Math 140A - Notes
Neil Donaldson
March 7, 2025
Introduction
Analysis is one of the major sub-disciplines of mathematics, concerned with continuity, limits, calcu-
lus, and accurate approximations.
Analytic ideas date back thousands of years. For instance, Archimedes (c. 287–212 BC) used limit-
type approaches to approximate the circumference of a circle and to compute the area under a
parabola.
1
Philosophical objections to such ideas are just as old: how can it make sense to sum
infinitely many infinitesimally small quantities? This was part of a deeper debate among the ancient
Greeks and other cultures: is the matter comprising the natural world atomic (consisting of minute,
discrete, indivisible objects) or continuous (arbitrarily and infinitely divisible). Several of Zeno’s fa-
mous paradoxes (5
th
C. BC) grapple with such difficulties: Achilles and the Tortoise is essentially an
argument that the infinite series
∞
∑
n=1
1
2
n
=
1
2
+
1
4
+
1
8
+
1
16
+ ··· is meaningless.
1
2
1
2
2
1
2
3
1
2
4
···
0 1
1
2
3
4
···
As the picture suggests, with modern definitions it makes sense for this sum to evaluate to 1.
The development of calculus by Newton, Leibniz and others in the late 1600s permitted the easy
application of infinitesimal ideas to important problems in the sciences, though they did not prop-
erly address the ancient philosophical concerns. The main subject of this course (and its sequel) is
the rigorous logical development of the foundations of calculus: the triumph of 18
th
–19
th
century
mathematics. The critical notions of limit and continuity only became settled in during the early
1800s (courtesy of Bolzano, Cauchy, Weierstrass and others), with another 50 years passing before
Riemann’s thorough description of the definite integral.
In this course we consider sequences, limits, continuity and infinite series, with power series, dif-
ferentiation and integration relegated to the sequel. We begin with something more basic: to nu-
merically measure continuous quantities, we need to familiarize ourselves with the real numbers. A
concrete description is difficult, so we build up to it via the natural numbers and the rationals. . .
1
Archimedes’ circle is reminiscent of Riemann sums; his parabola requires evaluation of the infinite series
∞
∑
n=0
1
4
n
=
4
3
.
1 Completeness
1.1 The Set N of Natural Numbers
You’ve been using the natural numbers N = {1, 2, 3, 4, 5, . . .} since you first learned to count. In
mathematics, these must be axiomatically described. Here is one approach.
Axioms 1.1 (Peano). The natural numbers are a set N satisfying the following properties:
1. (Non-emptiness) N is non-empty.
2. (Successor function) There exists a function f : N → N. This is usually denoted ‘+1’ so that
we may write,
n ∈ N =⇒ n + 1 ∈ N
3. (Initial element) The successor function f is not surjective. Otherwise said, there is an element
1 ∈ range f which is not the successor of any element.
2
4. (Unique predecessor/order) f is injective. Otherwise said,
m + 1 = n + 1 =⇒ m = n
5. (Induction) Suppose A ⊆ N is a subset satisfying
(a) 1 ∈ A, (b) n ∈ A =⇒ n + 1 ∈ A.
Then A = N.
Axioms 1–4 state that N is defined by repeatedly adding 1 to the initial element; for instance
3 := f
f (1)
= f (1 + 1) = (1 + 1) + 1
Parts (a) and (b) of axiom 5 are the familiar base case and induction step a standard induction: let P
n
be
the proposition ‘n ∈ A’ to recover the usual form of the Principle of Mathematical Induction.
Example 1.2. Prove that 7
n
−4
n
is divisible by 3 for all n ∈ N.
Let A be the set of natural numbers for which 7
n
− 4
n
is divisible by 3. It is required to prove that
A = N.
(a) If n = 1, then 7
1
−4
1
= 3, whence 1 ∈ A.
(b) Suppose n ∈ A. Then 7
n
−4
n
= 3λ for some λ ∈ N. But then
7
n+1
−4
n+1
= 7 ·7
n
−4
n+1
= 7(3λ + 4
n
) −4
n+1
= 3 ·7λ + (7 −4) ·4
n
= 3
7λ + 4
n
is divisible by 3. It follows that n + 1 ∈ A.
Appealing to axiom 5, we see that A = N, hence result.
2
By convention, the first natural number is 1; we could use 0, x, α, or any symbol you wish!
2
What about the integers? The integers satisfy only axioms 1, 2 and 4. For instance:
3. The function f : Z → Z : n 7→ n + 1 is surjective (indeed bijective/invertible). The ‘initial
element’ 1 ∈ N is the successor of 0 ∈ Z.
Reversing this observation provides an explicit construction of Z from N: simply extend the succes-
sor function f so that every element has a unique predecessor: 0 is the unique predecessor of 1, −1
the unique predecessor of 0, etc. In essence we are forcing f (n) = n + 1 to be bijective!
Exercises 1.1. Key concepts/results: Peano’s Axioms, Induction
Most of these exercises are to refresh your memory of induction. Use either the language of Peano’s
axiom 5, or the (possibly) more familiar base-case/induction-step formulation.
1. Prove that 1
2
+ 2
2
+ ··· + n
2
=
1
6
n(n + 1)(2n + 1) for all natural numbers n.
2. Prove that 3 + 11 + ··· + (8n −5) = 4n
2
−n for all n ∈ N.
3. (a) Guess a formula for 1 + 3 + ··· + (2n −1) by evaluating the sum for n = 1, 2, 3, and 4.
(For n = 1 the sum is simply 1)
(b) Prove your formula using mathematical induction.
4. Prove that 11
n
−4
n
is divisible by 7 for all n ∈ N.
5. The principle of mathematical induction can be extended as follows. A list P
m
, P
m+1
, . . . of
propositions is true provided (i) P
m
is true, (ii) P
n+1
is true whenever P
n
is true and n ≥ m.
(a) Prove that n
2
> n + 1 for all integers n ≥ 2.
(b) Prove that n! > n
2
for all integers n ≥ 4. (recall that n! = n(n −1) ···2 ·1)
6. Prove (2n + 1) + (2n + 3) + (2n + 5) + ··· + (4n −1) = 3n
2
for all n ∈ N.
7. For each n ∈ N, let P
n
denote the assertion “n
2
+ 5n + 1 is an even integer”.
(a) Prove that P
n+1
is true whenever P
n
is true.
(b) For which n is P
n
actually true? What is the moral of this exercise?
8. For n ∈ N, let n! denote the factorial function (0! = 1) and define the binomial coefficient
n
k
=
n!
k!(n − k)!
for k = 0, 1, . . . , n
The binomial theorem asserts that, for all n ∈ N,
(a + b)
n
=
n
∑
k=0
n
k
a
n−k
b
k
= a
n
+ na
n−1
b +
n(n −1)
2
a
n−2
b
2
+ ··· + nab
n−1
+ b
n
(a) Show that
(
n
k
)
+
(
n
k−1
)
=
(
n+1
k
)
for k = 1, 2, . . . , n.
(b) Prove the binomial theorem by induction.
9. Show that Peano’s induction axiom is false for the set of integers Z by exhibiting a proper subset
A ⊂ Z which satisfies conditions (a) and (b).
10. Consider Z
3
= {0, 1, 2} under addition modulo 3. That is,
0 + 1 = 1, 1 + 1 = 2, 2 + 1 = 0
Which of Peano’s axioms are satisfied?
3
1.2 The Set Q of Rational Numbers
The rational numbers may be defined in several ways. For instance, we could consider the set of
relatively prime ordered pairs
Q =
(p, q) : p ∈ Z, q ∈ N, gcd(p, q) = 1
⊆ Z ×N
Things seem more familiar if we write
p
q
instead of (p, q) and adopt the convention that
λp
λq
=
p
q
for
any non-zero λ ∈ Z. The usual operations (+, ·, etc.) are easily defined, consistently with those for
the integers (Exercise 6).
An alternative approach involves equations. Each linear equation qx − p = 0 where p, q ∈ Z and
q = 0 corresponds to a rational number. For example
13x + 27 = 0 ↭ x = −
27
13
Of course the equation 26x + 54 = 0 also corresponds to the same rational number!
Extending this process naturally leads us to consider higher degree polynomials.
Definition 1.3. A number x is algebraic if it satisfies an equation of the form
3
a
n
x
n
+ a
n−1
x
n−1
+ ··· + a
1
x + a
0
= 0 ( ∗)
for some integers a
0
, . . . , a
n
.
Examples 1.4. 1.
√
2 is algebraic since it satisfies the equation x
2
−2 = 0.
2. x =
5
p
7 +
√
3 is also algebraic:
x
5
−7 =
√
3 =⇒ (x
5
−7)
2
= 3 =⇒ x
10
−14x
5
+ 46 = 0
The next result is helpful for deciding whether a given number is rational and can assist with factor-
izing polynomials.
Theorem 1.5 (Rational Roots). Suppose a
0
, . . . , a
n
∈ Z and that x ∈ Q satisfies (∗). If x =
p
q
is
rational, written in lowest terms, then p | a
0
and q | a
n
.
Proof. Substitute x =
p
q
into the polynomial equation and multiply through by q
n
to see that
a
n
p
n
+ a
n−1
p
n−1
q + ··· + a
1
pq
n−1
+ a
0
q
n
= 0
This is an equation in integers. All terms except the last contain a factor of p, whence p | a
0
q
n
. Since
gcd(p, q) = 1, it follows that p | a
0
. The result for q is almost identical: all but the first term above
has a factor of q.
3
You should be alarmed by this! We seem to have given up constructing new numbers and instead are merely describing
their properties. No matter, a construction of the real numbers will come later.
4
Examples 1.6. 1. We prove that
√
2 is irrational. Plainly x =
√
2 satisfies the polynomial equation
x
2
−2 = 0. If
√
2 =
p
q
were rational in lowest terms, then the rational roots theorem forces
p | 2 and q | 1 =⇒
√
2 ∈ {±1, ±2}
Since none of the values ±1, ±2 satisfy x
2
−2 = 0, we have a contradiction.
2. y = (
√
3 −1)
1/3
satisfies ( y
3
+ 1)
2
= 3, whence y
6
+ 2y
3
− 2 = 0. If y =
p
q
were rational in
lowest terms, then p | 2 and q | 1, whence y = ±1, ±2; none of which satisfy the polynomial.
3. z =
4+
√
3
5
1/2
satisfies 5z
2
−4 =
√
3, from which 25z
4
−40z
2
+ 13 = 0. If z =
p
q
were rational
in lowest terms, then p | 13 and q | 25. There are twelve possibilities: it is tedious to check, but
none satisfy the required polynomial,
z = ±1, ±13, ±
1
5
, ±
13
5
, ±
1
25
, ±
13
25
In this case it is easier to bypass the theorem: if z ∈ Q then
√
3 = 5z
2
−4 would also be rational!
4. We use the theorem to factorize the polynomial 3x
3
+ x
2
+ x −2 = 0. If x =
p
q
is a rational root,
then p | 2 and q | 3 give several possibilities:
x ∈
±1, ±2, ±
1
3
, ±
2
3
It doesn’t take long to check that x =
2
3
is the only rational root. A factor of 3x − 2 may be
extracted by long division to obtain
3x
3
+ x
2
+ x −2 = (3x −2)(x
2
+ x + 1)
The quadratic has no real roots: absent complex numbers, the factorization is complete.
It is far from clear that non-algebraic (transcendental) numbers exist: e and π are the most famous.
These satisfy no polynomial equation with integer coefficients, though demonstrating such is tricky.
Exercises 1.2. Key concepts: Algebraic Numbers, Rational Roots Theorem/Testing for Irrationality
1. Describe all linear equations corresponding to the rational number
101
29
.
2. Show that
√
3,
√
5 and
√
24 are not rational numbers: what are the relevant polynomials?
3. Show that 2
1/3
and 13
1/4
are not rational numbers.
4. Show that ( 2 +
√
2)
1/2
and ( 5 −
√
3)
1/3
are irrational.
5. Explain why 4 −7b
2
must be rational if b is rational.
6. Given rational numbers (p, q), (r, s) as ordered pairs, what are (p, q) + (r, s) and (p, q) ·(r, s)?
7. Let n ∈ N. Use the rational roots theorem to prove that
√
n ∈ Q ⇐⇒
√
n ∈ N.
8. In the proof of the rational roots theorem, explain why the condition gcd(p, q) = 1 allows us to
conclude that p | a
0
q
n
=⇒ p | a
0
5
1.3 Ordered Fields
We have thus far formally constructed the natural numbers and used them to build the integers and
rational numbers. It is a significantly greater challenge to construct the real numbers. We start by
thinking about ordered fields, of which both Q and R are examples.
Axioms 1.7. A field F is a set with two binary operations +, · which satisfy (for all a, b, c ∈ F),
4
Addition Multiplication
Closure a + b ∈ F ab ∈ F
Associativity a + ( b + c) = (a + b) + c a(bc) = (ab)c
Commutativity a + b = b + a ab = ba
Identity ∃0 ∈ F such that a + 0 = a ∃1 ∈ F such that a ·1 = a
Inverse ∃− a ∈ F such that a + (−a) = 0 If a = 0, ∃a
−1
∈ F such that aa
−1
= 1
Distributivity a(b + c) = ab + ac
A field F is ordered if we also have a binary relation ≤ which satisfies (again for all a, b, c ∈ F):
O1 a ≤ b or b ≤ a
O2 a ≤ b and b ≤ a =⇒ a = b
O3 a ≤ b and b ≤ c =⇒ a ≤ c
O4 a ≤ b =⇒ a + c ≤ b + c
O5 a ≤ b and 0 ≤ c =⇒ ac ≤ bc
For an ordered field, the symbol < is used in the usual manner: x < y ⇐⇒ x ≤ y and x = y.
As with Peano’s axioms for the natural numbers, these are not worth memorizing. Instead you
should quickly check that you believe all of them for your current understanding of the real numbers;
you can’t prove anything since the real numbers are yet to be defined!
Example 1.8. It is worth considering the rational numbers in a little more detail. These inherit a
natural ordering from Z and N:
p
q
≤
r
s
⇐⇒ ps ≤ qr (remember that q, s > 0)
It is now possible, though tedious, to prove that each of the axioms of an ordered field holds for Q,
using only basic facts about multiplication, addition and ordering within the integers. For instance,
4
Write multiplication · as juxtaposition unless necessary, and use the common shorthand a
2
= a · a. The field axioms
are very easy to remember if you know some abstract algebra:
• The addition axioms say that
F, +
is an abelian group.
• The multiplication axioms say that
F \{0}, ·
is an abelian group.
• The distributive axiom describes how addition and multiplication interact.
6
Commutativity of Multiplication Given a =
p
q
and b =
s
t
rational, we have
ab =
ps
qt
=
sp
tq
= ba
since multiplication of integers (numerator and denominator) is commutative.
O3 Suppose a ≤ b and b ≤ c. Write a =
p
q
, b =
r
s
and c =
t
u
where all three denominators are
positive. By assumption,
ps ≤ qr and ru ≤ st =⇒ psu ≤ qru ≤ qst
=⇒ pu ≤ qt (divide by s = 0)
=⇒ a =
p
q
≤
t
u
= c
Basic Results about ordered fields
As with the axioms of an ordered field, these are not worth memorizing.
Theorem 1.9. Let F be a ordered field with at least two elements 0 = 1. Then:
1. a + c = b + c =⇒ a = b 2. a ·0 = 0
3. (−a)b = −(ab) 4. (−a)(−b) = ab
5. ac = bc and c = 0 =⇒ a = b 6. ab = 0 =⇒ a = 0 or b = 0
7. a ≤ b =⇒ −b ≤ −a 8. a ≤ b and c ≤ 0 =⇒ bc ≤ ac
9. 0 ≤ a and 0 ≤ b =⇒ 0 ≤ ab 10. 0 ≤ a
2
11. 0 < 1 12. 0 < a =⇒ 0 < a
−1
13. 0 < a < b =⇒ 0 < b
−1
< a
−1
All these statements should be intuitive for the fields Q and R. Try proving a few using only the
axioms; they are most easily done in the order presented. For instance, part 2 might be proved as
follows:
a ·0 + 0 = a · 0 = a · (0 + 0) = a · 0 + a ·0 (additive identity/distibutive axioms)
=⇒ 0 = a · 0 (part 1)
We finish with a final useful ingredient.
Definition 1.10. In an ordered field F, the absolute value of an element a is
|
a
|
:=
(
a if a ≥ 0
−a if a < 0
7
Theorem 1.11. In any ordered field:
1.
|
a
|
≥ 0
2.
|
ab
|
=
|
a
|
·
|
b
|
3.
|
a + b
|
≤
|
a
|
+
|
b
|
(△-inequality)
4.
|
a −b
|
≥
||
a
|
−
|
b
||
(reverse/extended △-inequality)
All parts are straightforward if you consider the ±-cases separately for a, b.
Exercises 1.3. Key concepts: Ordered Field (Q an example arising naturally from Z), △-inequality
1. Which of the axioms of an ordered field fail for N? For Z?
2. Prove parts 11 and 13 of Theorem 1.9.
(Hint: You can use any of the parts that come before. . . )
3. (a) Prove that
|
a + b + c
|
≤
|
a
|
+
|
b
|
+
|
c
|
for all a, b, c ∈ R.
(Hint: Apply the triangle inequality twice. Don’t consider eight separate cases!)
(b) For any a
1
, . . . , a
n
∈ R, use induction to prove
|
a
1
+ a
2
+ ··· + a
n
|
≤
|
a
1
|
+
|
a
2
|
+ ··· +
|
a
n
|
4. (a) Show that
|
b
|
< a ⇐⇒ −a < b < a.
(b) Show that
|
a −b
|
< c ⇐⇒ b −c < a < b + c.
(c) Show that
|
a −b
|
≤ c ⇐⇒ b −c ≤ a ≤ b + c.
5. Let a, b ∈ R. Show that if a ≤ b
1
for every b
1
> b, then a ≤ b.
(Hint: draw a picture if you’re stuck. This is a very important example!)
6. In an ordered field, suppose that 0 ≤ a and 0 ≤ b. Explain carefully why 0 ≤ a + b.
7. Following Example 1.8, prove that Q satisfies axiom O5.
(Hint: if a =
p
q
, etc., what is meant by ac ≤ bc?)
8. (Hard!) The complex numbers C = {x + iy : x, y ∈ R} form a field. The lexicographic ordering
of C is defined by
x + iy ≤ p + iq ⇐⇒
(
x < p or
x = p and y ≤ q
Which of the order axioms O1–O5 are satisfied by the lexicographic ordering?
(Provide a counter-example if an axiom is not satisfied; don’t prove your claims if an axiom is satisfied.)
8
1.4 The Completeness Axiom, or Least Upper Bound Principle
Though we haven’t provided an explicit definition of the real numbers, you should be comfortable
that both Q and R are ordered fields. We now ask how these might be distinguished axiomatically.
Perhaps surprisingly, only one additional axiom is required! We first need some terminology.
Definition 1.12 (Maxima, Minima & Boundedness). Let S ⊆ R be non-empty.
1. S is bounded above if it has an upper bound M:
∃M ∈ R such that ∀s ∈ S, s ≤ M
2. We write M = max S, the maximum of S, if M is an upper bound for S and M ∈ S.
3. S bounded below, a lower bound m, and the minimum min S are defined similarly.
4. S is bounded if it is bounded both above and below. It is bounded by M if
∀s ∈ S,
|
s
|
≤ M (M is an upper bound, −M a lower bound)
Examples 1.13. 1. If S is a finite set, then it is bounded and has both a maximum and a minimum.
For instance, S = {−3, π, 12} has min S = −3 and max S = 12.
2. N has minimum 1, but no maximum. Z and Q have neither: both are unbounded.
3. The half-open interval S = [0, 3) is bounded, e.g. by M = 5; it has minimum 0 but no maximum.
While this last is intuitive, it is worth giving an explicit argument, in this case by contradiction.
5
Suppose M = max S exists; necessarily 0 ≤ M < 3. We draw a picture to get the lay of the land:
since M ∈ S, we’ve placed it inside the interval, away from 3.
0 3
M
s =
M+3
2
The crux of the argument is to observe that there must be some s ∈ S which is larger than M,
the natural choice being the average s :=
1
2
(M + 3). Now observe that
3 − s = s − M =
1
2
(3 − M) > 0
In particular, s ∈ S and s > M. Since S contains an element larger than M, it follows that M
cannot be the maximum of S. In conclusion, S has no maximum.
Lemma 1.14. 1. If M is an upper bound for S, so is M + ε for any ε ≥ 0.
2. If M = max S exists, then it is unique.
Try proving these basic facts yourself.
5
S has a maximum means: ∃M ∈ S such that ∀s ∈ S, s ≤ M. We prove the negation ∀M ∈ S, ∃s ∈ S such that s > M.
9
Example 1.15. In a variation on the previous example, we show that the set
S = Q ∩ [0,
√
2) = {x ∈ Q : 0 ≤ x <
√
2}
has no maximum. The approach is similar to before: given a hypothetical maximum M, we find
some s ∈ S between M and
√
2. The challenge is that we can’t use the average
1
2
(M +
√
2) : this isn’t
rational (why?) and so doesn’t lie in S!
To fix this, we informally construct a sequence. Define s
n
to be
√
2 to n decimal places:
s
0
= 1, s
1
= 1.4 =
14
10
, s
2
= 1.41 =
141
100
, s
3
= 1.414 =
1414
1000
, . . .
Since any finite decimal is rational and 0 ≤ s
n
<
√
2, we see that s
n
∈ S. Moreover,
√
2 − s
n
≤ 10
−n
can be made arbitrarily small by choosing N sufficiently large.
Now suppose M = max S exists. Since M ∈ S, we have M <
√
2. Choose any N ∈ N large enough
so that 10
−N
<
√
2 −M (any integer N > −log
10
(
√
2 −M) will do!). Certainly s
N
∈ S and moreover,
√
2 − s
N
≤ 10
−N
<
√
2 − M =⇒ M < s
N
The purported maximum M is plainly not an upper bound for S: contradiction.
≈
0 1 M
√
2
s
0
s
N−1
s
N
10
−N
Suprema and Infima
We generalize the idea of maximum and minimum values to any bounded sets.
Definition 1.16. Let S ⊆ R be non-empty.
1. If S is bounded above, its supremum sup S is its least upper bound. Otherwise said,
(a) sup S is an upper bound: ∀s ∈ S, s ≤ sup S,
(b) sup S is the least such: if M is an upper bound, then sup S ≤ M.
2. Similarly, if S is bounded below, its infimum inf S is its greatest lower bound:
(a) inf S is a lower bound: ∀s ∈ S, inf S ≤ s,
(b) inf S is the greatest such: if m is a lower bound, then m ≤ inf S.
sup Sinf S sm M
S
upper bounds
lower bounds
10
Example 1.17. The interval S = [2, 5) has sup S = 5 and inf S = 2 (= min S). We verify these claims:
(a), (b) are the properties in the definition.
(a) Since s ∈ S ⇐⇒ 2 ≤ s < 5, we see that 5 is indeed an upper bound and 2 a lower bound.
(b) We demonstrate the contrapositive. Suppose M < 5 and define
6
s = max{
1
2
(M + 5), 4}. Then
M < s < 5 and s ∈ S. It follows that M is not an upper bound for S. The least upper bound is
therefore sup S = 5.
For the infimum: if m > 2, define t = min{
1
2
(m + 2), 4} to see that 2 < t < m and t ∈ S, whence
m is not a lower bound.
2 3 4 5
S
sup S
inf S
m
M
t s
Axiom 1.18 (Completeness of R). If S ⊆ R is non-empty and bounded above, then sup S exists (and
is a real number!).
It is precisely this property that distinguishes the real numbers from the rationals.
7
Certainly every
bounded set S of rational numbers has a supremum; the issue is that sup S need not be rational!
By reflecting across zero (Exercise 9), we obtain the same thing for the infimum.
Theorem 1.19 (Existence of Infima). If S ⊆ R non-empty and bounded below, then inf S ∈ R exists.
A Useful Contrapositive Part (b) of the Definition is plainly a biconditional: if sup S ≤ M, then M
is at least as large as an upper bound and is therefore also an upper bound for S (Lemma 1.14)! As in
Example 1.17, one often uses the contrapositive of part (b):
M < sup S if and only if M is not an upper bound for S.
Unpacking this further using the meaning of upper bound (and substituting x for M) we recover a
useful result that will be used repeatedly.
Lemma 1.20. 1. Let S be bounded above. Then x < sup S ⇐⇒ ∃s ∈ S such that x < s.
2. Let S be bounded below. Then y > inf S ⇐⇒ ∃t ∈ S such that t < y.
sup S
inf S
st
y
x
S
6
The number 4 is merely an arbitrary element to make sure s ∈ S in case M were huge and negative!
7
More formally (the details are too much for us): if F is an ordered field with 0 = 1 and which satisfies the completeness
axiom, then F is isomorphic to the real numbers.
11
Examples 1.21. We state the following without proof or calculation. You should be able to justify
everything using the definition, or by mirroring Example 1.17.
1. A bounded set has many possible bounds, but only one supremum or infimum.
2. If S has a maximum, then max S = sup S. Similarly, if a minimum exists, then min S = inf S.
3. (Example 1.15) S = Q ∩ [0,
√
2) has sup S =
√
2: this is a set of rational numbers whose supre-
mum is not rational.
4. S = Q ∩(π, 4) has sup S = 4, inf S = π, and no maximum nor minimum.
5. S = {
1
n
: n ∈ N} = {. . . ,
1
4
,
1
3
,
1
2
, 1} has sup S = max S = 1, inf S = 0, and no minimum.
6. S =
∞
S
n=1
[n, n +
1
2
) = [1, 1.5) ∪ [2, 2.5) ∪[3, 3.5) ∪ ··· has inf S = 1. It is not bounded above.
7. S =
∞
T
n=1
[
1
n
, 1 +
1
n
) has inf S = 1 = sup S since S = {1}.
The Archimedean Property and the Density of the Rationals
We finish this section by discussing a crucial property related to completeness, and of the distribution
of the rational numbers among the reals.
Theorem 1.22 (Archimedean Property). If b > 0 is a real number, then ∃n ∈ N such that n > b.
More generally: a, b > 0 =⇒ ∃n ∈ N such that an > b.
We assume nothing about R except that is an ordered field satisfying the completeness axiom and
where 0 = 1 (footnote 7). The natural numbers in this context are defined as the subset
N = {1, 1 + 1, 1 + 1 + 1, . . .} ⊆ R
and Peano’s axioms are a theorem.
Proof. Suppose the result were false. Then ∃b > 0 such that n ≤ b for all n ∈ N; that is, N is bounded
above! By completeness, sup N exists, and we trivially see that
0 < 1 =⇒ sup N < sup N + 1 =⇒ sup N −1 < sup N
By Lemma 1.20, ∃n ∈ N such that n > sup N −1. But then sup N < n + 1 which is clearly a natural
number! Thus sup N is not an upper bound for N: contradiction.
For the more general statement, simply replace b with
b
a
.
The use of completeness is necessary: there exist non-Archimedean ordered fields!
Example (1.15, cont.). The Archimedean property is precisely what is needed to justify the existence
of an integer N > −log
10
(
√
2 − M).
12
Corollary 1.23 (Density of Q in R). Between any two real numbers, there exists a rational number.
The idea is hopefully straightforward: given a < b, stretch the interval by an integer factor n until it
contains an integer m, before dividing by n to obtain a <
m
n
< b. We use the Archimedean property
to establish the existence of the scale factors m, n.
Proof. Suppose WLOG that 0 ≤ a < b, and apply the Archimedean property to
1
b−a
> 0:
∃n ∈ N such that n >
1
b−a
A second application (or trivially if a = 0) says ∃k ∈ N such that k > an. Now consider the set
J := {j ∈ N : an < j ≤ k}
and define m = min J: this exists since J is a finite non-empty set of natural numbers.
8
≈
0
a b an bn
k
m
m
n
J
Clearly m > an > m −1, since m = min J. But then m ≤ an + 1 < bn. We conclude that
an < m < bn =⇒ a <
m
n
< b
By iterating this result we see that any interval ( a, b) contains infinitely many rational numbers. It can
moreover be established that the irrational numbers are also dense in R (Exercise 6).
Exercises 1.4. Key concepts: Suprema, Completeness (distinguishes R), Contrapositive criterion,
Archimedean property/Density of Q ⊂ R
1. Decide whether each set is bounded above and/or below. If so, state its supremum and/or
infimum (no working is required).
(a) ( 0, 1) (b) {2, 7} (c) {0}
(d)
∞
S
n=1
[2n, 2n + 1] (e)
1 −
1
3
n
: n ∈ N
(f) {r ∈ Q : r
2
< 2}
(g)
∞
S
n=1
1 −
1
n
, 1 +
1
n
(h) {
1
n
: n ∈ N and n is prime} (i) {cos(
nπ
3
) : n ∈ N}
2. Modelling Example 1.15, sketch an argument that S = Q ∩(π, 4] has no minimum.
(Hint: let s
n
be π rounded up to n decimal places)
3. Let S be a non-empty, bounded subset of R.
(a) Prove that inf S ≤ sup S.
(b) What can you say about S if inf S = sup S?
8
This part of the argument is necessary since, in this context, we haven’t established the well-ordering property of N
(essentially Peano’s fifth axiom).
13
4. Let S and T be non-empty subsets of R with the property that s ≤ t for all s ∈ S and t ∈ T.
(a) Prove that S is bounded above and T bounded below.
(b) Prove that sup S ≤ inf T.
(c) Give an example of such sets S, T where S ∩ T is non-empty.
(d) Give an example of such sets S, T where S ∩ T is empty, and sup S = inf T.
5. Prove that if a > 0 then there exists n ∈ N such that
1
n
< a < n.
6. Let I = R \Q be the set of irrational numbers. Given real numbers a < b, prove that there exists
x ∈ I such that a < x < b.
(Hint: First show {r +
√
2 : r ∈ Q} ⊆ I)
7. Let A, B be non-empty bounded subsets of R, and let S be the set of all sums
S := {a + b : a ∈ A, b ∈ B}
(a) Prove that sup S = sup A + sup B.
(b) Prove that inf S = inf A + inf B.
8. Show that sup{r ∈ Q : r < a} = a for each a ∈ R.
9. We prove Theorem 1.19 on the existence of the infimum.
Let S ⊆ R be non-empty and let m be a lower bound for S. Define T = {t ∈ R : −t ∈ S} by
reflecting S across zero.
0
S
T
inf S = −sup T
m
sup T
−m s−s
(a) Prove that −m is an upper bound for T.
(b) By completeness (Axiom 1.18), sup T exists. Prove that inf S = −sup T by verifying Defi-
nition 1.16 parts 2(a) and (b).
14
1.5 The Symbols ±∞
Thus far the only subsets of the real numbers that have a supremum are those which are non-empty
and bounded above. In this very short section, we introduce the ∞-symbol to provide all subsets of the
real numbers with both a supremum and an infimum.
Definition 1.24. Let S ⊆ R be any subset. If S is bounded above/below, then sup S/inf S are as in
Definition 1.16. Otherwise:
1. We write sup S = ∞ if S is unbounded above, that is
∀x ∈ R, ∃s ∈ S such that s > x
2. We write inf S = −∞ if S is unbounded below,
∀y ∈ R, ∃t ∈ S such that t < y
3. By convention, sup ∅ := −∞ and inf ∅ := ∞, though these will rarely be of use to us.
The symbols ±∞ have no other meaning (as yet); in particular, they are not numbers. If one is willing
to abuse notation and write x < ∞ and y > −∞ for any real numbers x, y, then the conclusions of
Lemma 1.20 are precisely statements 1 & 2 above!
Examples 1.25. 1. sup R = sup Q = sup Z = sup N = ∞, since all are unbounded above. We also
have inf R = inf Q = inf Z = −∞ (recall that inf N = min N = 1).
2. If a < b, then any interval [a, b], (a, b), [a, b) or (a, b] has supremum b and infimum a, even if one
end is infinite. For example,
S = (7, ∞) = {x ∈ R : x > 7}
has sup S = ∞ and inf S = 7.
3. Let S = {x ∈ R : x
3
−4x < 0}. With a little factorization, we see that
x
3
−4x = x(x −2)(x + 2) < 0 ⇐⇒ x < −2 or 0 < x < 2
It follows that S = (−∞, −2) ∪ (0, 2) , from which sup S = 2 and inf S = −∞.
Exercises 1.5. Key concepts: ±∞ are shorthands for unboundedness: they are not numbers!
1. Give the infimum and supremum of each of the following sets:
(a) {x ∈ R : x < 0} (b) {x ∈ R : x
3
≤ 8}
(c) {x
2
: x ∈ R} (d) {x ∈ R : x
2
< 8}
2. Let S ⊆ R be non-empty, and let −S = {−s : s ∈ S}. Prove that inf S = −sup(−S).
3. Let S, T ⊆ R be non-empty such that S ⊆ T. Prove that inf T ≤ inf S ≤ sup S ≤ sup T.
4. If sup S < inf S, what can you say about S?
15
1.6 A Development of R (non-examinable)
The comment in footnote 7 constitutes a synthetic definition of the real numbers: there is essentially
just one set with the required properties. While this might satisfy an algebra-addict, it is nice to be
able to provide an explicit construction. The following approach uses so-called Dedekind cuts.
First one defines N, Z and Q. Use Peano’s axioms and proceed as in sections 1.1 and 1.2. The
operations +, · and ≤ are defined, first on N and then for Z and Q building on these concepts for the
integers.
Definition 1.26. A Dedekind cut α
∗
is a non-empty proper subset of Q with the properties:
1. (Closed downwards) If r ∈ α
∗
and s ∈ Q with s < r, then s ∈ α
∗
.
2. (No maximum) If M is an upper bound for α
∗
, then M ∈ α
∗
.
Define R to be the set of all Dedekind cuts!
The rough idea is that a real number α corresponds to the Dedekind cut α
∗
of all rational numbers less
than α.
Examples 1.27. 1. For any rational number r, the corresponding real number is the Dedekind cut
r
∗
= {x ∈ Q : x < r}
For instance 4
∗
= {x ∈ Q : x < 4} is the Dedekind cut definition of the real number 4.
2. It is a little trickier to explicitly define cuts corresponding to irrational numbers, though some
are relatively straightforward. For instance the real number
√
2 would be the set
√
2
∗
= {x ∈ Q : x < 0 or x
2
< 2}
It remains to prove that the set of Dedekind cuts satisfies the axioms of a complete ordered field. The
full details are too much, so here is a rough overview.
• Define the ordering of Dedekind cuts via
α
∗
≤ β
∗
⇐⇒ α
∗
⊆ β
∗
One can now prove axioms O1–O3 and that the ordering corresponds to that of Q.
• Define addition of cuts via
α
∗
+ β
∗
:= {a + b : a ∈ α
∗
, b ∈ β
∗
}
This suffices to prove the addition axioms and O4: a careful definition of −α
∗
is required.
• Multiplication is horrible: if α
∗
, β
∗
≥ 0
∗
then
α
∗
β
∗
:= {ab : a ≥ 0, a ∈ α
∗
, b ≥ 0, b ∈ β
∗
}∪{q ∈ Q : q < 0}
which may be carefully extended to cover situations when α
∗
or β
∗
< 0
∗
. Once this has been
done, one can then prove the multiplication axioms, the final order axiom O5, and the distribu-
tive axiom.
16
• The completeness axiom must also be verified, though it comes almost for free! If A ⊆ R (a set
of Dedekind cuts), then the supremum of A is simply
sup A =
[
α
∗
∈A
α
∗
Think about it. . .
An alternative approach to R using sequences of rational numbers will be given later.
Exercises 1.6. Key concepts: R is unnatural and difficult to construct in a logical manner
1. Show that if α
∗
, β
∗
are Dedekind cuts, then so is
α
∗
+ β
∗
= {r
1
+ r
2
: r
1
∈ α
∗
, r
2
∈ β
∗
}
2. Let α
∗
, β
∗
be Dedekind cuts and define the ‘product’:
α
∗
· β
∗
= {r
1
r
2
: r
1
∈ α
∗
, r
2
∈ β
∗
}
(a) Calculate some ‘products’ using the cuts 0
∗
, 1
∗
and (−1)
∗
.
(b) Discuss why this ‘product’ is unsatisfactory for defining multiplication in R.
3. We verify the Archimedean property (Theorem 1.22) using the Dedekind cut definition of R (it
is somewhat easier since the unboundedness of N and Q are baked in).
(a) Explain why every cut β
∗
is bounded above by some rational number.
(Hint: if β
∗
satisfies Definition 1.26 parts 1 & 2 but is unbounded above, then what is it?)
(b) If β
∗
> 0
∗
is a positive cut bounded above by
p
q
with p, q ∈ N, show that n := p + 1
corresponds to a cut for which n
∗
> β
∗
.
17
