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Unformatted text preview: e interval 0, 1 such that S 0, 1 but if U is the set of interior points
of S then U
0, 1 .
Once again, take the set of all rational numbers between 0 and 1.
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 set of real numbers, that H is a closed set and that S H, prove that S
We could argue that S H and that, because H is closed, H H. H. 4. Given two sets A and B of real numbers, 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 sets A and B of real numbers, 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?
Since A B A we have A B A and similarly that A B B. Thus
A B A B.
Now observe that if A 0, 1 and B 1, 2 then
A B 0, 1
1, 2 1
and
A B .
Of course, we could give more spectacular examples like A Q and B R Q.
6. Prove that if S is any set of real numbers then the set R S is the set of interior points of the set R S.
Most students should be encouraged to write two separate arguments here. The first task is to
show that every member of the set R S must be an interior point of R S. Then one should show
that every interior point of R S must belong to R S. 100 On the other hand, a strong student could be permitted to observe that if x is any given number
then the statement that x does not belong to S is the statement that there exists a number 0
such that x , x
S , and that the latter equation is just the condition that
x , x
R 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 true that if A and B are sets of real numbers and
ABR
then A B R?
The answer is no. Look at A Q and B R Q.
9. Prove that if A and B are open sets and
then A ABR
B R. What if only one of the sets A and...
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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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