1. INTRODUCTION TO REAL ANALYSIS 1
INDIVIDUAL TASK
EXERCISES 2.3
THE COMPLETENESS PROPERTY OF
By:
Muhammad Nur Chalim
4101414101
MATHEMATICS DEPARTMENT
MATHEMATICS AND NATURAL SCIENCES FACULTY
SEMARANG STATE UNIVERSITY
2016
2. EXERCISES 2.3
Problem
4. Let *
( )
+. Find and .
5. Let be a nonempty subset of that is bounded below.
Prove that * +
8. Let be nonempty. Show that if , then for every number the
number is not an upper bound of , but the number is an upper bound of .
(The converse is also true; see Exercise 2.4.3.)
9. Show that if and are bounded subsets of , then is a bounded set. Show that
( ) * +
10. Let S be bounded set in and let be nonempty subset of .
Show that
Solution
4. Let *
( )
+. Find inf and sup .
Solution:
Choose
( )
( )
,
( )
,
( )
,
( )
,
( )
and etc.
then we conclude that :
a. If we substitute by even number, then value of will be increased with the
minimum value of is as the lower bound.
3. b. If we substitute by odd number, then value of will be decreased with the
maximum value of is as the upper bound.
So, * +
Thus, inf and sup
5. Let be a nonempty subset of that is bounded below.
Prove that * +
Proof :
Given is bounded below, then based on Definition 2.3.2 (b) there exists .
Let * +
is bounded below then is bounded above and Supremum Property implies that there
exists is sup
Let ( ) .
We get
( ) ( ) (multiply both sides by )
( , then )
Based on the definition of infimum, we get
( )
( * +)
Thus, ( * +)
8. Given be nonempty,
It will be shown that number is not an upper bound of but the number is
an upper bound of for every number .
(a) Suppose that is an upper bound.
Based on 2.4.2 definition:
If is an upper bound of S, and let
Then we will obtain
4. ( ) ( ) ( ) ( Add by to both sides)
( ) ( A1 and A4)
( ) (A2)
(A3)
Since , it is not satisfy that equation. It is a contradiction.
We obtain .
Since , then is not an upper bound of S.
(i) Suppose that is not an upper bound.
Based on 2.4.2 definition:
If , is an upper bound of , and let
( ) ( ) ( ) ( Add by to both sides)
( ) ( A1 and A4)
( ) (A2)
(A3)
Since , it is not satisfy that equation. It is a contradiction.
We obtain .
Since , then is an upper bound of S.
9. Given if A and B are bounded subsets of , then is a bounded set
It will be shown that ( ) * +
For , then
( ) * +
( ) * +
5. So, we can conclude that is a bounded set.
Let and * +
and
is an upper bound of , because
if , then , and if , then .
We get .
If z is any upper bound of
then is an upper bound of and , so that
Hence . Therefore, ( ).
Thus, ( ) * +
10. Given be bounded set in and
It will be shown that
Let
To show that we divide this problem into 2 cases:
(i)
Because of , so that
and
From Definition 2.4.1 and 2.4.2 , we can conclude that , for S
is a nonempty subset of R, so we get
(ii)
Because of S0 S, so that
and
From Definition 2.4.1 and 2.4.2 , we can conclude that , for S is
a nonempty subset of R, so we get
Thus, from (i) and (ii), we can conclude that
