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Unformatted text preview: c. Give an example of a set S of real numbers such that if U is the set of interior points of S then U S.
We could take S to be a singleton like 3 or it could be the set of all integers. It could also be
the set of all rational numbers between 0 and 1.
d. Give an example of a subset S of the interval 0, 1 such that S 0, 1 but if U is the set of interior points
of S then U
0, 1 .
Take S to be the set of rational numbers that lie in the interval 0, 1 . The set of interior points of
S is empty.
2. Given that
1
n S n Z , evaluate S. Hint: Show that
1
n Z .
n
First show that 0 S. Then observe that every negative number belongs to the set
any positive number then x belongs to the interval
1 ,1
n1 n
for some positive integer n.
S 0 Þ Ý, 0 and that if x is 3. Given that S is a subset of a metric space, that H is a closed set and that S H, prove that S H.
From the elementary properties of closure we know that S H and we know from the relationship
between closure and the property of being closed that H H.
4. Given two subsets A and B of a metric space, prove that
A Þ B A Þ B. Solution: Since A A and B B we have A Þ B A Þ B.
and therefore, since the union of the two closed sets A and B is closed we have
A Þ B A Þ B.
On the other hand, since A is included in the closed set A Þ B we have
A AÞB
and, similarly we can see that
B AÞB
and so
A Þ B A Þ B.
Therefore
A Þ B A Þ B.
5. Given two subsets A and B of a metric space, prove that
A B A B.
Do the two sides of this inclusion have to be equal? What if A and B are open? What if they are closed?
From the fact that A B A we deduce that A B A and we see similarly that A B B.
Therefore A B A B.
In the metric space R, if A 0, 1 and B 1, 2 then A and B are open and A B A B.
On the other hand, if A and B are closed subsets of a metric space then 122 A BA BA B. 6. Prove that if S is any subset of a metric space X then the set X S is the set of interior points of the set X S.
Given any point x X, the condition x X S means that there exists a number 0 such that
B x,
S and this condition says that there exists a number 0 such that B x,
X S.
The latter assertion says that x is an interior point of the set X S.
7. Given that is an upper bound of a given set S of real numbers, prove that the following two conditions are
equivalent:
a. We have sup S.
b. We have S. To prove that condition a implies condition b we assume that sup S. We need to show that
S. Suppose that 0. Using the fact that is the least upper bound of S and that we
choose a member x of S such that x. Since x
,
S we have
,
S
.
To prove that condition b implies condition a we assume that S. We need to show that is the
least upper bound of S. Suppose that p . Since the set p, Ý is a neighborhood of we have
p, Ý
S
. Thus, since is an upper bound of S and since no number p can be an upper
bound of S we conclude that is the least upper bound of S.
8. Is it tru...
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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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