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Chapter I
The Real Line System
Readings:
 Elementary Analysis: The Theory of Calculus. (chapter 1)
 Introduction to Real analysis (chapter 2)
The set N of Natural Numbers:
We denote the set {1,2,3,……} of all numbers by N.
The set N has the following properties:
N1. 1 N .
N2. If n N , then its successor n 1 N .
N3. 1 is not the successor of any element in N.
N4. If n and m in N have the same successor, then n = m.
N5. A subset of N which contains 1 and which contains n+1 whenever it
contains n, must equals N.
(i.e. If B  N such that (i) 1  B
(ii) If n 1 Bn  B , then B = N)
Properties N1~N5 are known as the Peano Axioms.
Axiom N5 is the basis of mathematical induction.
The set Q of Rational Numbers:
Rational numbers are numbers of the form
1
m
where m , n  Z and
n
n  0.
The set Q of rational numbers is a very “nice” algebraic system until one tries to
solve equations like x2 = 2. No rational number satisfies this equation (we’ll prove
this later), but there is a number that satisfies this equation.
2 ,  , e  Q ( Do you know why?).
So there are “gaps” in Q.
Definition 1.1:
A number is called an algebraic number if it satisfies a polynomial equation
an xn+an-1 xn-1+..........+a1 x +ao = 0
where ao , a1, ………, an are all integers, an  0 , and n  1 .
Notes:
(i) Rational numbers are algebraic numbers. If r 
m
, then r satisfies the equation
n
nx-m=0.
(ii)
2 is an algebraic number.
Theorem 1.2: (The Rational Zeros Theorem)
Suppose that ao , a1, ………, an are integers and that r is a rational number
satisfying the equation
an xn+an-1 xn-1+..........+a1 x +ao = 0
n  1 , an  0 , ao  0 , and r 
(1) where
p
, p and q are integers having no common factor and
q
q  0 . Then q divides an and p divides ao.
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In other words: The only rational candidate for solutions of (1) have the form
r
p
where q divides an and p divides ao.
q
Proof: r 
p
satisfies equation (1). Then
q
n
 p
 p
an    ...............a1    ao  0
q
q
Multiplying by qn, we get
an pn+an-1 pn-1q+an-2 pn-2q2+……………+a1 pqn-1+aoqn= 0
 an pn=-q[an-1 pn-1+an-2 pn-2q+……………+a1 pqn-2+aoqn-1]
 q divides an pn
But p and q have no common factor, therefore q divides an .
Also, aoqn =-p[an pn-1+an-1 pn-2q+……………+a1 qn-1]
 p divides ao .
2 can’t represent a rational number.
Example 1.3:
By theorem 1.2 the only rational number that could possibly solve the equation
x 2- 2 = 0 are ±1, ±2 (since an = 1 and ao = -2  p = ±1, ±2 and q = ±1). We can
substitute each of the four numbers into the equation x2- 2 = 0 to eliminate them as
possible solutions. But
Example 1.4:
3
2 is a solution, therefore
2 can’t be a rational number.
6 can’t be a rational number.
By theorem 1.2 the only rational numbers that could possibly solve the equation
3
x 3- 6 = 0 are ±1, ±2, ±3, ±6 (since an = 1 and ao = -6  p = ±1, ±2, ±3, ±6 and
q = ±1). We know that none of these numbers is a solution of the equation x3- 6 =0.
But
3
6 is a solution, therefore
3
6 can’t be a rational number.
1
2
Example 1.5: b  (2  5 ) does not represent a rational number.
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b is a solution of the equation x6 -6x4 +12x2 -13 = 0. ( an = 1 and ao = -13  p = ±1,
±13 and q = ±1). By theorem1.2 the only possible rational solutions are ±1, ±13. We
know that none of these numbers is a solution. But b is a solution, therefore b can’t
be a rational number.
H.W. 1 : See Homework problems for chapter I
The Algebraic and Order Properties of R:
Algebraic Properties of R:
A1. a +b = b +a a, b  R .
A2. (a +b) +c = a +(b +c) a, b, c  R .
A3. a +0 = 0 +a = a a R .
A4. a R there is an element  a  R such that a +(-a ) = (-a ) +a = 0.
M1. a .b = b .a a, b  R .
M2. (a .b) .c = a .(b .c) a, b, c  R .
M3. a .1 = 1 .a = a a R .
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M4. a  R(a  0) there is an element
1
1
1
 R such that a .(
) = ( ) .a = 1.
a
a
a
D. a .(b +c) = (a .b) + (a .c)
(b +c) .a = (b .a) + (c .a) a, b, c  R
Definition 1.6:
A system that has more than one element and satisfies the nine above
properties is called a field .
The Order Properties of R:
O1. Given a and b, either a ≤ b or b ≤ a
O2. If a ≤ b and b ≤ a , then a = b
O3. If a ≤ b and b ≤ c , then a ≤ c
O4. If a ≤ b , then a + c ≤ b + c
O5. If a ≤ b and 0 ≤ c , then a . c ≤ b . c
Definition 1.7:
A field with an ordering satisfying O1 ~ O5 is called an ordered field.
Thus R is an ordered field.
Theorem 1.8:
If a  R is such that 0 ≤ a < ε for every ε > 0, then a = 0.
1
2
Proof: Suppose to the contrary that a > 0. Then take  o  a  0   o  a which is a
contradiction. Therefore a = 0.
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Absolute Value and the Real Line:
Definition 1.9:
The absolute value of a real number a, denoted by |a|, is defined by:
 a if

a   0 if
 a if

a  0

a  0
a  0

Theorem 1.10:
(i)
|ab| = |a| |b|
a, b  R
(ii)
|a|2 = |a2|
a R
(iii)
If c > 0, then |a| ≤ c if and only if –c ≤ a ≤ c
(iv)
-|a| ≤ a ≤ |a|
a R
Theorem 1.11: (Triangular Inequality)
If a, b  R , then |a + b| ≤ |a| + |b|.
Proof: -|a| ≤ a ≤ |a| and -|b| ≤ b ≤ |b|. Adding we get
(a  b)  ab  a  b
 ab  a  b
by theorem 1.10(iii).
Corollary 1.12: If a, b  R , then
(i) a  b  a  b
(ii) a  b  a  b
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Proof: (i) a = a - b + b
 a  a  b  b  a  b  b
 a  b  a b
b=b–a+a
 b  b  a   a  b  a  a
  a b  b  a  a  b
Therefore,  a  b  a  b  a  b
 a  b  a b
by theorem 1.10(iii)
(ii) a  b  a  (b)  a   b  a  b
Therefore, a  b  a  b .
Definition 1.13:
The distance between two elements a and b in R is dist (a , b) = a  b .
Definition 1.14:
Let a  R and ε > 0. Then the ε-neighborhood of a is the set
V a   x  R : x  a   
[ x V a     x  a    a    x  a   ]
Theorem 1.15:
Let a  R . If x V (a) for every ε > 0, then x = a.
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Proof: x V (a)   0
 xa 
  0
 x  a  0 by theorem 1.8
 x  a.
H.W. 2: See Homework problems for chapter I
___________________________________________________________________
The Completeness property of R:
Suprema and Infima:
Definition 1.16:
Let S be a nonempty subset of R.
(a) If S contains a largest element so (i.e., so  S and s  so s  S ), then we call so
the maximum of S and we write so = max S.
(b) If S contains a smallest element s1 (i.e., s1  S and s  s1 s  S ), then we call s1
the minimum of S and we write s1 = min S.
Example 1.17:
(a) Every finite nonempty subset of R has a maximum and a minimum.
max{ n  Z : 4  n  100}  100
min{ n  Z : 4  n  100}  3
(b) max [a , b] = b
min [a , b] = a
(a , b) has no maximum or minimum.
[a , b) has no maximum but its minimum is a.
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(c) The sets Z and Q have no maximum or minimum.
N has a minimum which is 1.
(d) min{ r  Q : 0  r  2}  0 . But the set has no maximum.
Note: Always min S  S and max S  S .
Definition 1.18:
Let S be a nonempty subset of R.
(a) The set S is said to be bounded above if there exists a number u  R such that
s  u s  S . Each number u is called an upper bound of S.
(b) The set S is said to be bounded below if there exists a number w R such that
s  w s  S . Each number w is called a lower bound of S.
(c) A set is said to be bounded if it is both bounded above and below. A set is said
to be unbounded if it is not bounded.
Example 1.19:
(a) The sets (a , b), (a , b], [a , b) and [a , b]are bounded above, the number b and
any number larger then b is an upper bound.
Also they are bounded below, the number a and any number less then a is a
lower bound.
(b) The sets Z, Q, N and R are unbounded. But N is bounded below, the number 1
and any number less then 1 is a lower bound.
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(c) The set {r  Q : 0  r  2} is bounded. The number
then
2 and any number larger
2 is an upper bound, the number 0 and any number less then 0 is a lower
bound.
(d) The set {n( 1) : n  N} is unbounded. It is not bounded above.
n
Note: If the set S is bounded above (or below) then the upper bounds (or the lower
bounds) don’t have to belong to S.
Definition 1.20:
Let S be a nonempty subset of R.
(a) If S is bounded above, then a number u is said to be a supremum (or a least
upper bound) of S if it satisfies the conditions:
(i)
u is an upper bound of S.
(ii)
If v is any upper bound of S, then u ≤ v.
We write u = sup S.
(b) If S is bounded below, then a number w is said to be an infimum (or a greatest
lower bound) of S if it satisfies the conditions:
(iii)
w is a lower bound of S.
(iv)
If t is any lower bound of S, then t ≤ w.
We write w = inf S.
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Example 1.21:
(a) sup [a , b] = sup (a , b) = sup (a , b] = sup [a , b) = b.
inf [a , b] = inf (a , b) = inf (a , b] = inf [a , b) = a.
(b) sup {r  Q : 0  r  2} =
2.
inf {r  Q : 0  r  2} = 0.
(c) inf {n( 1) : n  N} = 0.
n
Notes:
(i)
If a set S has a maximum, then max S = sup S.
Also if it has a minimum, then min S = inf S.
(ii)
sup S and inf S need not belong to the set S.
(iii)
There can only be only one supremum and only one infimum.
(iv)
Not every subset of R has a supremum or an infimum.
Lemma 1.22:
Let S be a nonempty set, and u = sup S. Then the following statements are
equivalent:
(i)
If v is any upper bound of S, then u ≤ v.
(ii)
If z < u, then z is not an upper bound of S.
(iii)
If z < u, then sz  S such that z < sz .
(iv)
If ε > 0, then s  S such that u – ε < sε .
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Exercise: Obtain the four equivalent statements above for w = inf S.
The Completeness Property of R:
Every nonempty set of real numbers that has an upper bound also has a
supremum in R.
Corollary 1.23: Every nonempty subset of R that has a lower bound also has an
infimum in R.
Proof: Let –S = { s : s  S } .
S is bounded below, therefore there exists a real number t such that
t  s s  S
 t   s s  S
 t is an upper bound of –S.
 By the completeness property of R, the set –S has a supremum u = sup (-S).
Then by definition 1.20:
(i)  s  u s  S
(ii) If v is any upper bound of –S, then u  v .
(i)   u  s s  S
(ii)  If -v is any lower bound of S, then  v  u .
Therefore, -u = inf S.
i.e. sup (-S) = -inf S .
H.W. 3: See Homework problems for chapter I
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Theorem 1.24: (The Archimedean Property):
If x  R , then nx  N such that x < nx .
Proof: Assume that n  x n N .  x is an upper bound of N. Therefore by the
completeness property, N has a supremum u  R .
u - 1 < u  u – 1 is not an upper bound of N
 m N such that u – 1 < m
 u<m+1
But m 1 N , and this contradicts the fact that u is an upper bound of N.
Hence, nx  N such that x < nx .
Corollary 1.25:
If t > 0, then nt  N such that 0 
1
t.
nt
1

Proof: inf  : n  N  = 0.
n

1

t > 0  t is not a lower bound of the set  : n  N  . Thus nt  N such that
n

0
1
t.
nt
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Density of Rational Numbers in R:
Theorem 1.26: (The Density Theorem):
If x and y are real numbers with x < y, then  a rational number r  Q such that
x < r < y. (In other words: Between any two real numbers there is a rational number).
Proof: Assume that x > 0.
-x + y > 0  by corollary 1.24 n  N such that
1
 yx
n
 nx + 1 < ny
 m N such that m 1  nx  m
nx > 0
 m  nx  1
Therefore, m < ny
Now, nx < m < ny  x 
m
 y.
n
Thus the rational number  r 
m
satisfies x < r < y.
n
What if x  0 ? (Exercise)
Corollary 1.27:
If x and y are real numbers with x < y, then  an irrational number z such that
x < z < y.
Proof:
x
2

y
2
 By theorem 1.26, r  Q such that
 xr 2  y.
Then z  r 2 is irrational and satisfies x < z < y.
14
x
2
r
y
2
The Extended Real Numbers.
It is often convenient to extend the system of real numbers by the addition of
two elements +∞ and -∞. This enlarges the set R to the set R  {,} . The set
R  R  {,} is called the set of extended real numbers.
One use of the extended real numbers is in the expression “sup S ”. If S is a
nonempty set of real numbers with an upper bound, we define sup S to be the least
upper bound of S. If S has no upper bound, we write sup S = +∞. Then sup S is
defined for all nonempty sets.
If S is bounded below then inf S is defined to be the greatest lower bound of S.
If S has no lower bound, we write inf S = -∞.
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Finite and Infinite Sets:
Definition 1.28:
(a) The empty set φ is said to have 0 elements.
(b) If n N , a set S is said to have n elements if there is a bijection (1-1 and onto
mapping) from the set Nn = {1,2,3,……..,n} onto S.
(c) A set is said to be finite if it is either empty or it has n elements.
(d) A set S is said to be infinite if it is not finite. i.e., if S has an infinite number of
elements.
Example 1.29:
(a) The sets N, Z, Q, and R are all infinite sets.
(b) All intervals are infinite sets.
(c) The set {r  Q : 0  r  2} is an infinite set.
(d) The set {n  Z : 10  n  20} is a finite set.
Theorem 1.30:
Suppose that S and T are sets such that T  S .
(i)
If S is a finite set, then T is a finite set.
(ii)
If T is an infinite set, then S is an infinites set.
Theorem 1.31:
The finite union of finite sets is also finite. (i.e., if A1 , A2 , ……,An are all finite
n
sets, then B   Ai is also finite).
i 1
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Countable Sets:
Definition1.32:
(a) A set S is said to be denumerable if there exists a bijection of N onto S. And
we write S = {s1 , s2 , s3 ,………}.
(b) A set S is said to be countable if it is either finite or denumerable.
(c) A set S is said to be uncountable if it is not countable
Sets
Finite Sets (Countable)
Infinite Sets
Denumerable Sets (Countable)
Uncountable Sets
Theorem 1.33:
(i)
The union of two denumerable sets is denumerable.(  countable)
(ii)
N × N is denumerable.(  countable)
(iii)
The set Q of rational numbers is denumerable.(  countable)
Theorem 1.34:
Suppose that S and T are sets such that T  S .
(i)
If S is a countable set, then T is a countable set.
(ii) If T is an uncountable set, then S is an uncountable set.
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Theorem 1.35:
The countable union of countable sets is countable. (i.e. if A1 , A2 , A3…… are

all countable sets, then B   Ai is also countable).
i 1
Example 1.36:
(a) The set E  {2n : n  N } of even natural numbers is denumerable since the
mapping f (n) = 2n for n  N is a bijection of N onto E.
Similarly, the set O  {2n  1 : n  N } is denumerable. (why?)
(b) The set Z of integers is denumerable since the mapping f : N → Z defined by

 0
 n
f ( n)  
 2
1  n
 2
if
n 1
if
nE
if
nO
is a bijection of N onto Z.
(c) All intervals are uncountable sets.
(d) The set {r  Q : 0  r  2} is a countable set.
(d) The set {n  Z : 10  n  20} is a countable set.
Exercise:
Is the countable union of finite sets also finite?
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