can contain more then three members of S , the
set S has no limit point.
c. Give an example of a compressed subset of the metric space 0, 1 that has no limit point in 0, 1 .
The set 1
n Z has the desired properties.
n
d. Prove that if a metric space X is compressed and x X then the set X
x is also compressed.
To obtain a contradiction, suppose that the set X
x fails to be compressed and choose
0 such that whenever a and b are different points in the set X
x we have d a, b
.
Since X is compressed and since the inequality d a, b never holds when a and b are
2
different points in the set X
x there must exist a point a X
x such that d a, x .
2
Choose such a point a.
Again, using the fact that X is compressed, choose a point b X
x such that
d b, x d a, x . We see at once that b a. Moreover
d a, b
d a, x d x, b
2
2
contradicting the choice of .
e. Prove that if a metric space X is compressed and F is a finite subset of X then the set X F is
compressed.
The set X F can be obtained by removing the points of F from X one at a time and applying
part d each time.
f. Prove that a metric space X is totally bounded if and only if every infinite subset of X is compressed.
The desired result follows from an earlier stated property of total boundedness.
20. Given that S is a subset of a metric space X and that every infinite subset of S has a limit point, prove that
every infinite subset of the set S must have a limit point.
Suppose that E is an infinite subset of S. Using the fact that E is infinite we choose a sequence x n
of points of E such that whenever m n we have x m x n .
For each positive integer n we choose a point y n S such that d x n , y n 1 .
n
In the event that the set y n n Z is finite there must be a point y S such that y n y for
infinitely many values of n; and we choose such a point y.
In the event that the set y n n Z is infinite, it must have a limit point and we choose such a
limit point and call it y.
In either event we have found a point y X such that whenever 0 there are infinitely many
values of n for which d y, y n . We complete the proof by showing that this point y must be a limit
point of E. Suppose that 0.
There are infinitely many integers n 2 for which d y, y n . For each of these integers n we
2
have
d xn, yn d yn, y 1
d xn, y
n
2
and since the sequence x n is oneone we conclude that the...
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 Fall '08
 STAFF
 Math, Calculus

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