1. INTRODUCTION TO REAL ANALYSIS 1
INDIVIDUAL TASK
EXERCISES 2.5
INTERVALS
By:
Muhammad Nur Chalim
4101414101
MATHEMATICS DEPARTMENT
MATHEMATICS AND NATURAL SCIENCES FACULTY
SEMARANG STATE UNIVERSITY
2016
2. EXERCISES 2.5
Problem
1. If , - and , - are closed intervals in , show that if and only if
and
2. If is nonempty, show that is bounded if and only if there exists a closed
bounded interval such that .
6. If is nested sequence of intervals and if , - show
that and
7. Let 0 1 for . Prove that ⋂ * +
8. Let . / for . Prove that ⋂
Solution
1. If , - and , - are closed intervals in , show that if and only if
and
Solution:
Given , - and , - are closed intervals in
( ) We have
It will be shown that and
Since , then
Consequently, and
So, it is proved that if then and
( ) We have and
It will be shown that
Since and , then for every , we have
Since , then we get
Therefore, for every . Then, we get
So, it is proved that if and , then
3. Thus, rom a and b, we can conclude that and .
2. If is nonempty, show that is bounded if and only if there exists a closed
bounded interval such that .
Solution:
is nonempty
We will show that S bounded if and only if there is an closed bounded interval such
that .
a. We will show that if bounded then there is an closed bounded interval such that
.
Suppose bounded by lower bound and upper bound . Then for an we have
. Because of that , -, so we get where , -.
So, if bounded then there is an bound closed interval such that .
b. We will show that if bounded then there is an closed bounded interval such that
then bounded.
Suppose where closed bound interval , -, but then for every
. Because of that bounded.
So, if there is an closed bounded interval such that then bounded.
6. If is nested sequence of intervals and if , - show
that and .
Solution:
Since, - , -
It follows as in Exercise 1 that
Therefore, we have and
7. Let 0 1 for . Prove that ⋂ * +
Solution:
Let 0 1 for
We get , - 0 1 , -
then , - 0 1 0 1
4. This is a closed nested interval
We can see that . It means that is not empty
We will prove that there is no other element of ⋂
Suppose ⋂
Then ,and
Let , then by the corrollary Archimedian Property 2.4.5 we get there exist
m
1
So that , -
It’s contradiction with := 0 1
Thus. ⋂
Therefore ⋂ * +
8. Let . / for . Prove that ⋂
Solution:
Let . /
We get ( ) . / ( )
Then ( ) ( ) ( )
every
Suppose that ⋂
Thus for every
Let , then by the corrollary Archimedian Property 2.4.5
There exist
m
1
It’s contradiction with 0 1
Thus, dan . /
Therefore, ⋂
