JAWABAN

1. INTRODUCTION TO REAL ANALYSIS 1
INDIVIDUAL TASK
EXERCISES 2.4
APPLICATION OF THE SUPREMUM PROPERTY
By:
Muhammad Nur Chalim
4101414101
MATHEMATICS DEPARTMENT
MATHEMATICS AND NATURAL SCIENCES FACULTY
SEMARANG STATE UNIVERSITY
2016

2. EXERCISES 2.4
Problem
2. If { }, find and .
4. Let S be a nonempty bounded sets in
(a) Let , and let . Prove that , .
(b) Let , and let . Prove that ,
.
6. Let A and B be bounded nonempty subsets of , and let .
Prove that and
12. Given any , show that there exists a unique such that .
13. If , show that there exists such that .
Solution
2. Given { }
and ?
{ }, { }, and { }
According to Corollary 4.4 then
According to problem 3 exercise 2.3 then
According to problem 4 exercise 2.3 then ,
So,
According to problem 4 then ,
So,
According to problem 6 then
Thus, { } and { }
4. Given , is bounded in
(a) and .
We will prove that , .

3. Proof of :
is bounded in , then based on the completeness property of , has infimum. Let
then is lower bound of
So,
, then
So, is lower bound of ...............................................(*)
Suppose is any lower bound of , we will prove that
is any lower bound of . It means that
and
Thus, is lower bound of
Based on definition , then and is lower bound of
Since , then ............................................. (**)
From (*) and (**) then based on definition
Thus,
Proof of :
is bounded in , then based on the completeness property of , has supremum.
Let then is upper bound of
So,
, then
So, is upper bound of ...............................................(*)
Suppose is any upper bound of , we will prove that
is any upper bound of . It means that
and
Thus, is upper bound of
Based on definition , then and is upper bound of
Since , then ............................................. (**)
From (*) and (**) then based on definition
Thus,

4. (b) and .
We will prove that , .
Proof of
is bounded in , then based on the completeness property of , has supremum.
Let then is upper bound of
So,
, then
So, is upper bound of ...............................................(*)
Suppose is any lower bound of , we will prove that
is any lower bound of . It means that
and
Thus, is upper bound of
Based on definition , then and is upper bound of
Since , then .............................................. (**)
From (*) and (**) then based on definition
Thus,
Proof of .
is bounded in , then based on the completeness property of , has supremum.
Let then is lower bound of
So,
, then
So, is lower bound of ...............................................(*)

5. Suppose is any upper bound of , we will prove that
is any upper bound of . It means that
and
Thus, is lower bound of
Based on definition , then and is lower bound of
Since , then .............................................. (**)
From (*) and (**) then based on definition
Thus,
6. Given and are bounded in
Suppose
We will prove that and
Proof:
are bounded in , it means that and have lower and upper bounds
and have lower and upper bounds, it means that and have supremum and
infimum.
(a) Suppose , it means that is upper bound of
So, ............................................... (1)
Suppose , it means that is upper bound of
So, ............................................... (2)
From (1) and (2) we will obtain ,
So. is upper bound of ..................... (*)
Let be any upper bound of , we will prove that
is any upper bound of . It means that ,
So, is upper bound of
Based on definition , then and is upper bound of A
,

6. So, is upper bound
Based on definition , then and is upper bound of .
So, for any is upper bound of ....................... (**)
From (*) and (**) and based on definition of supremum, then
Thus,
(b) Suppose , it means that ’ is lower bound of
So, .............................................. (1)
Suppose , it means that is lower bound of
So, .............................................. (2)
From (1) and (2) we will obtain ,
So. is lower bound of ................... (*)
Let be any lower bound of , we will prove that
is any lower bound of . It means that ,
So, is lower bound of
Based on definition , then and is lower upper bound of
,
So, is lower bound
Based on definition , then and is lower bound of .
So, for any ’ is lower bound of .................. (**)
From (*) and (**) and based on definition of infimum, then
Thus,
12. Given any , show that there exists a unique such that .
In usual case where then is the unique element in .
If then value based on Corrolary 2.4.5 there exist
If then we can apply the same argument in the case .

7. So, we get there some
It will shown is unique
Since then and where
We get inequality
It’s a contradiction
Assumption that there exist then is false.
So, there exist an unique u element in such that
13. Given
We will show that
For a , we have . so, we have
Based on Corollary 2.4.5 if we have . Sehingga
we have
Thus,