MATH 102 INTRODUCTION TO MATHEMATICAL ANALYSIS
1. Some Fundamentals
1.1. Properties of Real Numbers.
The whole course will be based entirely on the study of sequence of numbers and functions defined on
the real line. Thus a detail study of the real numbers is important. However, such a study will take up
quite a bit of time. So we shall take a quicker approach by drawing up some essential properties about
real numbers, and to deduce some consequences from them. (W. Rudin, Principles of Mathematical
Analysis, or E. Landau, Foundations of analysis, 2nd.)
Proposition 1. (Associative law of addition) If a, b and c are any numbers, then
a + (b + c) = (a + b) + c
Proposition 2. (Existence of additive identity) There is a number denoted by 0 such that for any
number a
0 + a = a + 0 = a.
Proposition 3. (Existence of additive inverse) For every number a there is a number denoted by −a
such that
a + (−a) = (−a) + a = 0.
Proposition 4. (Commutative law for addition) If a, b are any numbers, then
a + b = b + a.
Proposition 5. (Associative law for multiplication) If a, b and c are any numbers, then
a · (b · c) = (a · b) · c.
Proposition 6. (Existence of multiplication identity) There is a number denoted by 1, such that for
any number a,
a · 1 = 1 · a = a.
Moreover, 1 6= 0.
Proposition 7. (Existence of multiplicative inverse) For any number a 6= 0 there is a number denoted
by a−1 such that
a · a−1 = a−1 · a = 1.
Proposition 8. (Commutative law for multiplication) If a and b are any number, then
a · b = b · a.
Proposition 9. (Distributive law) If a, b and c are any numbers, then
a · (b + c) = a · b + a · c.
Proposition 10. (Trichotomy law) There is a collection of positive numbers. For every number a
one and only one of the following follows:
(i) a = 0
(ii) a is positive (written a > 0)
(iii) −a is positive.
Proposition 11. (Closure under addition) If a > 0 and b > 0, then a + b > 0.
Proposition 12. (Closure under multiplication) If a > 0 and b > 0, then a · b > 0.
1
2
MATH 102 INTRODUCTION TO MATHEMATICAL ANALYSIS
Definitions.
a − b = a + (−b)
a > b means a − b > 0
a < b means a − b < 0
a ≥ b means either a > b or a = b
a ≤ b means either a < b or a = b
From Propositions 1-12 we may deduce the following consequences:
(1) If x + a = b, then x = b − a.
(x + a) + (−a) = b + (−a)
=b−a
(Definition)
x + (a + (−a)) = b − a
(Proposition 1)
x+0=b−a
(Proposition 3)
x=b−a
(Proposition 2)
(2) If ab = ac and a 6= 0, then b = c
a−1 (ab) = a−1 (ac)
(Proposition 7)
(a−1 a)b = (a−1 a)c
(Proposition 5)
1b = 1c
b=c
(Proposition 6)
(3) a · 0 = 0 for any a.
a · 0 = (0 + 0)a
(Proposition 2)
= 0a + 0a
(Proposition 9)
0a − 0a = 0a + (0a − 0a)
0 = 0a + 0
0 = 0 · a.
(4) If a · b = 0, then either a = 0 or b = 0.
If a = 0 and b = 0, then by (3), a · b = 0. So there is nothing to prove. So we may assume,
say, a 6= 0:
a−1 (ab) = a−1 0
(a−1 a)b = 0
1b = 0
b = 0.
(5) (−a)b = −(ab).
(−a)b + ab = (−a + a)b
=0·b
=0
Therefore (−a)b = −(ab).
MATH 102 INTRODUCTION TO MATHEMATICAL ANALYSIS
3
(6) (−a)(−b) = ab.
(−a)(−b) − (ab) = (−a)(−b) + (−a)b
(5)
= (−a)(−b + b)
= −a · 0
= 0.
Therefore (−a)(−b) = ab.
(7) If c < 0, then −c > 0.
For c < 0 means 0 > c which means 0 − c > 0 by definition. That is, −c > 0.
(8) If a < 0 and b < 0, then ab > 0.
We have −a > 0 and −b > 0 from (7). From Proposition 12, (−a)(−b) > 0. By (6), ab > 0.
Remark. a2 > 0 for any a 6= 0. In particular, 1 = 12 = 1 × 1 > 0. If a > 0, then a · a > 0 by
Proposition 12. If a < 0, then −a > 0 and a2 = (−a)(−a) > 0.
Definition. Let N denotes the set of natural numbers. That is, the set {1, 2, 3, . . . }. Let Z denotes
the set of integers. That is, the set {. . . , −3, −2, −1, 0, 1, 2, 3, . . . }. Let Q denotes the set of rational
numbers or fractions. That is, the set of numbers of the form pq where p and q are integers with no
common factors and q 6= 0.
Clearly the Z and Q satisfy Proposition 1-12, and hence their consequences.
However, we shall show that Q is not “complete” in the following sense.
Example. If x ∈ Q, then x2 6= 2.
Proof. Suppose not, let x =
m
n
where m, n ∈ Z, n 6= 0, and x2 = 2. Then
m2
=2
or
m2 = 2n2
n2
and m is even. So we may write m = 2k for some k 6= 0. Hence
2n2 = m2 = 4k 2
or
n2 = 2k 2 .
That is, n is also even. This contradicts the fact that m and n have no common factors.
¤
If we consider a “bigger set”, R — the whole real line
√ then we may say that there exists a number
x ∈ R such that x2 = 2. This number is denoted by 2. It is called an irrational number. Note that
we can merely assume its existence and we are unable to prove it exists at the point. Even this pose a
problem, since there are many different algebraic equations, so it has many different sorts of irrational
numbers. Clearly such an approach is unsatisfactory. Hence we shall assume ione more property later,
and this makes all the differences.
1.2. Inequalities.
Definition. For any number a, we define the absolute value or modulus |a| of it as follows:
(
a
if a ≥ 0
|a| =
.
−a if a ≤ 0
Its graph looks like f (x) = |x|.
4
MATH 102 INTRODUCTION TO MATHEMATICAL ANALYSIS
Theorem 1.1.
(I) |a · b| = |a| · |b|
(II) |a + b| ≤ |a| + |b|
(III) |a| − |b| ≤ |a − b|
|b| − |a| ≤ |a − b|
Remark. We may rewrite (III) as ||a| − |b|| ≤ |a − b|.
Proof. We distinguish 4 cases:
1. a ≤ 0, b ≥ 0
2. a ≥ 0, b ≤ 0
3. a ≤ 0, b ≥ 0
4. a ≤ 0, b ≤ 0.
Case 1 (I) |ab| = ab = |a||b|
(II) |a + b| = a + b = |a| + |b|
Case 4 (I) |ab| = ab = (−a)(−b) = |a||b|
(II) |a + b| = −(a + b) = −a + (−b) = |a| + |b|
Case 2 (I) |ab| = −(ab) = a(−b) = |a||b|
To prove (II) in this case, we must prove that
|a + b| ≤ |a| + |b| = a − b.
If a + b ≥ 0, we need to show a + b = |a + b| ≤ a − b. That is, b ≤ −b which is the case.
If a + b ≥ 0, we need to show a + b = |a + b| ≤ a − b That is, −a − b ≤ a − b. That is, −a ≤ a which is
the case again.
For (III), we need only |a| = |a − b + b| ≤ |a − b| + |b|. That is, |a| − |b| ≤ |a − b| and |b| = |b − a + a| ≤
|b − a| + |a|. That is, |b| − |a| ≤ |a − b|.
¤
Proposition 13. (Principle of mathematical induction)
Let A be any collection (or set) of natural numbers with the properties (1) 1 ∈ A and (2) k + 1 ∈ A
whenever k ∈ A. Then A = N.
Bernoulli’s inequality. If h + 1 > 0 then (1 + h)n ≥ 1 + nh for n ∈ N.
Clearly, it is true if n = 1. So suppose it is valid for n. That is, suppose (1 + h)n ≥ 1 + nh.
(1 + h)n+1 = (1 + h)(1 + h)n ≥ (1 + h)(1 + nh)
= 1 + nh + h + nh2
= 1 + (n + 1)h + nh2 > 1 + (n + 1)h.
Hence Bernoulli’s inequality follows from the principle of induction.
Theorem 1.2. Every non-empty set A of natural numbers has a least member.
Proof. Suppose that this is not the case. In particular, 1 ∈
/ A. Let B be the set of natural numbers
1, . . . , n which are not in A. Clearly 1 ∈ B.
Suppose the natural number k ∈ B, then k + 1 must also belong to B. For suppose 1 + k ∈ A,
then k + 1 would be the least member of A. A contradiction. Hence k + 1 ∈ B. Hence B = N from
Proposition 13 and A = ∅.
¤
Remark. The empty set ∅ is the only set of natural number without a least member.
We note that Theorem 1.2. is equivalent to the principle of induction. That is, we can deduce
Proposition 13 from Theorem 1.2. We leave this this as an exercise.
MATH 102 INTRODUCTION TO MATHEMATICAL ANALYSIS
5
1.3. A useful inequality. The Greek letter epsilon ² is usually need in analysis for a small or arbitrary
quantity.
If a < b + ² for any given ² (That is, no matter how small) then it follows that a ≤ b.
For suppose it is not true that a ≤ b, then a > b must be true. Since ² is arbitrary, we choose
² = a − b > 0. Then a < b + ² = b + (a − b) = a. A contradiction. Hence a ≤ b must be true.
One can argue similarly that if |x − a| < ² is true for any ² > 0, where x is a variable and a is fixed.
Then x = a. (Try this!)
1.4. Least upper bounds.
Definition. A set A of real numbers is said to be bounded above if ∃ (there exists) a number a such
that x ≤ a for all x ∈ A. Such a number is called an upper bound for A.
Example. If A = {x|0 ≤ x ≤ 1} then 1, 2, 100, 101.1 are all upper bounds of A.
Similarly, A0 = {x|0 ≤ x < 1} has the same upper bounds. However, none of the upper bounds
actually belongs to A0 .
Definition. A number a is a least upper bound (or supremum) of A if
(1) a is an upper bound
(2) if b is an upper bound then a ≤ b.
Remark. Note that if a and b are both least upper bounds for A, then it is easy to see that a = b. So
we may speak of the least upper bound.
A set of real numbers is said to be bounded below if there is a number a such that x ≥ a for all
x ∈ a. A number a is the greatest lower bound (or infimum) of A if
(1) a is a lower bound
(2) if b is a lower bound then a ≥ b.
1.5. The least upper bound property of R.
Proposition 14. If A is a non-empty set of real numbers and A is bounded above, then A has a
supremum (sup A) or least upper bound (inf b).
We note that Proposition 14 cannot be deduced from Proposition 1-13. And that it is an assumption
only for R. For example, this assumption is not true for Q.
Let A be the set of all real numbers x satisfying x2 < 2, then there is no rational number q which is
an upper bound for A.
Theorem 1.3. The set of natural numbers N is not bounded above.
Proof. Suppose that N is bounded above. Note that N is not empty. Hence there exists a supremum
α of N by Proposition 14.
That is,
n≤α
for all n ∈ N
That is,
n+1≤α
for all n ∈ N since n + 1 ∈ N by Proposition 13
That is,
n≤α−1
for all n ∈ N
That is, α − 1 is an upper bound for N and which is less than α the supremum. A contradiction. ¤
Corollary 1.4. For any ² > 0 there is a natural number n with
1
n
< ².
Proof. Suppose this is not true, then there exists an ² > 0 such that
n ≤ 1² for all n ∈ N. This contradicts Theorem 1.3.
1
n
≥ ² for all n ∈ N. That is,
¤
Theorem 1.5. There exists a positive real number a such that a2 = 2.
Proof. We define S = {s ∈ R : s ≥ 0 and s2 < 2}. Clearly the set S is non-empty and bounded above.
Hence it has a supremum α by Proposition 14.
We shall show that α2 < 2 or α2 > 2 are both impossible.
6
MATH 102 INTRODUCTION TO MATHEMATICAL ANALYSIS
First we assume that α2 < 2. That is, 2 − α2 > 0. Also note that α > 1. Notice that
positive number and hence we may find a n ∈ N such that
2−α2
2α+1
is a
1
2 − α2
<
.
n
2α + 1
This follows from corollary 1.4. Consider
1
α
1
1
(α + )2 = α2 + 2 + 2 ≤ α2 + (2α + 1)
n
n n
n
< α2 + (2 − α2 ) = 2
Hence (α + n1 ) < 2 and this implies that α + n1 ∈ S, contradicting the fact that alpha is the supremum
of S Hence α2 ≥ 2.
The proof that “α2 > 2 is impossible” is left as an exercise.
¤
Theorem 1.6. Every interval (0, b) contains a rational numbers.
Proof. Since b − a > 0 it follows from Corollary 1.4. that there exists a rational number k such that
1
< b − a.
k
Define A = {n ∈ N| nk > a}. Since A is not empty (why?) and so it must have a least member n0 say,
by Theorem 1.2. That is,
n0
n0 − 1
>a
and
≤ a.
k
k
Notice that
n0
1
a<
≤ a + < b.
k
k
Hence nk0 is the required rational number.
¤
Corollary 1.7. Every interval (a, b) contains an irrational number.
Proof. Simply consider ( √a2 , √b2 ) then it contains a rational r by Theorem 1.6. That is,
a
b
√ <r< √
2
2
or
√
a < r 2 < b.
√
And it is easy to show that r 2 is an irrational.
¤
1.6. Further properties about supremum etc.
Definition. If a set S has a largest element M , we call M the maximum of the set S and write
M = max S. If S has a smallest element m, we call m the minimum of S and write m = min S.
Let us recall the example A = {x|0 ≤ x ≤ 1} and A0 = {x|0 ≤ x < 1}. Clearly max A = 1 = sup A
and min A = 0 = inf A.
However although min A0 exists and equals 0, max A doesn’t exist. But sup A0 exists and equal 1.
Example. B = { n1 : n ∈ N}. Find max B, min B, sup B, inf B.
max B = 1 = sup B. It is easy to see that there is no min B. For suppose min B = n10 for some n0
then n01+1 < n10 .
We show inf B = 0. First note that inf B ≥ 0. For suppose inf B < 0 then 0 > inf B2 > inf B is a
lower bound greater than inf B. Hence inf B ≥ 0. Suppose inf B > 0. Then by Corollary 1.4., one can
find n ∈ N such that
1
< inf B.
n
This contradicts the definition of inf B. Hence inf B = 0.
MATH 102 INTRODUCTION TO MATHEMATICAL ANALYSIS
7
Theorem 1.8. Suppose that S is a non-empty set of real numbers which is bounded above and ξ > 0.
Then
sup ξx = ξ sup x.
x∈S
x∈S
Proof. Let α = sup S. So
That is,
x≤α
ξx ≤ ξα
for all x ∈ S
for all x ∈ S
Therefore sup{ξx|x ∈ S} ≤ ξα. That is,
sup ξx ≤ ξα = ξ sup x.
x∈S
x∈S
Let β = sup ξx then
x∈S
ξx ≤ β
1
1
(ξx) ≤ β
ξ
ξ
1
x≤ β
ξ
for all x ∈ S
for all x ∈ S(ξ > 0)
That is,
1
α = sup x ≤ β ⇒ ξ sup x ≤ sup ξx.
ξ
x∈S
x∈S
x∈S
Hence result follows.
¤
We list some simple results.
(i) ξ > 0, and S is bounded above.
sup ξS = ξ sup S,
∈ ξS = ξ inf S.
(ii) ξ < 0,
sup ξS = ξ inf S,
inf ξS = ξ sup S.
In particular, if −S = {−x|x ∈ S} then
sup(−S) = − inf S
and
inf(−S) = − sup S.
(iii) Suppose both A and B are bounded above, then
sup A + sup B = sup(A + B),
where A + B = {a + b|a ∈ A, b ∈ B}.
1.7. Countable and uncountable sets. Some definitions. Two sets A and B are said to be
equivalent written as A ∼ B, if there exists a one-to-one correspondence f : A → B (that is, a
bijection).
A set A ⊂ R is finite if there is a n ∈ N such that A ∼ {1, 2, . . . , n}. A set which is not finite is
called infinite. A set A ⊂ R is called countably infinite (or denumerable) if A ∼ N.
A set is called countable if it is either finite or countably infinite. A set is called uncountable if it is
not countable.
Theorem 1.9. If f : A → B is an injection and B is countable then A is also countable.
8
MATH 102 INTRODUCTION TO MATHEMATICAL ANALYSIS
Proof. We may assume that A is infinite, otherwise there is nothing to prove. Since A ∼ f (A) and
hence f (A) ⊂ B is also infinite. This implies that B is countably infinite (that is, not finite).
Let ϕ : N → B be a bijection and ϕ(n) = bn . That is, we put B = {b1 , b2 , . . . }.
Now let k1 be the first integer such that bk ∈ f (A); let k2 ≥ k1 be the next integer so that bk2 ∈ f (A),
etc. Then
f (A) = {bk1 , bk2 , bk3 , . . . }
Now define g : f (A) → N by g(bki ) = i which is clearly a bijection. The composed mapping g◦f : A → N
shows that A ∼ N as required.
¤
Corollary 1.10. Every subset of a countable set is countable.
Theorem 1.11. If A and B are countably infinite then the Cartesian product A × B is countably
infinite.
Proof. We may write A = {a1 , a2 , a3 , . . . } and B = {b1 , b2 , b3 , . . . }. We write A × B in the following
way.
(a1 , b1 ) (a1 , b2 ) (a1 , b3 ) . . .
(a2 , b1 ) (a2 , b2 ) (a2 , b3 ) . . .
(a3 , b1 ) (a3 , b2 ) (a3 , b3 ) . . .
...
...
...
...
Define ϕ : A × B → N by φ(a1 , b1 ) = 1, φ(a1 , b2 ) = 2, φ(a2 , b1 ) = 3, φ(a1 , b3 ) = 4, φ(a2 , b2 ) = 5,
φ(a3 , b1 ) = 6, φ(a1 , b4 ) = 7, . . .
¤
Theorem 1.12. The set Q of all rational numbers is countably infinite.
Theorem 1.13. For any a, b ∈ R, the interval (a, b) is uncountable.
Proof. It is sufficient to consider the interval (0, 1). We assume on the contrary that the set is countable.
Recall that any number in (0, 1) can be written as
0.a1 a2 a3 a4 · · · = a1 × 10−1 + a2 × 10−2 + a3 × 10−3 + . . .
where a1 , a2 , a3 , . . . are digits taken from 0 to 9. So we may list (0, 1) as {a1 , x2 , x3 , . . . } such that
x1 = a11 a12 a13 . . . a1n . . .
x2 = a21 a22 a23 . . . a2n . . .
x3 = a31 a32 a33 . . . a3n . . .
..
..
.
.
xn = an1 an2 an3 . . . ann . . .
..
..
.
.
So any number in (0, 1) can be found from the above list. We now defined a number as follows
(
7 aii 6= 7
x = b1 b2 b3 . . . bn . . .
where bi =
8 aii = 7
Then clearly the x differs from any number of {a1 , x2 , . . . } by at least one digit. That is, x is not on
the list. A contradiction.
¤