MATH2241
Introduction to Mathematical Analysis
2022-2023
• Chapter 1 The Real Numbers
• Chapter 2 Sequences
• Chapter 3 Series
• Chapter 4 Limits and Continuity
• Chapter 5 Differentiation
• Chapter 6 The Riemann Integral
1
1
The Real Numbers
The foundation of calculus rests upon the concept of real numbers, so let us begin with a brief
description of the properties of the real numbers. Since you are probably familiar with most of such
properties, the aim of this chapter is to review the important features and persuade you that all the
properties could be traced back to only a few basic ones, namely, the field axioms, the order axioms
and the completeness axiom.
1.1
The Field Axioms
On the set R of real numbers there are two binary operations, denoted by + and ·, called addition
and multiplication respectively. By a binary operation ∗ on a set S we mean for every ordered pair
of elements (a, b) ∈ S × S we can associate a unique element a ∗ b in S. This suggests that for every
ordered pair (a, b) ∈ R × R, we can associate a unique real number a + b, called the sum of a and b,
and a unique real number a · b, called the product of a and b. We will assume that these two binary
operations on R satisfy the axioms listed in this section.
We begin with the axioms for addition:
(A1) a + b = b + a for all a, b ∈ R (Commutative Law for Addition)
(A2) a + (b + c) = (a + b) + c for all a, b, c ∈ R (Associative Law for Addition)
(A3) There exists an element of R, called an additive identity and denoted by 0, such that a + 0 = a
for any a ∈ R (Existence of an Additive Identity)
(A4) For any a ∈ R, there exists an element −a ∈ R, called an additive inverse or a negative of a,
such that a + (−a) = 0 (Existence of an Additive Inverse/Negative)
We will first establish some basic facts which are easy consequences of (A1)-(A4):
Proposition 1.1. (Uniqueness of Additive Identity) If x is an element of R such that a + x = a for
any a ∈ R, then x = 0.
Proof.
x = x + 0 by (A3)
= 0 + x by (A1)
= 0
by assumption with a = 0
Proposition 1.2. (Uniqueness of Additive Inverse) Let a ∈ R. If x is an element of R such that
a + x = 0, then x = −a.
Proof.
x =
=
=
=
=
=
=
x+0
x + (a + (−a))
(x + a) + (−a)
(a + x) + (−a)
0 + (−a)
(−a) + 0
−a
2
by
by
by
by
by
by
by
(A3)
(A4)
(A2)
(A1)
assumption
(A1)
(A3)
Proposition 1.1 and 1.2 allow us to say “the additive identity” of R and “the additive inverse”
(or “the negative:) of an element of R.
Proposition 1.3. Let a ∈ R. We have −(−a) = a.
Proof. Since
(−a) + a = a + (−a) = 0
by (A1) and (A4)
and
(−a) + (−(−a)) = 0
by (A4),
we conclude from Proposition 1.2 that −(−a) = a.
Exercise
(i) Let a ∈ R. Prove that if x, y are elements of R such that a + x = a + y, then x = y.
(ii) Let a, b ∈ R. Prove that −(a + b) = (−a) + (−b).
(iii) Prove that −0 = 0.
Next we list the axioms for multiplication:
(M1) a · b = b · a for all a, b ∈ R (Commutative Law for Multiplication)
(M2) a · (b · c) = (a · b) · c for all a, b, c ∈ R (Associative Law for Multiplication)
(M3) There exists an element of R, called a multiplicative identity and denoted by 1, such that
a · 1 = a for any a ∈ R (Existence of a Multiplicative Identity)
(M4) For any a 6= 0 ∈ R, there exists an element a−1 ∈ R, called a multiplicative inverse or a
reciprocal of a, such that a · a−1 = 1 (Existence of a Multiplicative Inverse/Reciprocal)
Here are some simple consequences of (M1)-(M4):
Proposition 1.4. (Uniqueness of Multiplicative Identity) If x is an element of R such that a · x = a
for any a ∈ R, then x = 1.
Proposition 1.5. (Uniqueness of Multiplicative Inverse) Let a 6= 0 in R. If x is an element of R
such that a · x = 1, then x = a−1 .
Proposition 1.6. Let a 6= 0 ∈ R. We have (a−1 )−1 = a.
We omit the proofs for these propositions which can be easily translated from those for Propositions 1.1, 1.2 and 1.3. It follows from Propositions 1.4 and 1.5 that we can talk about “the multiplicative identity” of R and “the multiplicative inverse” (or “the reciprocal”) of a nonzero element
of R.
Exercise
(i) Let a 6= 0 ∈ R. Prove that if x, y are elements of R such that a · x = a · y, then x = y.
(ii) Let a 6= 0, b 6= 0 and a · b 6= 0 ∈ R. Prove that (a · b)−1 = a−1 · b−1 . (Remark: In fact there is no
need to assume a · b 6= 0 as we shall see later in Proposition 1.10)
(iii) Prove that 1−1 = 1.
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The following axiom, known as the distributive law, connects the two operations:
(D) For any a, b, c ∈ R, a · (b + c) = a · b + a · c (Distributive Law)
Here are more consequences involving the distributive law:
Proposition 1.7. For any a, b, c ∈ R, (a + b) · c = a · c + b · c.
Proof.
(a + b) · c = c · (a + b) by (M1)
= c · a + c · b by (D)
= a · c + b · c by (M1)
Proposition 1.8. For any a ∈ R, a · 0 = 0.
Proof. Since
a+a·0 =
=
=
=
a·1+a·0
a · (1 + 0)
a·1
a
by
by
by
by
(M3)
(D)
(A3)
(M3),
we conclude from Proposition 1.1 that a · 0 = 0.
Proposition 1.9. We have (−1) · (−1) = 1.
Proof. Since
(−1) + (−1) · (−1) =
=
=
=
(−1) · 1 + (−1) · (−1)
(−1) · (1 + (−1))
(−1) · 0
0
by
by
by
by
(M3)
(D)
(A4)
Proposition 1.8,
we conclude from Proposition 1.2 with a = −1 that (−1) · (−1) = −(−1), which is equal to 1 by
Proposition 1.3 with a = 1.
The following shows an important property of multiplication of real numbers - the product of two
real numbers is zero only when one (or both) of these numbers is zero.
Proposition 1.10. For any a, b ∈ R, if a · b = 0, then a = 0 or b = 0.
Proof. It suffices to assume a 6= 0 and prove that b = 0. If a 6= 0, then there exist a unique
multiplicative inverse a−1 ∈ R. Multiplying a · b by a−1 gives
a−1 · (a · b) = (a−1 · a) · b by (M2)
= 1·b
by (M4)
= b
by (M3).
On the other hand, since we have assumed that a · b = 0, a−1 · (a · b) = a−1 · 0 = 0 by Proposition
1.8. Hence we have b = 0.
4
Exercise
(i) Prove that a · (−b) = −(a · b) for any a, b ∈ R.
(ii) Prove that (−1) · a = −a for any a ∈ R.
Remarks
(i) The additive identity 0 and the multiplicative identity 1 are normally required to be distinct.
What would happen if 0 = 1?
(ii) Subtraction and Division are defined respectively by
a − b := a + (−b)
and
a/b := a · b−1
provided that b 6= 0. Note that division by 0 is excluded by (M4).
(iii) From now on we will drop the use of the dot to indicate multiplication and write ab for a · b.
(iv) We will write a1 = a, a2 = aa, a3 = (a2 )a, and in general, an+1 = (an )a for any n ∈ N. Moreover,
if a 6= 0, then we write a0 = 1 and a−n = 1/(an ) for any n ∈ N.
(v) Let S be a set with two binary operations “+” and “·”. Substitute S in place of R in (A1)-(A4),
(M1)-(M4) and (D), and assume that “+” and “·” satisfy all these axioms together with 0 6= 1. Then
S is called a field. Hence the set R of real numbers is an example of a field. This is why the axioms
(A1)-(A4), (M1)-(M4) and (D) are known as the field axioms.
1.2
The Order Axioms
Next we would like to capture the idea that the set of real numbers has a natural order with some
axioms. We assume that there is a nonempty subset P of R, called the set of positive real numbers,
satisfying
(P1) If a, b ∈ P, then a + b ∈ P.
(P2) If a, b ∈ P, then ab ∈ P.
(P3) If a ∈ R, then exactly one of the following holds:
a ∈ P,
a = 0,
−a ∈ P.
The above axioms are known as the order axioms. Note that 0 ∈
/ P by axiom (P3). We introduce
the following notations:
• a > b means a − b ∈ P and a < b means b − a ∈ P.
• a ≥ b means a − b ∈ P ∪ {0} and a ≤ b means b − a ∈ P ∪ {0}.
With these notations the notion of inequality between two real numbers is defined in terms of the
set P. Moreover, the order axioms can now be rewritten as
(P1) If a > 0 and b > 0, then a + b > 0.
(P2) If a > 0 and b > 0, then ab > 0.
(P3) If a ∈ R, then exactly one of the following holds:
a > 0,
a = 0,
5
a < 0.
From this version of the order axioms, we see that (P3) divides R into three distinct types of
elements and hence it is usually called the Trichotomy Law. The Trichotomy Law also implies that
for any a, b ∈ R, exactly one of the following holds:
a > b,
a = b,
a < b.
The following are some consequences of the order axioms:
Proposition 1.11. We have 1 > 0.
Proof. Suppose it is not true that 1 > 0. Since we have assumed 1 6= 0, it follows from (P3) that
−1 > 0. Then
−1 > 0 ⇒ (−1)(−1) > 0 by (P2)
⇒ 1>0
by Proposition 1.9.
But −1 > 0 and 1 > 0 cannot be both true by (P3) and hence we arrive at a contradiction.
Proposition 1.12. For any a, b ∈ R, we have a > b if and only if −a < −b.
Proof.
a>b ⇔
⇔
⇔
⇔
a−b>0
−(−a) − b > 0
−b − (−a) > 0
−a < −b
by
by
by
by
definition
Proposition 1.3
(A1)
definition.
Proposition 1.13. For any a, b, c ∈ R, we have
(i) a ≤ b or b ≤ a
(ii) a ≤ b and b ≤ a ⇒ a = b
(iii) a ≤ b and b ≤ c ⇒ a ≤ c
Proof. The proofs for (i) and (ii) are left as exercise. For (iii), if a = b or b = c, then there is nothing
to prove. Hence we need only consider the case where a < b and b < c. Applying (A1)-(A4) we see
that
c − a = (c − b) + (b − a)
and it follows from (P1) that c − a > 0 as required.
Proposition 1.14. For any a, b, c ∈ R,
(i) if a < b, then a + c < b + c.
(ii) if a < b and 0 < c, then ca < cb.
(iii) if a < b and c < 0, then ca > cb.
Proof. (i) For any a, b, c ∈ R, if a < b, then (b + c) − (a + c) = b − a > 0. Hence a + c < b + c as
required. (ii) For any a, b, c ∈ R, if a < b and 0 < c, then c(b − a) > 0 by (P2). But from (D) and the
exercise, we have c(b − a) = cb + c(−a) = cb − ca. Hence ca < cb as required. (iii) Similar to (ii).
Exercise
(i) Prove that a2 ≥ 0 for any a ∈ R.
(ii) Prove that if a ≤ b and c ≤ d, then a + c ≤ b + d.
(iii) Prove that if a > 0, then a−1 > 0.
(iv) Prove that if a, b > 0 and a < b, then b−1 < a−1 .
6
By the Trichotomy Law (P3), given any a ∈ R,
Thus the function | · | : R → R given by

 a
0
|a| =

−a
exactly one of a > 0, a = 0 or a < 0 will hold.
if a > 0
if a = 0
if a < 0
is well-defined. This is known as the absolute value function. We leave some basic properties of the
absolute value function as exercise:
Exercise
(i) Prove that | − a| = |a| for any a ∈ R.
(ii) Prove that |ab| = |a||b| for any a, b ∈ R.
(iii) Prove that for any a ∈ R and for any c ≥ 0, |a| ≤ c if and only if −c ≤ a ≤ c.
(iv) Prove that −|a| ≤ a ≤ |a| for any a ∈ R.
The following inequality will be of particular importance:
Theorem 1.1. (Triangle Inequality) For any a, b ∈ R, we have |a + b| ≤ |a| + |b|.
Proof. One can prove the inequality case by case. Alternatively, from the exercise, we have
−|a| ≤ a ≤ |a| and
− |b| ≤ b ≤ |b|.
Also these inequalities can be added and hence we have
−(|a| + |b|) ≤ a + b ≤ |a| + |b|.
It follows that |a + b| ≤ |a| + |b| (see the above exercise).
Exercise
(i) Prove that |a − b| ≤ |a| + |b| for any a, b ∈ R.
(ii) Prove that | |a| − |b| | ≤ |a − b| for any a, b ∈ R.
1.3
The Completeness Axiom
Recall that a set with two binary operations satisfying the field axioms is called a field. Any field
that satisfies the order axioms is called an ordered field. Thus R is an ordered field. Another example
of an ordered field is given by the set Q of rational numbers - note that Q also satisfies the field and
order axioms. However, the set Q is not a nice ordered field to work with in the sense that it has
many missing points or gaps - for example, there is no r ∈ Q such that r2 = 2. Thus Q is lacking
in some ways and the goal of this section is to look at an essential property of R - a property that
is not shared by the rational numbers. In order to describe this special property and to handle the
notion of gaps better, we begin with the following definitions.
Definition 1.1. Let A be a nonempty subset of R.
(i) The set A is bounded above if there exists u ∈ R such that a ≤ u for all a ∈ A. Each such u
is called an upper bound of A.
(ii) The set A is bounded below if there exists l ∈ R such that l ≤ a for all a ∈ A. Each such l is
called a lower bound of A.
(iii) The set A is bounded if it is bounded above and below, otherwise it is said to be unbounded.
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Example
(i) Let A = {a ∈ R : 1 ≤ a < 5}. Then A is bounded above and 5, 9, 20.123 are examples of upper
bounds; it is also bounded below and 1, -3 are examples of lower bounds. It follows that the set A
is bounded.
(ii) The set B = {b ∈ R : b < n} is bounded above but not below, and hence it is unbounded.
(iii) The set N of natural numbers is bounded below. The “obvious” fact that N is not bounded
above will be discussed later.
Definition 1.2. Let A be a nonempty subset of R.
(i) The number u ∈ R is a supremum (or a least upper bound) of A if u is an upper bound of A,
and u ≤ u0 for any upper bound u0 of A.
(ii) The number l ∈ R is an infimum (or a greatest lower bound) of A if l is an lower bound of
A, and l0 ≤ l for any lower bound l0 of A.
Although a set can have many upper bounds (or lower bounds), it can have only one supremum
(or infimum).
Theorem 1.2. Let A be a nonempty subset of R. If A has a supremum or an infimum, then these
numbers are unique.
Proof. If u1 and u2 are both suprema of A, then by Definition, we have u1 ≤ u2 and u2 ≤ u1 . Hence
u1 = u2 . One can use a similar argument to show the uniqueness for infimum.
Theorem 1.2 suggests that we can refer to the supremum and the infimum of a set A. We shall
write these numbers as sup A and inf A respectively.
Example Let A = {a ∈ R : 0 ≤ a < 1}. Clearly 1 is an upper bound of A. Now suppose that
there is a smaller upper bound b. Then b < 1, and as it is an upper bound, we have b ≥ 0. Hence
b+1
b+1
b+1
1
2 ≤ 2 < 1, which means 2 ∈ A. But 2 is greater than b, contradicting the fact that b is an
upper bound. We conclude that sup A = 1. Similar argument shows that inf A = 0.
Observe that sup A and inf A may or may not be elements of A. If it happens that the supremum
belongs to the set A, then it is called the maximum of A, and is denoted by max A. Similarly, if
the infimum belongs to A, then it is called the minimum of A, and is denoted by min A. Thus the
supremum (or infimum) of a set can exist and not be the maximum (or minimum), but when the
maximum (or minimum) exists then it must also be the supremum (or infimum). For example, the
set A in the above example has no maximum, but 0 is its minimum (which is also its infimum).
Remark If A is a finite subset of R, then both the maximum and minimum of A exist.
By definition, to show that u = sup A, one has to show that u is an upper bound of A and that
any other upper bound of A must be greater than or equal to u. The following provides an equivalent
way of characterizing suprema – no number smaller than the supremum can be an upper bound of
the set.
Theorem 1.3. Let A be a nonempty subset of R and let u ∈ R be an upper bound of A. Then
u = sup A if and only if for any > 0, there exists a ∈ A such that u − < a.
Proof. Suppose u = sup A. For any > 0, consider the number u − . Since u − < u, it implies
by definition, that u − is NOT an upper bound of A. It follows that there must be some element
a ∈ A such that u − < a.
Conversely, assume u is an upper bound of A satisfying the condition that for any > 0, there
exists a ∈ A such that u − < a. If v is any number less than u, then we let = u − v > 0. Hence
there exists a ∈ A such that v = u − < a. This implies that v cannot be an upper bound of A.
Thus if u0 is an upper bound, then we must have u ≤ u0 . This proves that u = sup A.
8
The analogous characterization for infima is the following exercise.
Theorem 1.4. Let A be a nonempty subset of R and let l ∈ R be a lower bound of A. Then l = inf A
if and only if for any > 0, there exists a ∈ A such that a < l + .
Example Let A be a nonempty subset of R. Define
−A = {−a : a ∈ A} = {b : b = −a for some a ∈ A}.
Prove that
(i) A is bounded above by u if and only if −A is bounded below by −u.
(ii) sup A exists in R if and only if inf(−A) exists in R. When one of these numbers exists, we have
inf(−A) = − sup A.
Proof (i)
A is bounded above by u
⇔
⇔
⇔
⇔
∀a ∈ A, a ≤ u
∀a ∈ A, −a ≥ −u
∀b ∈ −A, b ≥ −u
−A is bounded below by −u.
(ii) Suppose sup A exists in R. Since sup A is an upper bound of A, − sup A is a lower bound of
−A by (i). To show that inf(−A) exists and equals − sup A, we are going to apply Theorem 1.4,
that is, we will show that for any > 0, there exists b ∈ −A such that b < − sup A + . But from
Theorem 1.3, we have for any > 0, there exists a ∈ A such that sup A − < a, or equivalently,
−a < − sup A + . Thus we are done if we take b = −a. Similar argument shows the other direction.
Let us turn our attention back to the essential property which distinguishes Q and R. The
following axiom is our final axiom about R concerning the existence of suprema:
Completeness Axiom: Let A ⊆ R be a nonempty set which is bounded above. Then sup A exists
in R.
In other words, the Completeness Axiom asserts that every nonempty subset of R that has an
upper bound also has a supremum in R. The analogous statement for the existence of infima can be
deduced from the Completeness Axiom (there is no need to introduce another axiom):
Theorem 1.5. Let A ⊆ R be a nonempty set which is bounded below. Then inf A exists in R.
Proof. Suppose that A is a nonempty subset of R that is bounded below. Then the nonempty set
−A is bounded above. By the Completeness Axiom, sup(−A) exists in R, and it follows that inf A
exists in R (and is equal to − sup(−A)).
Example Let A be a nonempty subset of R which is bounded above, and let c ∈ R. Define the sets
c + A and cA by
c + A := {c + a : a ∈ A} and cA := {ca : a ∈ A}.
(i) Prove that the set c + A is bounded above and that sup(c + A) = c + sup A.
(ii) If c ≥ 0, prove that the set cA is bounded above and sup(cA) = c sup A.
(iii) What can we say about the set cA when c < 0? Is cA bounded above/below ? What can we say
about its supremum/infimum ?
9
Proof (i) For any a ∈ A, we have a ≤ sup A. Hence c + a ≤ c + sup A for any a ∈ A and so the set
c + A is bounded above by c + sup A. By the definition of supremum, we have sup(c + A) ≤ c + sup A.
On the other hand, for any x ∈ c + A, we have x ≤ sup(c + A). Then x − c ≤ sup(c + A) − c for any
x ∈ c + A. But each x ∈ c + A can be written as c + a for some a ∈ A. So the last inequality becomes
(c + a) − c ≤ sup(c + A) − c for any c + a ∈ c + A, or equivalently, a ≤ sup(c + A) − c for any a ∈ A.
It follows that sup(c + A) − c is an upper bound for A and hence sup A ≤ sup(c + A) − c, that is,
c + sup A ≤ sup(c + A). We have shown that sup(c + A) ≤ c + sup A and c + sup A ≤ sup(c + A).
The result follows.
Alternatively, for any > 0, it implies that there exists a ∈ A such that sup A − < a. Adding
c to both sides of the inequality yields (c + sup A) − < c + a. Since each element x ∈ c + A can
be written as c + a for some a ∈ A, it follows that for any > 0, there exists x ∈ c + A such that
(c + sup A) − < x, and hence sup(c + A) = c + sup A.
(ii) First of all, the statement is certainly true for the case c = 0. Thus we assume in the following
that c > 0. For any a ∈ A, we have a ≤ sup A and so ca ≤ c sup A as c > 0. Hence the set cA
is bounded above by c sup A. By the definition of supremum, we have sup(cA) ≤ c sup A. On the
other hand, for any ca ∈ cA, we have ca ≤ sup(cA). Since c > 0, multiplying c−1 to both sides of
the inequality yields c−1 (ca) ≤ c−1 sup(cA) for any ca ∈ cA, or equivalently, a ≤ c−1 sup(cA) for any
a ∈ A. It follows that c−1 sup(cA) is an upper bound for A and hence sup A ≤ c−1 sup(cA), that is,
c sup A ≤ sup(cA). The result follows.
Alternatively, for any > 0, we have c−1 > 0 as c > 0. It implies that there exists a ∈ A such
that sup A − c−1 < a. Multiplying c to both sides yields c sup A − < ca. Hence we have proved
that for any > 0, there exists x = ca ∈ cA such that c sup A − < x. So sup(cA) = c sup A.
(iii) For any a ∈ A, we have a ≤ sup A and so c sup A ≤ ca as c < 0 in this case. Hence the set
cA is bounded below by c sup A. Now we claim that inf(cA) = c sup A. For any > 0, we have
−c−1 > 0 as c < 0. It implies that there exists a ∈ A such that sup A + c−1 = sup A − (−c−1 ) < a.
Multiplying c to both sides yields ca < c sup A + . Hence we have proved that for any > 0, there
exists x = ca ∈ cA such that x < c sup A + . So inf(cA) = c sup A.
Example Let A be a nonempty subset of R and let f, g : A → R be functions with the same domain
A. Denote by ranf and rang the range of f and g respectively, that is,
ran f = {f (x) : x ∈ A} and ran g = {g(x) : x ∈ A}.
Suppose that ran f and ran g are bounded subsets of R.
(i) Prove that ran (f + g) = {f (x) + g(x) : x ∈ A} is also a bounded subset of R.
(ii) Prove that
sup ran (f + g) ≤ sup ran f + sup ran g,
that is,
sup{f (x) + g(x) : x ∈ A} ≤ sup{f (x) : x ∈ A} + sup{g(x) : x ∈ A}.
(iii) Prove that
inf ran f + inf ran g ≤ inf ran (f + g),
that is,
inf{f (x) : x ∈ A} + inf{g(x) : x ∈ A} ≤ inf{f (x) + g(x) : x ∈ A}.
(iv) For each of (ii) and (iii), give examples to show that the inequality can be either the equality or
strict inequality.
Proof (i) Since it is given that ran f and ran g are bounded subsets of R, there exist m, m0 , M, M 0 ∈
R such that m ≤ f (x) ≤ M and m0 ≤ g(x) ≤ M 0 for any x ∈ A. Therefore, we have m + m0 ≤
10
f (x) + g(x) ≤ M + M 0 for any x ∈ A, which means ran (f + g) = {f (x) + g(x) : x ∈ A} is also a
bounded subset of R.
(ii) First of all, since ran f , ran g and ran (f + g) are bounded subsets of R, their suprema and
infima exist in R. Let α = sup ran f and β = sup ran g. Then we have f (x) ≤ α and g(x) ≤ β for
any x ∈ A. This implies f (x) + g(x) ≤ α + β for any x ∈ A and hence α + β is an upper bound for
ran (f + g). It follows that sup ran (f + g) ≤ α + β and we are done.
(iii) The proof for this part is completely analogous to the proof for part (ii). Just replace each
supremum by infimum, upper bound by lower bound, and reverse the direction of each inequality.
(iv) Let A = R and let f (x) = g(x) = 1 for any x ∈ A. Then sup ran (f + g) = 2 = 1 + 1 =
sup ran f + sup ran g, and inf ran f + inf ran g = 1 + 1 = 2 = inf ran (f + g). The next is an
example with strict inequalities. Let A = R and let
0
if x 6= 0
0
if x 6= 0
f (x) =
and g(x) =
1
if x = 0
−1
if x = 0
Then we have sup ran f = 1, inf ran f = 0, sup ran g = 0 and inf ran g = −1. Furthermore, we
have f (x) + g(x) = 0 for any x ∈ A and hence sup ran (f + g) = 0 = inf ran (f + g). It follows that
sup ran (f + g) = 0 < 1 = 1 + 0 = sup ran f + sup ran g, and inf ran f + inf ran g = 0 + (−1) =
−1 < 0 = inf ran (f + g).
We mentioned earlier that Q is missing some numbers and is therefore not a nice ordered field
to work with. We can now say more about these missing numbers in terms of the suprema of some
subsets of Q - there are subsets of Q that are bounded above but for which no suprema exist in Q.
It is exactly these suprema that are missing from Q. As an example, we shall show later that the set
{x ∈ Q : x2 < 2}
is one such subset. It follows that the ordered field Q does not satisfy the Completeness Axiom, and
hence the completeness property is a property that we can use to distinguish Q from R. We say that
R is a complete ordered field but Q is not.
1.4
Consequences of Completeness
We will discuss in this section some important applications of the Completeness Axiom on R. The
first one is a result that guarantees the existence of square roots.
Theorem 1.6. (The Existence of Square Roots) There exists a unique positive real number a such
that a2 = 2.
Proof. Consider the set A = {x ∈ R : x2 < 2}. Since 12 < 2, so 1 ∈ A and A is nonempty. Moreover,
A is bounded above by 2 since if x > 2, then x2 > 4 and hence x ∈
/ A. It then follows from the
Completeness Axiom that A has a supremum in R and we write a = sup A. We remark that a ≥ 1 > 0
is positive. Now we are going to prove a2 = 2 by ruling out the possibilities a2 < 2 and a2 > 2.
(i) Assume that a2 < 2. The idea here is to find a positive real number h which is small enough
so that (a + h)2 < 2. This will then suggest that a + h ∈ A and hence contradict the fact that a is
an upper bound of A. To see how to choose h, we note that
(a + h)2 = a2 + 2ah + h2 < a2 + 2ah + ah = a2 + 3ah
if we require that 0 < h < a. Furthermore,
a2 + 3ah < a2 + (2 − a2 ) = 2
11
if we require that 0 < h <
Therefore, if we choose
2−a2
3a
(note that we are assuming a2 < 2 and hence
h=
then we have h > 0, h < a and h <
2−a2
3a .
2−a2
3a
is positive).
2 − a2
1
min a,
,
2
3a
It then follows from our construction of h that
(a + h)2 < 2,
which means we have constructed a number a + h ∈ A, contradicting the fact that a is an upper
bound of A. Hence the case a2 < 2 is not possible.
(ii) Assume that a2 > 2. This time the idea is to find a positive real number h which is small
enough so that (a − h)2 > 2. Since a is the supremum of A, by Theorem 1.3, there exists b ∈ A such
that a − h < b. This implies (a − h)2 < b2 < 2, contradicting our choice of h. To see how to choose
h, we note that
(a − h)2 = a2 − 2ah + h2 > 2 + h2 > 2
if we require that a2 − 2ah > 2, that is, h <
is positive). Therefore, if we choose
a2 −2
2a
h=
then we have 0 < h <
a2 −2
2a .
1
2
(note that we are assuming a2 > 2 and hence
a2 − 2
2a
a2 −2
2a
It then follows from our construction of h that
(a − h)2 > 2.
Now since h > 0 and a = sup A, by Theorem 1.3, there exists b ∈ A such that a − h < b. But then
this implies (a − h)2 < b2 < 2, contradicting (a − h)2 > 2. Hence the case a2 > 2 is also not possible.
Since it is not possible to have a2 < 2 and a2 > 2, by the Trichotomy Law, we must have a2 = 2.
The uniqueness part follows from Proposition 1.10. Suppose c is a positive real number such that
c2 = 2, then
a2 = c2 ⇒ a2 − c2 = 0 ⇒ (a + c)(a − c) = 0.
We then conclude that c = a or c = −a. But a > 0 and so −a < 0, meaning that a is the unique
positive number satisfying a2 = 2.
Remarks √
(i) We write 2 or 21/2 for this unique a.
(ii) A small modification of the above proof can be applied to show that for any a ≥ 0, there exists
√
√
a unique real number, denoted by a or a1/2 , whose square is a. The number a is known as the
positive square root of a.
(iii) Using the binomial theorem one can also show that for any a ≥ 0, there exists a unique positive
√
n-th root of a, denoted by n a or a1/n , for each n ∈ N.
What happens if we replace the set A in Theorem 1.6 by the set of rational numbers
B = {x ∈ Q : x2 < 2}?
The set B consists only of rational numbers whose squares are all less than 2. It is certainly nonempty
and bounded above, and hence b = sup B exists by the Completeness Axiom. If b is rational, then
the same argument as in Theorem 1.6 suggests that b2 = 2. But this is impossible (why ?) 1 and
1
Suppose b = p/q where p and q have no common factors. It implies that p2 = 2q 2 . Thus p2 is even and also p itself
is even. It implies that q 2 is even, and also q is even. p and q have a common factor. This is contradiction.
12
hence the set B does not have a supremum which belongs to Q. Thus we have found a nonempty
subset of Q which is bounded above but for which the supremum does not exist in Q. In other words,
the ordered field Q does not satisfy the Completeness Axiom, as we have mentioned in the last part
of the previous section.
Another important consequence of the Completeness Axiom is known as the Archimedean Property which basically states that the set N of natural numbers is not bounded above in R. Although
this property may seem obvious, it depends on the Completeness Axiom (together with the property
that N is closed under addition). After studying the Archimedean Property, we will also look at how
the set Q is sitting inside R – the density of Q in R.
Theorem 1.7. (Archimedean Property) For any x ∈ R, there exists an n ∈ N such that x < n, i.e.,
N is not bounded above.
Proof. Assume to the contrary that N is bounded above. By the Completeness Axiom, N should
have a supremum. We set a = sup N. Now consider the number a − 1, which is smaller than the
supremum and so by Theorem 1.3, there exists an n ∈ N such that a − 1 < n. But then we have
a < n + 1. Since n + 1 ∈ N, we have a contradiction to the fact that a is an upper bound for N.
Corollary 1.1. For any real number r > 0, there exists an n ∈ N such that
1
n
< r.
Proof. This follows easily from Theorem 1.7 by letting x = 1/r.
Remark The above corollary is often referred to as the Archimedean Property as well.
Example Let
A={
1
1
: n ∈ N} = {a : a = for some n ∈ N}.
n
n
Prove that inf A = 0.
Proof. Clearly, A is a nonempty subset of R bounded below by 0. We will use Theorem 1.3 to show
that 0 is the infimum of A. For any > 0, the Archimedean Property (the corollary) implies that
there exists n ∈ N such that n1 < . Thus for any > 0, there exists a ∈ A such that a < . It then
follows from Theorem 1.4 that inf A = 0.
Exercise
(i) Let A ⊆ N be a nonempty set. Prove that if A is bounded above, then sup A ∈ A (and hence
max A exists and is equal to sup A).
(ii) Let A ⊆ N be a nonempty set. Prove that inf A ∈ A (and hence min A exists and is equal to
inf A).
(iii) For any x ≥ 0, prove that there exists m ∈ N such that m − 1 ≤ x < m. (Hint: Consider the set
A = {n ∈ N : x < n}. Use the Archimedean Property (Theorem 1.7) to show that A is nonempty
and then use (ii) to construct m.)
The next result will concern how the set Q fits inside R:
Theorem 1.8. (Density of Q in R) For any x, y ∈ R with x < y, there exists a rational number
r ∈ Q such that x < r < y.
Proof. Without loss of generality, we may assume that x ≥ 0. The case where x < 0 follows quickly
from this proof. The main idea here is to find m, n ∈ N so that
x<
m
< y.
n
13
Since x < y, we have y − x > 0 and it follows from the Archimedean Property that there exists an
n ∈ N such that n1 < y − x, or equivalently, there exists an n ∈ N such that
nx + 1 < ny.
Next we apply Exercise (iii) to nx and so there exists m ∈ N such that m − 1 ≤ nx < m. It follows
that
nx < m ≤ nx + 1 < ny,
so that nx < m < ny and therefore
x<
m
< y,
n
as desired.
Theorem 1.8 shows the “betweenness property” for Q in the sense that given any two real numbers
there is a rational number between them. We say that Q is dense in R. Without working too hard,
we can use this result to show that the set of irrational numbers is also dense in R:
Corollary 1.2. For any x, y ∈ R with x < y, there exists an irrational number s such that x < s < y.
√
Proof. Exercise (one may use the fact that 2 is irrational).
There is a natural class of subsets of R given by intervals. If a, b ∈ R with a < b, then we have
(a, b) = {x ∈ R : a < x < b},
[a, b] = {x ∈ R : a ≤ x ≤ b},
[a, b) = {x ∈ R : a ≤ x < b},
(a, b] = {x ∈ R : a < x ≤ b}.
Intervals of the type (a, b) are called open intervals with the endpoints being excluded in the set.
Intervals of the type [a, b] are called closed intervals with the endpoints being included in the set.
Intervals using both square and round brackets are called half-closed intervals or half-open intervals.
In the special case a = b, then [a, b] = [a, a] = a is a singleton set, whereas the others are all equal to
empty set. So far these intervals are all bounded (both above and below). If a ∈ R, then
(a, ∞) = {x ∈ R : a < x},
[a, ∞) = {x ∈ R : a ≤ x},
(−∞, a) = {x ∈ R : x < a},
(−∞, a] = {x ∈ R : x ≤ a}.
These are all unbounded intervals: (a, ∞) and [a, ∞) have no upper bounds, and (−∞, a) and (−∞, a]
have no lower bounds. We sometimes write (−∞, ∞) for R. Note that ∞ and −∞ are just convenient
symbols, and they are NOT real numbers. Do not apply any theorem or property that is stated for
real numbers to ∞ and −∞.
Now suppose we have a collection of intervals I1 , I2 , I3 , ... such that
I1 ⊇ I2 ⊇ I3 ⊇ · · · ⊇ Ik ⊃ Ik+1 ⊇ · · ·
that is, In+1 is a subset of In for each n ∈ N. Such a collection of intervals is called a nested sequence
of intervals.
14
For example, if we let In = [0, 1/n] for each n ∈ N. Then we have In ⊇ In+1 for each n ∈ N
and hence this is a nested sequence of intervals. We note that 0 is a point which belongs to every
In . Moreover, 0 is the only common point to all In (this follows from the Archimedean Property, see
Corollary 1.1). We can write this as
∞
\
In = {0}.
n=1
Thus the intersection of all the In is nonempty (and is a single point in the above example). However,
the situation is completely different if the intervals in the sequence are not closed. For example, if
In = (0, 1/n) for each n ∈ N. Then the intersection of all the In in this nested sequence will be
empty.
Now we would like to know under what assumptions on the intervals will the intersection of all
intervals be nonempty? The following result will provide us the answer, and the Completeness Axiom
plays an important role in the proof of it:
Theorem 1.9. (Nested Intervals Property) For each n ∈ N, if In = [an , bn ] is a nested sequence of
closed and bounded intervals, then there exists x ∈ R such that x ∈ In for all n ∈ N, that is, we have
∞
\
In 6= ∅.
n=1
Proof. Since In = [an , bn ] is a nested sequence of intervals, we have
a1 ≤ a2 ≤ a3 ≤ · · · ≤ an ≤ · · · ≤ bn ≤ · · · ≤ b3 ≤ b2 ≤ b1 .
Consider the set A = {an : n ∈ N} of left endpoints of the intervals In :
Thus each bn is an upper bound for A. By the Completeness Axiom, the supremum x = sup A exists.
We now claim that x ∈ In for all n ∈ N. Consider a particular In = [an , bn ]. Since x is an upper
bound for A, we have x ≥ an . But bn is an upper bound for A and x is the least upper bound. Thus
we have x ≤ bn . Hence x ∈ In and since this is true for every n ∈ N, we conclude that
x∈
∞
\
In
n=1
and the intersection is nonempty.
Remark (and Exercise) We have seen an example earlier that if the nested intervals are not closed,
then the intersection is empty. Thus the closedness assumption cannot be removed from the theorem.
Similarly, the theorem is not true if the boundedness assumption is removed. Construct an example
of a nested sequence of unbounded intervals such that the intersection of all of them is empty.
Exercise Construct an example of a nested sequence of closed and bounded intervals such that the
intersection of all of them is not a single point.
Remark Use the Nested Intervals Property to prove that [0, 1] is uncountable (and hence R is
uncountable).
15
Proof. Recall that a set A is uncountable if there is no bijective function f : N → A.
Assume to the contrary that there exists a bijective function f : N → [0, 1]. We construct a
nested sequence of closed and bounded intervals as follows:
• First consider f (1). Define I1 to be any nondegenerate closed and bounded interval in [0, 1]
(i.e., it is of the form [a, b] with a < b) which does NOT contain f (1).
• Next, define I2 to be any nondegenerate closed and bounded interval in I1 which does NOT
contain f (2). Since I2 ⊆ I1 , so automatically it will NOT contain f (1) as well.
• Repeating this process gives us a nested sequence of nondegenerate closed and bounded intervals
I1 ⊇ I2 ⊇ I3 ⊇ · · · ⊇ Ik ⊇ Ik+1 ⊇ · · ·
Now by the Nested Intervals Property, there exists x ∈ In for all n ∈ N. It is not hard to see that
x 6= f (n) for any n ∈ N. Note that if x = f (n0 ) for some n0 ∈ N, then x ∈
/ In0 by the construction of
In0 (contradiction). Hence we have found an element x in [0, 1] such that x 6= f (n) for any n ∈ N ,
contradicting the assumption that f : N → [0, 1] is bijective. We conclude that there is no bijective
function f : N → [0, 1], and hence [0, 1] is uncountable.
16