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Unformatted text preview: the set Z of integers.
2. Prove that in the metric space R we have L Q R.
Suppose that x is any real number. To show that x is a limit point of Q, suppose that 0. Since
there are rational numbers in the interval x, x we know that the set x , x
Q
x
.
3. Prove that in the metric space R we have L 1
n Z 0 .
n
If x is any negative number then the interval Ý, 0 is a neighborhood of x that fails to contain any
members of the set 1
n Z . Thus a negative number can not be a limit point of
n
1
nZ.
n 128 If x is any positive number then the interval
infinitely many members of the set 1
n
n
then the condition , Ý is a neighborhood of x which fails to contain
Z . To see why, note that if n is a positive integer
x
2 1
x ,Ý
n
2
can hold only if n 2 . Therefore no positive number can be a limit point of 1
n Z .
x
n
1
Finally, we need to explain why 0 must be a limit point of n
n Z . Suppose that 0.
Choose an integer k 1 and observe that
1
1
0 , 0
n Z
0
n
k
1
0 must be nonempty.
n Z
from which we deduce that the set 0 , 0
n
4. a. Give an example of an infinite subset of R that has no limit point.
As we saw in Exercise 1, the infinite set Z has no limit point.
b. Give an example of a bounded subset of R that has no limit point.
A finite set like 2 will not have any limit points. We could also look at the empty set .
c. Give an example of an unbounded subset of R that has no limit point.
As we saw in Exercise 1, the infinite set Z has no limit point.
d. Give an example of an unbounded subset of R that has exactly one limit point.
The unbounded set Z Þ 1
n Z has only the limit point 0.
n
e. Give an example of an unbounded subset of R that has exactly two limit points.
The set
ZÞ 1
n Z Þ 1 1
n Z
n
n
has the two limit points 0 and 1. We can see this directly or we can use the assertion proved in
Exercise 6 below. 5. Prove that if A and B are subsets of a metric space X and if A B then L A
LB.
Suppose that A and B are subsets of a metric space X and that A B. Suppose that x is a limit
point of A. We need to explain why x has to be a limit point of B. Suppose that 0. Since the set
B x,
A
x is nonempty and since
B x,
A
x
B x,
B
x
we deduce that the set B x,
B
x is nonempty.
6. Prove that if A and B are subsets of a metric space X then
L AÞB L A ÞL B . Solution: Since A
LB L A Þ B . Thus A Þ B we know that L A L A Þ B and similarly we know that L A ÞL B
L AÞB .
Now suppose that a point x fails to belong to the set L A Þ L B . Choose a number 1 0 such that the
ball B x, 1 contains only finitely many members of the set A. Choose a number 2 0 such that the ball
B x, 2 contains only finitely many members of the set B. We now define to be the smaller of the two
numbers 1 and 2 and we observe that, although 0, the interval B x, contains only finitely many
members of the set A Þ B. Therefore no number that lies outside the set L A Þ L B can be a limit point
of A Þ B and w...
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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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