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.

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

C. BC) grapple with such difﬁculties: Achilles and the Tortoise is essentially an

argument that the inﬁnite series

∞

∑

n=1

+ ··· is meaningless.

···

0 1

···

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

–19

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. . .

Archimedes’ circle is reminiscent of Riemann sums; his parabola requires evaluation of the inﬁnite series

∞

∑

n=0

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.

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

the proposition ‘n ∈ A’ to recover the usual form of the Principle of Mathematical Induction.

Example 1.2. Prove that 7

−4

is divisible by 3 for all n ∈ N.

Let A be the set of natural numbers for which 7

− 4

is divisible by 3. It is required to prove that

A = N.

(a) If n = 1, then 7

−4

= 3, whence 1 ∈ A.

(b) Suppose n ∈ A. Then 7

−4

= 3λ for some λ ∈ N. But then

n+1

−4

n+1

= 7 ·7

−4

n+1

= 7(3λ + 4

) −4

n+1

= 3 ·7λ + (7 −4) ·4

= 3



7λ + 4



is divisible by 3. It follows that n + 1 ∈ A.

Appealing to axiom 5, we see that A = N, hence result.

By convention, the ﬁrst natural number is 1; we could use 0, x, α, or any symbol you wish!

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

+ ··· + n

n(n + 1)(2n + 1) for all natural numbers n.

2. Prove that 3 + 11 + ··· + (8n −5) = 4n

−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

−4

is divisible by 7 for all n ∈ N.

5. The principle of mathematical induction can be extended as follows. A list P

, P

m+1

, . . . of

propositions is true provided (i) P

is true, (ii) P

n+1

is true whenever P

is true and n ≥ m.

(a) Prove that n

> n + 1 for all integers n ≥ 2.

(b) Prove that n! > n

for all integers n ≥ 4. (recall that n! = n(n −1) ···2 ·1)

6. Prove (2n + 1) + (2n + 3) + (2n + 5) + ··· + (4n −1) = 3n

for all n ∈ N.

7. For each n ∈ N, let P

denote the assertion “n

+ 5n + 1 is an even integer”.

(a) Prove that P

n+1

is true whenever P

is true.

(b) For which n is P

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





k!(n − k)!

for k = 0, 1, . . . , n

The binomial theorem asserts that, for all n ∈ N,

(a + b)

∑

k=0





n−k

= a

+ na

n−1

b +

n(n −1)

n−2

+ ··· + nab

n−1

+ b

(a) Show that

(

)

(

k−1

)

(

n+1

)

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

= {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?

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

instead of (p, q) and adopt the convention that

λp

λq

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 = −

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

+ a

n−1

+ ··· + a

x + a

= 0 ( ∗)

for some integers a

, . . . , a

Examples 1.4. 1.

√

2 is algebraic since it satisﬁes the equation x

−2 = 0.

2. x =

7 +

√

3 is also algebraic:

−7 =

√

3 =⇒ (x

−7)

= 3 =⇒ x

−14x

+ 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

, . . . , a

∈ Z and that x ∈ Q satisﬁes (∗). If x =

rational, written in lowest terms, then p | a

and q | a

Proof. Substitute x =

into the polynomial equation and multiply through by q

to see that

+ a

n−1

q + ··· + a

n−1

+ a

= 0

This is an equation in integers. All terms except the last contain a factor of p, whence p | a

. Since

gcd(p, q) = 1, it follows that p | a

. The result for q is almost identical: all but the ﬁrst term above

has a factor of q.

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.

Examples 1.6. 1. We prove that

√

2 is irrational. Plainly x =

√

2 satisﬁes the polynomial equation

−2 = 0. If

√

2 =

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 = 0, we have a contradiction.

2. y = (

√

3 −1)

1/3

satisﬁes ( y

+ 1)

= 3, whence y

+ 2y

− 2 = 0. If y =

were rational in

lowest terms, then p | 2 and q | 1, whence y = ±1, ±2; none of which satisfy the polynomial.

3. z =



√



1/2

satisﬁes 5z

−4 =

√

3, from which 25z

−40z

+ 13 = 0. If z =

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, ±

, ±

In this case it is easier to bypass the theorem: if z ∈ Q then

√

3 = 5z

−4 would also be rational!

4. We use the theorem to factorize the polynomial 3x

+ x

+ x −2 = 0. If x =

is a rational root,

then p | 2 and q | 3 give several possibilities:

x ∈



±1, ±2, ±

, ±



It doesn’t take long to check that x =

is the only rational root. A factor of 3x − 2 may be

extracted by long division to obtain

+ x

+ x −2 = (3x −2)(x

+ 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

2. Show that

√

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 +

√

1/2

and ( 5 −

√

1/3

are irrational.

5. Explain why 4 −7b

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

=⇒ p | a

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),

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:

≤

⇐⇒ 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,

Write multiplication · as juxtaposition unless necessary, and use the common shorthand a

= 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.

Commutativity of Multiplication Given a =

and b =

rational, we have

ab =

= ba

since multiplication of integers (numerator and denominator) is commutative.

O3 Suppose a ≤ b and b ≤ c. Write a =

, b =

and c =

where all three denominators are

positive. By assumption,

ps ≤ qr and ru ≤ st =⇒ psu ≤ qru ≤ qst

=⇒ pu ≤ qt (divide by s = 0)

=⇒ a =

≤

= 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

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 if a ≥ 0

−a if a < 0

Theorem 1.11. In any ordered ﬁeld:

≥ 0

a + b

≤

(△-inequality)

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

≤

for all a, b, c ∈ R.

(Hint: Apply the triangle inequality twice. Don’t consider eight separate cases!)

(b) For any a

, . . . , a

∈ R, use induction to prove

+ a

+ ··· + a

≤

+ ··· +

4. (a) Show that

< a ⇐⇒ −a < b < a.

(b) Show that

a −b

< c ⇐⇒ b −c < a < b + c.

a −b

≤ c ⇐⇒ b −c ≤ a ≤ b + c.

5. Let a, b ∈ R. Show that if a ≤ b

for every b

> 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 =

, 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.)

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,

≤ 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.

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

s =

M+3

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 :=

(M + 3). Now observe that

3 − s = s − M =

(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.

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.

Example 1.15. In a variation on the previous example, we show that the set

S = Q ∩ [0,

√

2) = {x ∈ Q : 0 ≤ x <

√

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

(M +

√

2) : this isn’t

rational (why?) and so doesn’t lie in S!

To ﬁx this, we informally construct a sequence. Deﬁne s

to be

√

2 to n decimal places:

= 1, s

= 1.4 =

, s

= 1.41 =

141

100

, s

= 1.414 =

1414

1000

, . . .

Since any ﬁnite decimal is rational and 0 ≤ s

√

2, we see that s

∈ S. Moreover,

√

2 − s

≤ 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

(

√

2 −M) will do!). Certainly s

∈ S and moreover,

√

2 − s

≤ 10

−N

√

2 − M =⇒ M < s

The purported maximum M is plainly not an upper bound for S: contradiction.

≈

0 1 M

√

N−1

−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

upper bounds

lower bounds

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

s = max{

(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{

(m + 2), 4} to see that 2 < t < m and t ∈ S, whence

m is not a lower bound.

2 3 4 5

sup S

inf S

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.

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

The number 4 is merely an arbitrary element to make sure s ∈ S in case M were huge and negative!

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.

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 = {

: n ∈ N} = {. . . ,

, 1} has sup S = max S = 1, inf S = 0, and no minimum.

6. S =

∞

n=1

[n, n +

) = [1, 1.5) ∪ [2, 2.5) ∪[3, 3.5) ∪ ··· has inf S = 1. It is not bounded above.

7. S =

∞

n=1

[

, 1 +

) 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

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

(

√

2 − M).

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 <

< 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

b−a

> 0:

∃n ∈ N such that n >

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.

≈

a b an bn

Clearly m > an > m −1, since m = min J. But then m ≤ an + 1 < bn. We conclude that

an < m < bn =⇒ a <

< 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)

∞

n=1

[2n, 2n + 1] (e)



1 −

: n ∈ N



(f) {r ∈ Q : r

< 2}

(g)

∞

n=1



1 −

, 1 +



(h) {

: n ∈ N and n is prime} (i) {cos(

nπ

) : n ∈ N}

2. Modelling Example 1.15, sketch an argument that S = Q ∩(π, 4] has no minimum.

(Hint: let s

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?

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).

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.

(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

< 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.

inf S = −sup T

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).

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

−4x < 0}. With a little factorization, we see that

−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

≤ 8}

: x ∈ R} (d) {x ∈ R : x

< 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?

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

∗

= {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

√

∗

= {x ∈ Q : x < 0 or x

< 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

∗

≤ β

∗

⇐⇒ α

∗

⊆ β

∗

One can now prove axioms O1–O3 and that the ordering corresponds to that of Q.

• Deﬁne addition of cuts via

∗

+ 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

∗

:= {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.

• 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 =

[

∗

∈A

∗

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

∗

+ β

∗

= {r

+ r

: r

∈ α

∗

, r

∈ β

∗

}

2. Let α

∗

, β

∗

be Dedekind cuts and deﬁne the ‘product’:

∗

· β

∗

= {r

: r

∈ α

∗

, r

∈ 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

with p, q ∈ N, show that n := p + 1

corresponds to a cut for which n

∗

> β

∗

2 Sequences

Sequences are the fundamental tool in our approach to analysis.

2.7 Limits of Sequences

Deﬁnition 2.1. A sequence of real numbers is a list indexed by the natural numbers

) = (s

, s

, . . .)

We call s

the initial term/element.

More formally, we could view a sequence as a function s

: N → R. Other letters may be used

, b

, etc.), though s

is typical in the abstract. It is also common to have sequences which start with

a different initial term (e.g., n = 0). If you need to be explicit, write, e.g., (s

)

∞

n=0

Examples 2.2. 1. Explicit sequences are often deﬁned by providing a formula for the n

term. For

instance, s



1 +



deﬁnes a sequence whose ﬁrst three terms are

= 2, s

, s

, . . .

Since each term is a rational number, (s

) could be described as a rational sequence.

2. Sequences can be deﬁned inductively. For instance, if t

= 1 and t

n+1

= 3t

−1, then

) = (1, 2, 5, 14, 41, . . .)

3. u

−4

deﬁnes a sequence with initial term u

)

∞

n=3



, . . .



Limits We are typically most interested in what happens to the terms of a sequence when n gets

large (one reason it is common to be non-explicit as to the initial term). In elementary calculus you

should have become used to examples such as

lim

+ 3n −1

−2

which encapsulates the idea that the expression s

+3n−1

−2

gets ‘close to’

when n is large. We

can easily convince ourselves of this with a calculator/computer: to 4 decimal places,

) =



4, 1.3, 1.04, 0.9348, 0.8767, 0.8396, 0.8138, 0.7947, . . .



, s

1000

= 0.6677

Our primary business is to make this idea logically watertight, the major issue being what is meant

by ‘close to.’ In Section 2.8 we will do so by developing the formal deﬁnition of limit. Before seeing

this, we quickly refresh a few simple examples. All these examples can be made formal later, but for

the present just rely on your intuition and experience: it is essential to have a good idea of the correct

answer before you try to prove it!

If there are multiple letters around, writing lim

n→∞

with a subscript can aid the reader.

Examples 2.3. 1. lim

= 0. Our instinct is that s

becomes arbitrarily small as n becomes large.

2. lim

7n+9

2n−4

. To convince yourself of this, you might write

7n+9

2n−4

2−

and observe that the

terms become tiny as n increases.

3. The sequence with n

term s

= (−1)

does not converge to anything: it diverges.

)

∞

n=0

= (1, −1, 1, −1, 1, −1, . . .)

4. If c

cos



πn



, then lim c

= 0. To see this, observe that the cosine term lies between ±1,

while

has limit 0.

5. The sequence deﬁned inductively by s

= 2, s

n+1

+ 3 begins

) =



2, 4, 5,

, . . .



This appears to have limit lim s

= 6. It is not hard to spot the pattern s

= 6 −

, which may

easily be veriﬁed by induction: for the induction step, observe that

+ 3 =



6 −



+ 3 = 6 −

n+1

Exercises 2.7. Key concepts: Sequences, Use your intuition!

1. Decide whether each sequence converges; if it does, state the limit. No proofs are required; if

you’re unsure what’s going on, try writing out the ﬁrst few terms.

(a) a

3n + 1

(b) b

3n + 1

4n −1

(d) d

= sin



nπ



2. Repeat the previous question for sequences whose n

term is as follows:

(a)

+ 3

−3

(b) 1 +

1/n

(d) (−1)

n (e)

+ 8n

−31

(f) sin



nπ



(g) sin



2nπ



(h)

n+1

+ 5

−7

(i)



1 +



(j)

6n + 4

+ 7

3. Give an example of:

(a) A sequence (x

) of irrational numbers having a limit lim x

that is a rational number.

(b) A sequence (r

) of rational numbers having a limit lim r

that is an irrational number.

4. Prove by induction that the sequence deﬁned in Example 2.2.2 has n

term t



n−1

+ 1



5. In future courses, you’ll meet sequences of functions. For instance, we could deﬁne a sequence

( f

) of functions f

: R → R inductively via

(x) ≡ 1, f

n+1

(x) := 1 +

(t) dt

Compute the functions f

, f

and f

. The sequence ( f

) should seem familiar if you think back

to elementary calculus; why?

2.8 The Formal Deﬁnition of Limit

While intuition regarding limits is essential, mathematics is about proving things rigorously; for this

one needs a formal deﬁnition. The essential difﬁculty is that ‘close to’ is poorly deﬁned without

context; we get round this by considering all possible measures of closeness simultaneously.

Deﬁnition 2.4. To say that a sequence (s

) converges to a limit s ∈ R means,

∀ϵ > 0, ∃N such that n > N =⇒

−s

< ϵ

We write lim s

= s or simply s

→ s; both are read “s

approaches (or tends to) s.”

A sequence converges if it has a limit, and diverges otherwise.

Try reading it this way: given any measure of closeness, we can make n large enough so that (by our

given measure) s

is close to s. This isn’t as hard as it looks, though a lot of examples will likely be

necessary for it to sink in. . .

Example 2.5. We prove that the sequence with

= 2 −

√

converges to s = 2.

By plotting the sequence, we see how ϵ controls

the distance (closeness) from s

to the limit s = 2;

no matter the size of ϵ, we must show that (s

) has

some tail whose terms are closer to s.

0 20 40 60

s + ϵ

s − ϵ

s = 2

n larger than this. . .

. . . guarantees s

here

Proving such ‘for all, there exists’ statements requires a speciﬁc argument structure:

• Suppose ϵ > 0 has been given and describe N as a function of ϵ (ϵ smaller means N larger).

• Verify algebraically that n > N =⇒

−s

< ϵ.

Scratch work. To ﬁnd a suitable N, start with what you want to be true and let it inspire you.

Whenever n > N, we want ϵ >

−s





2 −

√



−2



√



Choosing N =

should be enough to complete the proof!

Warning! “N =

” is not the correct conclusion! To make it clear that we’ve satisﬁed the deﬁnition,

we rearrange our scratch work: these last three lines are all you need to write!

Formal argument. Suppose ϵ > 0 is given and let N =

. Then

n > N =⇒

−s



2 −

√

−2



√

= ϵ

Thus lim s

= 2, as required.

N may be a real number or a natural number (equivalent by the Archimedean property, Theorem 1.22). It tends to be

easier to use N ∈ R for convergence and N ∈ N when directly proving divergence (see Deﬁnition 2.9 and Examples 2.10).

Lemma 2.6 (Uniqueness of Limit). If (s

) converges, then its limit is unique.

The proof structure should be familiar from other uniqueness arguments: assume there are two limits

s = t and obtain a contradiction. The picture explains the strategy: by choosing ϵ =

s−t

in the

deﬁnition we obtain a tail of the sequence which must be simultaneously close to both limits.

s + et − e

s+t

For all n > N, s

must lie both here and here!

Proof. Suppose s = t are two limits. Take ϵ =

s−t

and apply Deﬁnition 2.4 twice: ∃N

, N

such that

n > N

=⇒

−s

s −t

and n > N

=⇒

−t

s −t

Deﬁne N := max(N

, N

). Taking any n > N quickly yields a contradiction:

n > N =⇒

s −t

s −s

+ s

−t

≤

−s

−t

(△-inequality)

s −t

Theorem 2.7. If k > 0 is constant, then lim

= 0.

Proof. In this, and the examples that follow, only the formal arguments are required; scratch work is

included to show the thought process.

Scratch work. We want n > N =⇒

< ϵ. That is, n >

1/k

. It should be enough to choose N =

1/k

Formal argument. Suppose ϵ > 0 is given and let N =

1/k

. Then

n > N =⇒



−0



= ϵ

We conclude that lim

= 0.

Examples 2.8. 1. We prove that lim



√

n + 4 −

√



= 0.

Scratch work. We use a (hopefully) familiar trick for manipulating surd expressions:

√

n + 4 −

√

n =

√

n + 4 +

√

Formal argument. Suppose ϵ > 0 is given and let N =

. Then

n > N =⇒



√

n + 4 −

√



√

n + 4 +

√

= ϵ

Thus lim



√

n + 4 −

√



= 0.

2. We prove that lim

3n+1

n−7

= 3.

Scratch work. Given ϵ > 0, we want to choose N such that

n > N =⇒



3n + 1

n −7

−3



(3n + 1) −3(n −7)

n −7



n −7



< ϵ (∗)

For large n (n > 7) everything is positive, so it is sufﬁcient to have n −7 >

. . .

Formal argument 1. Suppose ϵ > 0 is given and let N = 7 +

. Then

n > N =⇒



3n + 1

n −7

−3



n −7

N −7

= ϵ

The absolute values are dropped since n > 7. We conclude that lim

3n+1

n−7

= 3.

Scratch work 2. An alternative approach is available if we play with (∗) a little. By insisting that

n ≥ 14, we can simplify the denominator n −7 ≥

n =⇒

n−7

≤

Formal argument 2. Suppose ϵ > 0 is given and let N = max



14,



. Then

n > N =⇒



3n + 1

n −7

−3



n −7



≤

(since n ≥ 14)

≤ ϵ (since N ≥

)

We again conclude that lim

3n+1

n−7

= 3.

The plot illustrates the two choices of N as functions of ϵ.

The second is always larger than the ﬁrst: if N = N

(ϵ)

works in a proof, then any larger choice N

(ϵ) will work

also; use this to your advantage to produce simpler argu-

ments!

0 1 2 3 4 5

= 7 +

= max



14,



3. Given s

−3n+1

, we prove that lim s

Scratch work. We want to conclude that



−3n+1

−



−2n

−9n−5

3(3n

+4)



< ϵ. Attempting to solve

for n is crazy! Instead we simplify the fraction using n ≥ 1 and the △-inequality:



−2n

−9n −5

3( 3n

+ n

+ 4)



△

≤

16n

3( 3n

+ n

+ 4)

16n

(1 ≤ n ≤ n

and n

+ 4 > 0)

The ﬁnal simpliﬁcation is merely for additional cleanliness.

Formal argument. Suppose ϵ > 0 is given and let N =

. Then,

n > N =⇒



−3n + 1

+ n

+ 4

−



−2n

−9n −5

3( 3n

+ n

+ 4)



△

≤

16n

3( 3n

+ n

+ 4)

(n ≥ 1)

16n

= ϵ

Other choices of N are feasible (e.g., Exercise 3); everything depends on how you want to

simplify things in your scratch work.

Divergent sequences

By negating Deﬁnition 2.4, we obtain explicit criteria for divergence.

Deﬁnition 2.9. A sequence (s

) does not converge to s ∈ R if,

∃ϵ > 0 such that ∀N, ∃n > N with

−s

≥ ϵ

A sequence is divergent if it does not converge to any limit s ∈ R. Otherwise said,

∀s ∈ R, ∃ϵ > 0 such that ∀N, ∃n > N with

−s

≥ ϵ

Examples 2.10. 1. We prove that the sequence with s

does not converge to s = 2.

Scratch work. Of course the limit is really 0.

If ϵ is anything smaller than 2, then s

will

eventually be further than ϵ from s = 2.

Choosing ϵ = 1 should be enough. In-

deed, since we only care about large n,

−2



−2



= 2 −

≥ 1 ⇐⇒ n ≥ 7

0 5 10 15

s = 2

s + ϵ

s − ϵ

Every tail contains some

that do not lie here

Direct proof. Let ϵ = 1. Given N ∈ N, let n = max(7, N + 1). Then n > N and

−2



−2



= 2 −

≥ 1 = ϵ

Contradiction proof. For an alternative approach, suppose lim s

= 2 and let ϵ = 1 in Deﬁnition

2.4. Then ∃N such that

n > N =⇒



−2



< 1 =⇒ 1 <

< 3 =⇒

< n < 7

This last is false for large n, in particular for n := max(7, N + 1). Contradiction.

While the two arguments are similar, the contradiction approach has the advantage in that you

need only remember one deﬁnition!

2. We prove (just by contradiction this time) that s

= (−1)

deﬁnes a divergent sequence.

Suppose that lim s

= s and let ϵ = 1 in the deﬁnition

of limit. Then ∃N ∈ N such that

n > N =⇒

(−1)

−s

< 1

=⇒ (−1)

−1 < s < 1 + (−1)

Choosing any even n > N forces 0 < s; any odd

n > N forces s < 0: contradiction.

−1

s + ϵ

s − ϵ

The choice of ϵ = 1 was suggested by the picture: if ϵ = 1, then, regardless of N, there exist

n > N with

−s

> 1.

3. We prove directly that the sequence deﬁned by s

= ln n is divergent.

Scratch work. Since logarithms increase unboundedly,

for large n we should have ln n ≥ s + 1,

for any purported limit s ∈ R.

Proof. Suppose s ∈ R is given and let ϵ = 1. Given N ∈ N, deﬁne n = max(N + 1, e

s+1

). Then

n > N and ln n ≥ ln(e

s+1

) = s + 1 (ln is increasing)

=⇒

−s

= ln n − s ≥ 1 = ϵ

We conclude that (s

) is divergent.

Bounded Sequences

As a ﬁrst taste using the limit deﬁnition abstractly, we consider several related results regarding the

boundedness of sequences.

Theorem 2.11. Suppose (s

) is convergent: lim s

= s.

1. If s

≥ m for all large

n, then lim s

≥ m.

2. If s

≤ M for all large n, then lim s

≤ M.

3. (s

) is bounded (∃M such that ∀n,

≤ M).

Proof. 1. We prove the contrapositive. Suppose s < m and let ϵ =

m−s

> 0. Then ∃N such that

n > N =⇒

−s

m −s

=⇒ s

−s <

m −s

=⇒ s

−m <

s −m

< 0 =⇒ s

< m

2. This is almost identical to part 1.

3. The picture shows the strategy: taking ϵ = 1 in the limit

deﬁnition bounds an inﬁnite tail of the sequence; the

ﬁnitely many terms that come before are a non-issue.

Let ϵ = 1 in the deﬁnition of limit. Then ∃N such that

n > N =⇒

−s

< 1 =⇒ s −1 < s

< s + 1

=⇒

< max{

s −1

s + 1

}

It follows that every term of the sequence is bounded by

M := max



s −1

s + 1

: n ≤ N



s + 1

s −1

bounded tail

Deﬁnition 2.19 will state what it means for a sequence to diverge to ∞: this isn’t (yet) what we’re trying to demonstrate.

Equivalently s

≥ m for all but ﬁnitely many n. In the language of the proof, for all n expect perhaps when n ≤ N.

Since we typically care only about large n, this caveat is sometimes left unstated: e.g., s

≥ m =⇒ lim s

≥ m.

Theorem 2.12 (Squeeze Theorem). Suppose sequences satisfy a

≤ s

≤ b

(for all large n), where

the two outer sequences converge to s. Then lim s

= s.

Proof. Subtract s from the assumed inequality to obtain

−s ≤ s

−s ≤ b

−s =⇒

−s

≤ max



−s



Suppose ϵ > 0 is given. Since lim a

= s = lim b

, there exist N

, N

such that

n > N

=⇒

−s

< ϵ and n > N

=⇒

−s

< ϵ

Now let N = max(N

, N

) to see that

n > N =⇒

−s

≤ max



−s



< ϵ

Example 2.13. Since 0 ≤

1+sin n

≤

for all n, the squeeze theorem forces lim

1+sin n

= 0.

Exercises 2.8. Key concepts: ϵ–N deﬁnition, divergence, boundedness, squeeze theorem

1. For each sequence, determine the limit and prove your claim.

(a) a

(b) b

7n−19

3n+7

4n+3

7n−5

(d) d

2n+4

5n+2

(e) e

sin n (f) f

+n−1

−10

2. Prove:

(a) lim[

√

+ 1 −n] = 0 (b) lim[

√

+ n −n] =

√

+ n −2n] =

3. (a) Show that n ≥ 2 =⇒ 2n

+ 9n + 5 ≤ 9n

(b) (Example 2.8.3) Give another proof that lim

−3n+1

by choosing N = max



√



4. (a) Prove that the sequence with n

term s

does not converge to −1.

(b) Prove that (s

) does not converge to 1.

5. Prove that the sequence deﬁned by t

= n

diverges.

6. (Example 2.10.3) Prove by contradiction that (ln n) diverges.

7. True or false: if (s

) is bounded, then it is convergent. Explain.

8. Let (t

) be bounded, and let ( s

) satisfy lim s

= 0. Prove that lim(s

) = 0.

9. We extend Theorem 2.11.

(a) Suppose lim s

= s where every s

> m. Can we conclude that s > m? Explain.

(b) Let (s

) be convergent and suppose lim s

> a. Prove that ∃N such that n > N =⇒ s

> a.

10. Suppose s ∈ R. Prove:

(a) lim s

= s ⇐⇒ lim(s

−s) = 0

(b) lim s

= s =⇒ lim

2.9 Limit Theorems for Sequences

We’d like to develop rules for limits so that we don’t have to resort to ϵ–N proofs every time. The

rough idea of these results is that limits respect the basic rules of algebra. Rather than dive straight

into abstract proofs, we start with two special cases that illustrate commonly encountered tricks.

Examples 2.14. Suppose that (s

) converges to s. We prove that lim 5s

= 5s and that lim s

= s

1. The sequence (5s

) is obtained by multiplying the original terms by 5. To prove lim 5s

= 5s

we must show:

∀ϵ > 0, ∃N such that n > N =⇒

−5s

< ϵ (∗)

This last amounts to observing that

−s

. Since

is simply another small number, (∗) is

just the statement lim s

= s in disguise! Here is a formal argument.

Let ϵ > 0 be given. Since lim s

= s, we know that

∃N such that n > N =⇒

−s

=⇒

−5s

< ϵ

2. The challenge is to make



−s



−s

+ s

small. The ﬁrst term can be made arbitrarily

small by the hypothesis lim s

= s. To control the second, we use the triangle-inequality:

+ s

−s + 2s

≤

−s

+ 2

Assuming

−s

≤ 1 gives a bound

s + s

≤ 1 + 2

. We now have enough for a proof.

Suppose lim s

= s, that ϵ > 0 is given and let δ = min(1,

1+2

). Then ∃N such that,

n > N =⇒

−s

< δ ≤ 1

=⇒



−s



−s

+ s

△

≤

−s

(

−s

+ 2

)

< δ(1 + 2

) ≤ ϵ

Theorem 2.15 (Limit laws). Limits respect algebraic operations: ±, ×, ÷ and rational roots.

More speciﬁcally, if (s

) converges to s and (t

) to t, then,

1. lim(s

±t

) = s ± t

2. lim(s

) = st; in particular, if k is constant, then lim ks

= ks

3. If t = 0, then lim

4. If k ∈ N, then lim

√

s, provided the roots exist (s

, s ≥ 0 if k even)

By part 4 and induction on part 2, lim s

= s

for any q ∈ Q.

Examples 2.14 are special cases of part 2 (k = 5 and then t

= s

It is non-standard, but if this approach makes you squeamish, you can introduce a second

ϵ:

Given ϵ > 0, let

ϵ =

, then ∃N such that n > N =⇒

−s

ϵ =⇒

−5s

< ϵ.

Rigorously proving the limit laws takes some work. Before engaging in some of this, we advertise

their beneﬁt by performing some calculations as you might have seen in elementary calculus.

Examples 2.16. 1. We evaluate lim

√

n−1

−2

using the limit laws.

lim

+ 2

√

n −1

−2

= lim

3 +

3/2

−

5 −

(n > 0)

lim



3 +

3/2

−



lim



5 −



(part 3)

lim 3 + lim

3/2

−lim

lim 5 −lim

(part 1)

3 + 0 − 0

5 −0

(parts 2, 4 and Theorem 2.7))

The calculation involves some generally accepted sleight of hand: none of the limits should

really be written until one knows they exist. To be completely logical would require us to

rewrite the argument upside down, though to sacriﬁce readability in this manner would be

ill-advised.

2. Suppose (s

) is deﬁned inductively via s

= 2 and s

n+1

) =



577

408

, . . .



This sequence in fact converges, though a proof requires ideas from Section 2.10. Given this

fact, the limit laws allow us to compute the limit s:

s = lim s

n+1



lim s





s +



=⇒

s =

=⇒ s

= 2

Since s

is plainly always positive, we conclude that lim s

√

We now commence our assault on the limit laws.

Proof. 1. We use a trick similar to that in Example 2.14.1: control both sequences so that both

−s

−t

, then sum.

Suppose ϵ > 0 is given. Since lim s

= s and lim t

= t, we see that ∃N

, N

such that

n > N

=⇒

−s

and n > N

=⇒

−t

Let N = max(N

, N

), then

n > N =⇒

+ t

−( s + t)

△

≤

−s

−t

= ϵ

The argument for s

−t

is almost identical.

2. Exercise 2.8.8 deals with (and extends) the case when s = 0. We therefore suppose s = 0. The

approach is similar to part 1, we just need to be a bit cleverer to break up

−st

Suppose ϵ > 0 is given. By Theorem 2.11, (t

) is bounded:

∃M such that ∀n,

≤ M

We make assume, WLOG, that M > 0. Since lim s

= s and lim t

= t, ∃N

, N

such that

n > N

=⇒

−s

and n > N

=⇒

−t

Let N = max(N

, N

), then

−st

+ st

−st

△

≤

−s

−t

M +

= ϵ

3 & 4.: See Exercise 7.

With a few general examples, the limit laws allow us to rapidly compute a great variety of limits.

Theorem 2.17. 1. If

< 1 then lim a

= 0

2. If a > 0 then lim a

1/n

= 1

3. lim n

1/n

= 1

Examples 2.18. 1. lim( 3n)

2/n

= (lim 3

1/n

)

(lim n

1/n

)

= 1.

2. Observe from the squeeze theorem (2.12) that



sin n



≤

→ 0. We conclude:

lim

2/n



3 − n

−1

sin n



1/5

−3/2

+ 7

(lim n

1/n

)

−



3 −lim

sin n



1/5

4 lim

3/2

+ 7

1 −

√

Proof. 1. The a = 0 case is trivial. Otherwise, given ϵ > 0, let N = log

ϵ, then

n > N =⇒



= ϵ

2. Suppose a ≥ 1 and let s

= a

1/n

−1. Since s

> 0, the binomial theorem

shows that

a = (1 + s

)

≥ 1 + ns

=⇒ 0 < s

≤

a −1

The squeeze theorem shows that lim s

= 0, whence lim a

1/n

= 1.

The a < 1 case and part 3 are in Exercise 8.

(1 + x)

∑

k=0

(

)

= 1 + nx +

n(n−1)

n(n−1)(n−2)

2·3

+ ···+ x

Divergence to ±∞ and the ‘divergence laws’

We consider unbounded sequences and provide a positive deﬁnition of a type of divergence.

Deﬁnition 2.19. We say that (s

) diverges to ∞ if,

∀M > 0, ∃N such that n > N =⇒ s

> M

We write lim s

= ∞, or s

→ ∞, and say that s

‘tends to’ ∞. The deﬁnition for s

→ −∞ is similar.

If (s

) neither converges nor diverges to ±∞, we say that it diverges by oscillation: you likely wrote

lim s

= DNE (“does not exist”) in elementary calculus.

Consider how M describes “closeness” to inﬁnity analogously to how ϵ measures closeness to s in

the original deﬁnition of limit (2.4).

Examples 2.20. As with convergence arguments, some scratch work might be helpful.

1. We show that lim(n

+ 4n) = ∞.

Let M > 0 be given, and let N =

√

M. Then

n > N =⇒ n

+ 4n > n

> N

= M

2. We prove that s

= n

−n

−2n + 1 diverges to ∞ .

The negative terms cause some trouble, though our solution should be familiar from previous

calculations:

⇐⇒ n

> 2(n

+ 2n −1) ⇐⇒ n > 2 +

−

Certainly this holds if n > 6. We can now complete the proof.

Let M > 0 be given, and let N = max( 6,

√

2M). Then

n > N =⇒ s

(2M) = M

3. We prove that s

= n

−n

diverges to −∞.

First observe that

= n

(1 −n) < −

⇐⇒ 1 − n < −

n ⇐⇒ n ≥ 2

Now let M > 0 be given,

and deﬁne N = max(2,

√

2M). Then

n > N =⇒ n > 2 =⇒ s

< −

≤ −M

You may prefer to phrase lim s

= −∞ instead as

∀m < 0, ∃N such that n > N =⇒ s

< m (in our argument M = −m)

Several of the limit laws can be adapted to sequences which diverge to ±∞.

Theorem 2.21. Suppose lim s

= ∞.

1. If t

≥ s

for all (large) n, then lim t

= ∞

2. If lim t

exists and is ﬁnite, then lim s

+ t

= ∞.

3. If lim t

> 0 then lim s

= ∞.

4. lim

= 0

5. If lim t

= 0 and t

> 0 for all (large) n, then lim

= ∞

Similar statements when lim s

= −∞ should be clear.

Proof. We prove two parts; try the rest yourself.

2. Since (t

) converges, it is bounded (below): ∃m such that ∀n, t

≥ m. Let M be given. Since

lim s

= ∞, ∃N such that

n > N =⇒ s

> M −m =⇒ s

+ t

> M −m + m = M

4. Let ϵ > 0 be given, and let M =

. Then ∃N such that

n > N =⇒ s

> M =

=⇒

< ϵ

Rational Sequences

We can now describe the limit of any sequence

where (p

), ( q

) are polynomials in n.

Example 2.22. By applying Theorem 2.21 (part 3) to

:= 3n + 4n

−2

→ ∞ and t

2 − n

−2

→

we see that

lim

+ 4

−1

= lim

3n + 4n

−2

2 − n

−2

= lim(3n + 4n

−2

) ·lim

2 − n

−2

= ∞

More generally, you should be able to conﬁrm a familiar result from elementary calculus:

Corollary 2.23. If p

, q

are polynomials in n with leading coefﬁcients p, q respectively then

lim











0 if deg(p

) < deg(q

)

if deg(p

) = deg(q

)

sgn(

)∞ if deg(p

) > deg(q

)

Exercises 2.9. Key concepts: Divergence to ±∞, Limit/divergence laws, lim a

1/n

= 1 = lim n

1/n

1. Suppose lim x

= 3, lim y

= 7 and that all y

are non-zero. Determine the following:

(a) lim(x

+ y

) (b) lim

−x

+ 4

2. Suppose s ∈ R. Prove that lim s

= s =⇒ lim s

= s

by mimicking Example 2.14.2.

3. Let s

= (100n)

100

. Describe s

and s

(1 followed by how many zeros?). Now compute lim s

4. Deﬁne (s

) inductively via s

= 1 and s

n+1

√

+ 1 for n ≥ 1.

(a) List the ﬁrst four terms of (s

(b) It turns out that (s

) converges. Assume this and prove that lim s

(1 +

√

5) .

5. Prove:

(a) lim(n

−98n) = ∞ (b) lim



√

n −n +



= −∞

6. Let x

= 1 and x

n+1

= 3x

for n ≥ 1.

(a) Show that if (x

) converges with limit a, then a =

or a = 0.

(b) What is lim x

? Prove your assertion and explain what is going on.

7. We prove parts 3 and 4 of the limit laws (Theorem 2.15). Assume lim s

= s and lim t

= t.

(a) Suppose t = 0. Explain why ∃N

such that n > N

=⇒

(b) Let ϵ > 0 be given. Since lim t

= t, ∃N

such that n > N

=⇒

−t

ϵ. Combine

and N

to prove that lim

(d) Use the following inequality (valid when s

, s > 0) to construct a proof for part 4



1/k

−s

1/k



−s

k−1

+ s

k−2

+ ··· + s

k−1

≤

−s

k−1

8. We ﬁnish the proof of Theorem 2.17.

(a) Suppose 0 < a < 1. By considering b =

, prove that lim a

1/n

= 1.

(b) Let s

= n

1/n

−1. Apply the binomial theorem to n = (1 + s

)

to prove that s

n−1

Hence conclude that lim n

1/n

= 1.

9. Prove the remaining parts of Theorem 2.21.

10. Assume s

= 0 for all n and that the limit L = lim



n+1



exists.

(a) Show that if L < 1, then lim s

= 0.

(Hint: if L < a < 1, obtain N so that n > N =⇒

< a

n−N

)

(b) Show that if L > 1, then lim

= +∞.

(Hint: apply (a) to the sequence t

)

11. Let p > 0 and a ∈ R be given. How does lim

n→∞

depend on the value of a?

2.10 Monotone and Cauchy Sequences

The deﬁnition of limit (Deﬁnition 2.4) has a major weakness: to demonstrate the convergence of

a sequence we must already know its limit! What we’d like is a method for determining whether a

sequence converges without ﬁrst guessing a suitable limit.

In this section we consider two important

classes of sequence for which this can be done.

Deﬁnition 2.24. A sequence (s

) is said to be:

• Monotone-up

if s

n+1

≥ s

for all n.

• Monotone-down if s

n+1

≤ s

for all n.

• Monotone if either of the above is true.

Examples 2.25. 1. The sequence with n

term s

+ 4 is (strictly) monotone-down:

n+1

n + 1

= s

2. A constant sequence (s

) = (s, s, s, s, . . .) is both monotone-up and monotone-down.

Theorem 2.26 (Monotone Convergence).

Every bounded monotone sequence is convergent.

Speciﬁcally:

• If ( s

) is bounded above and monotone-up,

then lim s

exists and equals sup{s

• If (s

) is bounded below and monotone-down,

then lim s

exists and equals inf{s

sup{s

}

In fact the conclusion lim s

= sup{s

} holds for all monotone-up sequences: if unbounded above,

then the result is ∞ (Exercise 5).

Proof. If (s

) is bounded above, then s := sup{s

} exists by the completeness axiom (s is ﬁnite!).

Let ϵ > 0 be given. By Lemma 1.20, there exists some s

> s − ϵ. Since (s

) is monotone-up,

n > N =⇒ s

≥ s

> s − ϵ =⇒ 0 ≤ s −s

< ϵ =⇒

s −s

< ϵ

The monotone-down case is similar.

This gets at a typical application of sequences: given a sequence whose elements have a useful property, one demon-

strates the existence of a new object (the limit) to which (hopefully!) the useful property transfers. For instance, if ( f

)

is a sequence of differentiable functions, we’d like to know if lim f

(x) exists and is itself differentiable with derivative

lim f

′

(x). Discussions of this ilk dominate Math 140B.

Some authors describe a monotone-up sequence as either non-decreasing or increasing. We prefer monotone-up since

it directly describes the direction of any possible movement in the sequence and prevents confusion over whether the

inequality is strict. If necessary, a sequence with s

n+1

> s

may be described as strictly increasing or strictly monotone-up.

Examples 2.27. 1. Deﬁne (s

) via s

= 1 and s

n+1

+ 8):

) =



1, 1.8, 1.96, 1.992, 1.9984, 1.99968, . . .



This sequence certainly appears to be monotone-up and converging to 2. Here is a proof:

Bounded above: s

< 2 =⇒ s

n+1

[

2 + 8

]

= 2. By induction, (s

) is bounded above by 2.

Monotone-up: s

n+1

−s

[

2 − s

]

> 0 since s

< 2.

Convergence: By monotone convergence, s = lim s

exists. Now use the limit laws to ﬁnd s:

s = lim s

n+1

(

lim s

+ 8

)

(s + 8) =⇒ s = 2

2. (Example 2.16.2, cont.) Let s

= 2 and s

n+1





Bounded below: The sequence is plainly always positive and thus bounded below by zero.

Monotone-down: We ﬁrst obtain an improved lower bound:

n+1





= 2 +



−



≥ 2

shows

that s

≥ 2 for all n. It follows that

n+1



1 +



≤ 1 =⇒ s

n+1

≤ s

Convergence: By monotone convergence, s = lim s

exists. Example 2.16.2 provides the limit:

s =



s +



=⇒ s =

√

This shows the necessity of completeness: (s

) is a monotone, bounded sequence of rational

numbers, but its limit is irrational.

3. A decimal number d

. . . is the limit of a monotone-up sequence of rational numbers:

. . . = d

+ lim

n→∞

∑

k=1

This is bounded above (by d

+ 1 ∈ Z) and so converges.

4. The sequence with s



1 +



is particularly famous. In Exercise 10 we show that (s

) is

monotone-up and bounded above. The limit provides, arguably, the oldest deﬁnition of e:

e := lim



1 +



This is the famous AM–GM inequality

x+y

≥

√

xy with x = s

and y =

Limits Superior and Inferior

One interpretation of lim s

is that it approximately describes s

for large n. Even when a sequence

does not have a limit, it is useful to be able to describe its long-term behavior.

Deﬁnition 2.28. Let (s

) be a sequence and deﬁne two related sequences (v

) and ( u

:= sup{s

: n ≥ N}, u

:= inf{s

: n ≥ N}

1. The limit superior of (s

) is

lim sup s

(

lim

N→∞

if (s

) bounded above

∞ if (s

) unbounded above

2. The limit inferior of (s

) is

lim inf s

(

lim

N→∞

if (s

) bounded below

−∞ if ( s

) unbounded below

lim sup s

lim inf s

The original sequence (s

) is wedged between (v

) and (u

) in a manner reminiscent of the squeeze

theorem (though lim sup and lim inf need not be equal). The next result summarizes the situation

more formally; we omit the proof since these claims should be clear from the deﬁnition and previous

results, particularly the monotone convergence theorem.

Lemma 2.29. 1. (v

) is monotone-down and ( u

) monotone-up.

2. lim sup s

and lim inf s

exist for any sequence (they might be inﬁnite).

3. If n ≥ N, then u

≤ s

≤ v

4. lim inf s

≤ lim sup s

Examples 2.30. 1. If s

, then v

= s

and u

= 0, whence lim sup s

= lim inf s

= 0.

2. The picture shows the sequences (s

), ( u

) and ( v

) when s

= 6 + (−1)



1 +



We won’t compute everything precisely, but the

picture suggests (s

) has two “sub”-sequences:

the odd terms increase while the even terms de-

crease towards, respectively

lim inf s

= 5, lim sup s

= 7

Here is one value from each derived sequence:

= inf{s

: n ≥ 3} = s

≈ 3.333

= sup{s

: n ≥ 13} = s

≈ 7.357

0 10

n,N

3 13

= inf{s

: n ≥ 3}

3. Let s

= (−1)

. This time the calculation is easy: for any N,

= inf{s

: n ≥ N} = −1 and v

= sup{s

: n ≥ N} = 1

Therefore lim sup s

= 1 and lim inf s

= −1.

Theorem 2.31. For any sequence (s

lim sup s

= lim inf s

⇐⇒ lim s

exists

In such a case all three values are equal.

Note that the limits can be ±∞.

lim s

Proof. (⇒) Suppose s := lim sup s

= lim inf s

• If s is ﬁnite, apply the squeeze theorem to u

≤ s

≤ v

(both extremes converge to s).

• If s = ∞, then u

≤ s

for all n. Theorem 2.21.1 shows that lim s

= ∞ = s.

• If s = −∞, instead use s

≤ v

(⇐) We could prove this now, but it will come almost for free a little later. . .

Cauchy Sequences

We now come to a class of sequences whose analogues will dominate your future studies.

Deﬁnition 2.32. A sequence (s

) is Cauchy

∀ϵ > 0, ∃N such that m, n > N =⇒

−s

< ϵ

A sequence is Cauchy when terms in the tails of the sequence are constrained to stay close to one

another. As we’ll see shortly, this will provide an alternative way to detect and describe convergence.

Examples 2.33. 1. Let s

. Let ϵ > 0 be given and let N =

. Then

m > n > N =⇒

−s

−

= ϵ (WLOG m > n)

Thus (s

) is Cauchy. A similar argument works for any s

for positive k.

2. Suppose s

= 5 and s

n+1

= s

n(n+1)

. As before, let ϵ > 0 be given and let N =

. Then,

n+1

−s

n(n + 1)

−

n + 1

=⇒

−s

△

≤

n+1

−s

+ ··· +

−s

m−1

−

= ϵ

Again we have a Cauchy sequence.

Augustin-Louis Cauchy (1789–1857) was a French mathematician, responsible (in part) for the ϵ-N deﬁnition of limit.

3. Deﬁne (s

)

∞

n=0

inductively:

= 1, s

n+1

(

+ 3

−n

if n even

−2

−n

if n odd

) =



1, 2,

107

, . . .



Since

n+1

−s

≤ 2

−n

, we see that

0 2 4 6 8 10 12 14

m > n =⇒

−s

△

≤

n+1

−s

+ ··· +

−s

m−1

∑

k=n

k+1

−s

≤

m−1

∑

k=n

−k

−n

−2

−m

1 −2

−1

< 2

1−n

where we used the familiar geometric sum formula from calculus:

b−1

∑

k=a

−r

1−r

Suppose ϵ > 0 is given, and let N = 1 −log

ϵ = log

. Then

m > n > N =⇒

−s

< 2

1−n

< 2

1−N

= ϵ

We conclude that (s

) is Cauchy.

The last picture illustrates the essential point of Cauchy sequences: (s

) appears to converge. . .

Theorem 2.34 (Cauchy Completeness). A sequence of real numbers is convergent if and only if it is

Cauchy.

Proof. (⇒) Suppose lim s

= s (is ﬁnite). Given ϵ > 0, we may choose N such that

m, n > N =⇒

−s

and

−s

=⇒

−s

−s + s −s

△

≤

−s

s −s

< ϵ

Otherwise said, (s

) is Cauchy.

(⇐) To discuss the convergence of (s

) we need a potential limit. In view of Theorem 2.31, the

obvious candidates are lim sup s

and lim inf s

. We have two goals: show that (s

) is bounded

whence the limits superior and inferior are ﬁnite; then show that these are equal.

(Boundedness of (s

)) Take ϵ = 1 in Deﬁnition 2.32:

∃N such that m, n > N =⇒

−s

< 1

It follows that

n > N =⇒

−s

N+1

< 1 =⇒ s

N+1

−1 < s

< s

N+1

+ 1

whence (s

) is bounded. It follows that lim sup s

and lim inf s

are both ﬁnite.

(lim sup s

= lim inf s

) Suppose ϵ > 0 is given. Since (s

) is Cauchy,

∃N ∈ N such that m, n > N =⇒

−s

< ϵ =⇒ s

< s

+ ϵ

Take the supremum over all n > N: since v

N+1

= sup{s

: n ≥ N + 1}, we see that

m > N =⇒ v

N+1

≤ s

+ ϵ

Now take the inﬁmum of the right hand side over all m > N to obtain

N+1

≤ u

N+1

+ ϵ (since u

N+1

= inf{s

: m ≥ N + 1})

Since (v

N+1

) is monotone-down and ( u

N+1

) monotone-up, we see that

lim sup s

≤ v

N+1

≤ u

N+1

+ ϵ ≤ lim inf s

+ ϵ =⇒ lim sup s

≤ lim inf s

+ ϵ

Since ϵ > 0 was arbitrary, we conclude that lim sup s

≤ lim inf s

. By Lemma 2.29 we

have equality.

By Theorem 2.31, we conclude that (s

) converges to lim sup s

= lim inf s

By the Theorem, Examples 2.33 all converge. All three limits can be found precisely (for instance,

see Exercise 7). With a small modiﬁcation to the second example, however, we obtain something

genuinely new:

Example (2.33.2 cont). Let s

= 5 and, for each n, deﬁne s

n+1

:= s

sin n

n(n+1)

. Since

sin n

≤ 1, the

computation proceeds almost the same as before:

n+1

−s

sin n

n(n + 1)

≤

n(n + 1)

= ···

The new sequence is Cauchy and thus convergent, though good luck explicitly ﬁnding its limit!

The main point is easy to miss: the Cauchy condition is a powerful tool for determining whether a

sequence converges without ﬁrst guessing a limit. While the proof depends on monotone convergence

(via limit superior/inferior), Cauchy completeness is more powerful in that it applies even to non-

monotone sequences.

An Alternative Deﬁnition of R Cauchy sequences suggest a deﬁnition of the real numbers which

does not rely on Dedekind cuts (Section 1.6).

Deﬁne an equivalence relation ∼ on the collection C of all Cauchy sequences of rational numbers:

) ∼ (t

) ⇐⇒ lim(s

−t

) = 0

Now deﬁne R :=



∼

to be the set of equivalence classes. All this is done without reference to

Cauchy completeness, though it certainly informs our intuition that (s

) and (t

) have the same limit

(as real numbers). Signiﬁcant work is still required to properly deﬁne +, ·, ≤, etc., and to verify the

axioms of a complete ordered ﬁeld—we won’t pursue this.

We don’t need real numbers to deﬁne the limit of the rational sequence (s

−t

): ∀ϵ ∈ Q

is enough. . .

Exercises 2.10. Key concepts: Monotone sequences & Convergence, Cauchy sequences & completeness,

Limits superior/inferior

1. Use Deﬁnition 2.32 to show that the sequence with s

is Cauchy. Repeat for t

n(n−2)

2. Let s

= 1 and s

n+1

for n ≥ 1.

(a) Find s

, s

and s

(b) Show that lim s

exists and hence prove that lim s

= 0.

3. Let s

= 1 and s

n+1

+ 1) for n ≥ 1.

(a) Find s

, s

and s

(b) Use induction to show that s

for all n, and conclude that (s

) is monotone-down.

exists and ﬁnd lim s

4. (a) Let (s

) be a sequence such that ∀n,

n+1

−s

≤ 3

−n

. Prove that (s

) is Cauchy.

(b) Let s

= 10 and, for each n, let s

n+1

= s

cos n

. Explain why (s

) is convergent.

n+1

−s

≤

for all n ∈ N?

5. Suppose (s

) is unbounded and monotone-up. Prove that lim s

= ∞.

(Thus lim s

= sup{s

} for any monotone-up sequence)

6. Let s

(−1)

. Find the sequences (u

), (v

) and explicitly compute lim sup s

and lim inf s

7. Consider the sequence in Example 2.33.3. Explain why s

= s

2n−2

−

Now use the geometric sum formula to evaluate lim s

(Since (s

) converges, this means the original sequence has the same limit)

8. Let S be a bounded nonempty set for which sup S /∈ S. Prove that there exists a monotone-up

sequence (s

) of points in S such that lim s

= sup S.

(Hint: for each n, use sup S −

to build s

)

9. Let ( s

) be a monotone-up sequence of positive numbers and deﬁne σ

+ s

+ ···+ s

Prove that (σ

) is monotone-up.

10. (Hard!) We prove that the sequence deﬁned by s



1 +



is convergent.

(a) Show that

1 +

n+1

1 +

= 1 −

(n + 1)

and

1 +

n+1

= 1 +

n(n + 2)

(b) Prove Bernoulli’s inequality by induction:

For all real x > −1 and n ∈ N

we have ( 1 + x)

≥ 1 + nx.

n+1

, use parts (a) and (b) to prove that (s

) is monotone-up.

(d) Similarly, show that t



1 +



n+1



1 +



deﬁnes a monotone-down sequence.

(e) Prove that (s

) and ( t

) converge, and to the same limit (this is Bernoulli’s deﬁnition of e).

(f) Prove that lim(1 −

)

= e

−1

2.11 Subsequences

The overall behavior of a sequence is often hard to describe, but if we delete some of its terms we

might obtain a subsequence with much simpler behavior.

Deﬁnition 2.35. Let (s

) be a sequence. A subsequence (s

) is a subset ( s

) ⊆ (s

), where

< n

< ···

A subsequence is simply an inﬁnite subset whose order is inherited from the original sequence.

Example 2.36. Take s

= (−1)

(recall Example 2.10.2) and let

= 2k. Then s

= 1 for all k. Note two important facts:

• The subsequence (s

)

∞

k=0

is indexed by k, not n.

• The subsequence is constant and thus convergent.

−1

Our main goal in this section is to prove the famous Bolzano–Weierstraß theorem (illustrated in the

example): that every bounded sequence has a convergent subsequence.

Lemma 2.37. If lim

n→∞

= s, then every subsequence (s

) satisﬁes lim

k→∞

= s.

Proof. Suppose s is ﬁnite and suppose ϵ > 0 is given. Then ∃N such that n > N =⇒

−s

< ϵ.

Since n

≥ k for all k, we see that

k > N =⇒ n

> N =⇒

−s

< ϵ

The case where s = ±∞ is an exercise.

Lemma 2.38. Every sequence has a monotonic subsequence.

Proof. Given (s

), we call the term s

‘dominant’ if m > n =⇒ s

< s

. There are two cases:

1. If there are inﬁnitely many dominant terms, then the subsequence of such is monotone-down.

2. If there are ﬁnitely many dominant terms, choose s

after all such. Since s

is not dominant,

∃n

> n

such that s

≥ s

. Induct to obtain a monotone-up subsequence.

dominant terms

Case 1: monotone-down subsequence Case 2: monotone-up subsequence

Theorem 2.39. Given a sequence (s

), there exist subsequences (s

) and ( s

) such that

lim s

= lim sup s

and lim s

= lim inf s

By the lemmas, we may moreover assume that these subsequences are monotonic.

Example 2.40. The picture shows the sequence with

term

(

(−1)

when n is even

1 −

when n is odd

Monotonic subsequences with limits lim sup s

= 1

and lim inf s

= 0 are indicated.

−1

10 20

Proof. We prove only the lim sup claim, since the other is similar. There are three cases to consider;

visualizing the third is particularly difﬁcult and may take several readings.

(lim sup s

= ∞) Since (s

) is unbounded above, for any

k > 0 there exist inﬁnitely many terms s

> k. We may

therefore inductively choose a subsequence (s

) via

= min{n ∈ N : s

> 1}

= min{n ∈ N : n

> n

k−1

, and s

> k}

Choosing the minimum isn’t necessary, though it

keeps the subsequence explicit. Clearly

> k =⇒ lim

k→∞

= ∞ = lim sup s

Example: lim sup

√

(1 + sin n) = ∞

(lim sup s

= −∞) Since lim inf s

≤ lim sup s

= −∞, Lemma 2.31 says that lim s

= −∞, whence

) itself is a suitable subsequence.

(lim sup s

= v ﬁnite) Let n

= 1 and deﬁne s

for k ≥ 2 inductively:

• Since (v

) is monotone-down and converges to v, take ϵ =

to see that

∃N

> n

k−1

such that v ≤ v

< v +

• Since v

= sup{s

: n ≥ N

}, Lemma 1.20 says

∃n

≥ N

such that s

> v

−

But then

v −s

△

≤

v −v

−s

. The squeeze theorem says that lim

k→∞

= v.

) being monotone-down is crucial: if N satisﬁes v

−v <

, so does N

:= max(N, 1 + n

k−1

Example (2.40 cont.). The example shows why the two-step construction is necessary. It may seem

that we should simply be able to choose subsequences of (u

) and ( v

). Indeed,

) =



−1, −1, −1, −1

| {z }

, −

| {z }

, −

| {z }

, . . .



contains a subsequence of (s

) converging to lim inf s

= 0. Unfortunately, (v

) = (2, 2, 1, 1, 1, . . .)

does not contain a subsequence of (s

). Taking n

= 2k + 1 (k ≥ 2) results in the displayed sequence:

= 1 −

2k + 1

> 1 −

= v

−

There are two immediate corollaries. The ﬁrst (see Exercise 3) fully establishes Theorem 2.31

lim sup s

= lim inf s

⇐⇒ lim s

exists (could be ±∞)

The second is the main goal of this section.

Theorem 2.41 (Bolzano–Weierstraß). Every bounded sequence has a convergent subsequence.

Proof 1. Lemma 2.38 says there exists a monotone subsequence. This is bounded and thus converges

by the monotone convergence theorem.

Proof 2. By Theorem 2.39, there exists a subsequence converging to the ﬁnite value lim sup s

For a third proof(!) we present the classic ‘shrinking-interval’ argument which has the beneﬁt of

easily generalizing to higher dimensions.

Proof 3. Suppose (s

) is bounded by M. One of the intervals [−M, 0] or [0, M] must contain inﬁnitely

many terms of the sequence (perhaps both do!). Call this interval E

and choose any n

∈ E

Split E

into left- and right half-intervals, one of which must contain inﬁnitely many terms of the

sequence for which n > n

;

call this half-interval E

and choose any s

∈ E

with n

> n

Repeat this ad inﬁnitum to obtain a subsequence (s

) and a family of nested intervals

[−M, M] ⊃ E

⊃ E

⊃ ··· of width

with s

∈ E

It remains only to see that (s

) converges; we leave this to Exercise 5.

Example 2.42. (s

) = (sin n) is bounded and therefore has a

convergent subsequence! Its limit s must lie in the interval [−1, 1] .

The picture shows the ﬁrst 1000 terms—remember that n is mea-

sured in radians. It is not at all clear from the picture what s or

our mystery subsequence should be! There is a reason for this, as

we’ll see momentarily. . .

−1

Only ﬁnitely many terms in (s

) could come before s

. . .

Subsequential Limits, Divergence by Oscillation & Closed Sets

Recall Deﬁnition 2.19: a sequence (s

) diverges by oscillation if it neither converges nor diverges to

±∞. We can now give a positive statement of this idea.

) diverges by oscillation

Thm 2.31

⇐⇒ lim inf s

= lim sup s

Thm 2.39

⇐⇒ ( s

) has subsequences tending to different limits

The word oscillation comes from the third interpretation: if s

= s

are limits of two subsequences,

then any tail of the sequence {s

: n > N} contains inﬁnitely many terms arbitrarily close to s

and

inﬁnitely many (other) terms arbitrarily close to s

. The original sequence (s

) therefore oscillates

between neighborhoods of s

and s

. Of course there could be many other subsequential limits. . .

Deﬁnition 2.43. We call s ∈ R ∪ {±∞} a subsequential limit of a sequence (s

) if there exists a

subsequence (s

) such that lim

k→∞

= s.

Examples 2.44. 1. The sequence deﬁned by s

has only one subsequential limit, namely zero.

Recall Lemma 2.37: lim s

= 0 implies that every subsequence also converges to 0.

2. If s

= (−1)

, then the subsequential limits are ±1.

3. The sequence s

= n

(1 + ( −1)

) has subsequential limits 0 and ∞.

4. All positive even integers are subsequential limits of (s

) = (2, 4, 2, 4, 6, 2, 4, 6, 8, 2, 4, 6, 8, 10, . . .).

5. (Hard!) Recall the countability of Q from a previous class: the standard argument enumerates

the rationals by constructing a sequence

) =



, −

| {z }

+q=2

, −

| {z }

+q=3

, −

| {z }

+q=4

, −

| {z }

+q=5

, . . .



We claim that the set of subsequential limits of (r

) is in fact the set R ∪{±∞ }!

To see this, let a ∈ R be given and choose a subsequence (r

) inductively:

• By the density of Q in R (Corollary 1.23), the set S

= Q ∩(a −

, a +

) contains inﬁnitely

many rational numbers and thus inﬁnitely many terms of the sequence (r

• Choose any r

∈ S

and, for each k ≥ 2, choose any

∈ S

such that n

> n

k−1

• Since

− a

≤

, we conclude that lim

k→∞

= a.

An argument for the subsequential limits ±∞ is in the Exercises. Somewhat amazingly, the

speciﬁc sequence (r

) is irrelevant: the conclusion is the same for any sequence enumerating Q!

6. (Even harder—Example 2.42, cont.) We won’t prove it, but the set of subsequential limits of

) = (sin n) is the entire interval [−1, 1]! Otherwise said, for any s ∈ [−1, 1] there exists a

subsequence ( sin n

) such that lim

k→∞

sin n

= s.

Theorem 2.45. Let (s

) be a sequence in R and let S be its set of subsequential limits. Then

1. S is non-empty (as a subset of R ∪ {±∞}).

2. sup S = lim sup s

and inf S = lim inf s

3. lim s

exists iff S has only one element: namely lim s

Proof. 1. By Theorem 2.39, lim sup s

∈ S.

2. By part 1, lim sup s

≤ sup S. For any convergent subsequence (s

) we have n

≥ k, whence

∀N, {s

: k ≥ N} ⊆ {s

: n ≥ N} =⇒ lim s

= lim sup s

≤ lim sup s

Since this holds for every convergent subsequence, we have sup S ≤ lim sup s

and therefore

equality. The result for inf S is similar.

3. Applying Theorem 2.31, we see that lim s

exists if and only if

lim sup s

= lim inf s

⇐⇒ sup S = inf S ⇐⇒ S has only one element

Closed Sets You should be comfortable with the notion of a closed interval (e.g. [0, 1]) from elemen-

tary calculus. Sequences allow us to make a formal deﬁnition.

Deﬁnition 2.46. Let A ⊆ R.

• We say that s ∈ R is a limit point of A if there exists a sequence (s

) ⊆ A converging to s.

• The closure A is the set of limit points of A. Plainly A ⊆ A for any set.

• A is closed if it equals its closure: A = A.

Examples 2.47. 1. The interval [0, 1] is closed. If (s

) ⊆ [0, 1] has lim s

= s, then

0 ≤ s

≤ 1

Thm 2.11

=⇒ s ∈ [0, 1]

More generally, every ‘closed interval’ [a, b] is closed, as are ﬁnite unions of closed intervals, for

instance [1, 5] ∪ [7, 11].

2. The ‘half-open’ interval (0, 1] is not closed: its closure is (0, 1] = [0, 1]. In particular, the se-

quence s

lies in ( 0, 1], but lim s

= 0 ∈ (0, 1].

3. Example 2.44.5 shows that the closure of the rational numbers is the reals: Q = R.

Theorem 2.48. If (s

) is a sequence, then its set of (ﬁnite) subsequential limits is closed.

We omit the proof since it involve unpleasantly many subscripts (subsequences of subsequences. . . ).

Exercises 2.11. Key concepts: Subsequences of a convergent sequence all have the same limit,

Existence of (monotone) subsequences tending to lim sup s

/ lim inf s

, Subsequential Limits,

Bolzano–Weierstraß: boundedness =⇒ ∃ convergent subsequence

1. Consider the sequences with the following n

terms:

= (−1)

= n

6n + 4

7n −3

For each sequence: state whether it converges, diverges to ±∞, or diverges by oscillation; give

an example of a monotone subsequence; state the set of subsequential limits; state the limits

superior and inferior.

2. Prove the case of Lemma 2.37 when lim s

= ∞

3. Suppose that lim s

= s (could be ±∞). Use Theorem 2.39 and Lemma 2.37 to prove that

lim sup s

= s = lim inf s

(This completes the proof of Theorem 2.31)

4. Suppose that L = lim s

exists and is ﬁnite.

(a) Given an example of such a sequence where (s

) is divergent.

(b) Prove that (s

) contains a convergent subsequence. What are the possible limits of this

subsequence? Why?

(Hint: use Bolzano–Weierstraß)

5. Complete the third proof of Bolzano–Weierstraß (Theorem 2.41) by proving that the constructed

subsequence (s

) is Cauchy.

6. (a) Show that the closed interval [a, b] is a closed set in the sense of Deﬁnition 2.46.

(b) Is there a sequence (s

) such that (0, 1) is its set of subsequential limits?

7. By considering Example 2.47.2, or otherwise, show that an inﬁnite union of closed intervals

need not be closed.

8. Let (r

) be any sequence enumerating of the set Q of rational numbers. Show that there exists

a subsequence (r

) such that lim

k→∞

= +∞.

(Hint: modify the argument in Example 2.44.5)

9. (Hard) Let (s

) be the sequence of numbers deﬁned

in the ﬁgure, listed in the indicated order.

(a) Find the set S of subsequential limits of (s

(b) Determine lim sup s

and lim inf s

···

2.12 Lim sup and Lim inf

In this short section we collect a couple of useful results, mostly for later use. First we observe that

the limit laws do not work as tightly for limits superior and inferior.

Theorem 2.49. Let (s

), (t

) be bounded sequences. Then:

1. lim sup(s

+ t

) ≤ lim sup s

+ lim sup t

2. If, in addition, (s

) is convergent to s, then we have equality

lim sup(s

+ t

) = s + lim sup t

Natural modiﬁcations can be made for inﬁma and products of sequences (see Exercise 3).

Example 2.50. To convince yourself that equality is unlikely, consider s

= (−1)

= −t

. Plainly

lim sup(s

+ t

) = 0 < 2 = lim sup s

+ lim sup t

Proof. 1. For each N, the set {s

+ t

: n ≥ N} is bounded above by

sup{s

: n ≥ N} + sup{t

: n ≥ N}

from which

sup{s

+ t

: n ≥ N} ≤ sup{s

: n ≥ N} + sup{t

: n ≥ N}

Take limits as N → ∞ for the ﬁrst result.

2. By part 1, we know that

lim sup(s

+ t

) ≤ s + lim sup t

For the other direction, rearrange and apply part 1 again:

lim sup t

= lim sup



+ t

) −s



≤ lim sup(s

+ t

) + lim sup(−s

)

= lim sup(s

+ t

) −s

The next result will be critical when we study inﬁnite series, particularly the ratio and root tests.

Theorem 2.51. Let (s

) be a non-zero sequence. Then

lim inf



n+1



≤ lim inf

1/n

≤ lim sup

1/n

≤ lim sup



n+1



In particular, lim



n+1



= L =⇒ lim

1/n

= L. (†)

Examples 2.52. 1. Here is a quick proof that lim n

1/n

= 1 (recall Theorem 2.17). Let s

= n, then

lim



n+1



= lim

n + 1

= 1 =⇒ lim n

1/n

= lim

1/n

= 1

2. Apply the corollary to s

= n! to see that

lim(n!)

1/n

= lim



n+1



= lim(n + 1) = ∞

Proof. We prove the third inequality. Assume lim sup



n+1



= L = ∞ (otherwise the inequality is

trivial). Suppose ϵ > 0 is given, and denote a = L + ϵ. Then

L = lim

N→∞

sup





n+1



: n ≥ N



< a =⇒ ∃N such that sup





n+1



: n ≥ N



< a

Now let b = a

−N

. For any n ≥ N, we therefore have



n+1



< a, whence

n > N =⇒

< a

n−N

=⇒

1/n

< a



−N



1/n

= ab

1/n

=⇒ lim sup

1/n

≤ a lim b

1/n

= a = L + ϵ

Since ϵ > 0 was arbitrary, we conclude the third inequality: lim sup

1/n

≤ L.

The second inequality is trivial and the ﬁrst is similar to the third.

Exercises 2.12. Key concepts: “Limit laws” for lim sup and lim inf, lim



n+1



= lim

1/n

1. Compute lim

(n!)

1/n

(Hint: let s

in Theorem 2.51 and recall that lim



1 +



= e)

2. Evaluate lim



(2n)!

(n!)



1/n

3. Let (s

) and ( t

) be non-negative, bounded sequences.

(a) Prove that lim sup(s

) ≤

(

lim sup s

) (

lim sup t

)

(b) Give an example which shows that we do not expect equality in part (a).

= s, prove that lim sup(s

) = s lim sup t

4. Consider the sequence with s

= s

2m+1

= 2

−m

)

∞

n=0



1, 1,

, . . .



Compute

1/n

and



n+1



when n is even and then when it is odd. Thus ﬁnd all expressions

in Theorem 2.51 and conclude that the converse of (†) is false.

3 Series

3.14 Inﬁnite Series and the Series Tests

For millennia, certainly since Zeno’s paradoxes of c. 430 BC, mathematicians have been interested in

the meaning and evaluation of inﬁnite sums such as

∞

∑

n=1

= 1 +

+ ···

The standard approach in modern mathematics is to outsource the deﬁnition to that of limits.

Deﬁnition 3.1. The n

partial sum s

of a sequence (a

)

∞

n=m

is the ﬁnite sum

∑

k=m

= a

+ a

m+1

+ ··· + a

• The (inﬁnite) series

∞

∑

n=m

is the limit lim s

of the sequence (s

) of partial sums.

• A series converges, s to ±∞ or diverges by oscillation as does the sequence (s

•

∑

converges absolutely if

∑

converges.

•

∑

converges conditionally if it converges but not absolutely (

∑

diverges to ∞).

We don’t (yet) know whether our motivating example converges, but at least we have a meaning:

∞

∑

n=1

= lim s

where s

∑

k=1

= 1 +

+ ··· +

Theorem 3.2 (Basic Series Laws). Inﬁnite series behave nicely with respect to addition and scalar

multiplication. For instance:

1. If

∑

is convergent and k is constant, then

∑

= k

∑

is convergent.

2. If

∑

and

∑

are convergent, then

∑

±b

) =

∑

are also convergent.

3. If

∑

= ∞ and k > 0, then

∑

= ∞.

4. If

∑

= ∞ and

∑

converges, then

∑

+ b

) = ∞.

Proof. Simply apply the limit/divergence laws to the sequence of partial sums. E.g. for 1,

∑

= lim

n→∞

∑

j=m

ﬁnite

sum

= lim

n→∞

∑

j=m

limit

laws

= k lim

n→∞

∑

j=m

= k

∑

The others may be proved similarly.

Series do not behave nicely with respect to multiplication (see also Exercise 3):

+ a

+ ··· =

∑

=



∑



∑





+ a

+ ···



+ b

+ ···



If the initial term is understood or is irrelevant to the situation, it is common to write

∑

Series which may be evaluated exactly

Given a series

∑

, our primary goal is to answer a simple question: “Does it converge?” Even when

the answer is yes, a precise computation of the limit will usually be beyond us. We instead develop

techniques (the upcoming series tests) which typically rely on comparing

∑

to some ‘standard’

series whose properties are completely understood: in particular. . .

Deﬁnition 3.3 (Geometric series). A sequence (a

) is geometric if the ratio of successive terms is

constant: a

= ba

for some constants a, b. A geometric series is the sum of a geometric sequence.

The computation of the sequence of partial sums should be familiar (for simplicity assume b = 1)

(1 − a)s



+ a

m+1

+ ··· + a



−



m+1

+ a

m+2

+ ··· + a

+ a

n+1



= a

− a

n+1

from which we quickly conclude:

Theorem 3.4. Suppose a is constant. Then

∑

k=m







− a

n+1

1 − a

if a = 1

n + 1 −m if a = 1

=⇒

∞

∑

n=m











converges to

1 − a

< 1

diverges to ∞ if a ≥ 1

diverges by oscillation if a ≤ −1

In particular,

∑

converges absolutely if

< 1 and diverges otherwise.

Examples 3.5. 1.

∞

∑

n=−1



−



= 2



−



−1

1 +

= −

2. Consider the series

∑

∞

∑

n=3





+ 2

. If this were convergent, then

∑

−

∑





would converge (Theorem 3.2); a contradiction.

Telescoping series A rarer type of series can be evaluated using the algebra of partial fractions.

Example 3.6. To compute

∞

∑

n=1

n(n+1)

, ﬁrst observe that

∑

k=1

k(k + 1)

∑

k=1

−

k + 1

−

+ ··· +

−

n + 1

= 1 −

n + 1

It follows that

∞

∑

n=1

n(n + 1)

= lim



1 −

n + 1



= 1

Similar arguments can be made for other series such as

∑

n(n+2)

The Cauchy Criterion

The starting point for our general series tests uses Cauchy completeness.

Example 3.7. Consider again the series

∑

. We show that the sequence of partial sums (s

) is

Cauchy. Suppose ϵ > 0 is given and let N =

. Then,

m > n > N =⇒

−s

∑

k=n+1

∑

k=n+1

k(k −1)

∑

k=n+1

k −1

−

= ϵ

where most terms cancel analogous to the telescoping series approach. By Cauchy completeness

(Theorem 2.34), (s

) converges and we conclude

∑

converges

Computing the value of this series is signiﬁcantly harder: a sketch argument for why

∑

∞

n=1

is in Exercise 10.

Theorem 3.8 (Cauchy criterion for series). A series

∑

converges precisely when

∀ϵ > 0, ∃N such that m > n > N =⇒

−s



∑

k=n+1



< ϵ

The previous example essentially veriﬁed the Cauchy criterion for

∑

Proof. Let (s

) be the sequence of partial sums. Then

∑

converges ⇐⇒ (s

) converges

⇐⇒ (s

) is a Cauchy sequence (Theorem 2.34)

⇐⇒



∀ϵ > 0, ∃N such that m > n > N =⇒

−s

< ϵ



Example 3.9. For contradiction, suppose that the harmonic series

∑

converges. Take ϵ =

in the

Cauchy criterion to observe that

∃N such that m > n > N =⇒



∑

k=n+1



However, taking m = 2n (plainly m > n since n > N ≥ 1) results in a contradiction:



∑

k=n+1



n + 1

+ ··· +



≥

m −n

= 1 −

We conclude that the harmonic series diverges to ∞.

The Series Tests

For the remainder of this section we develop tests for the convergence/divergence of an inﬁnite

series: the n

-term, comparison, root and ratio tests. The ﬁrst follows quickly from the Cauchy

criterion.

Theorem 3.10 (Divergence/n

-term test). If lim a

= 0 then

∑

is divergent.

Proof. We prove the contrapositive. Suppose

∑

is convergent, and that ϵ > 0 is given. Take

m = n + 1 in the Cauchy criterion. Then

∃

N such that m >

N =⇒

< ϵ (let

N = N + 1)

Otherwise said, lim a

= 0.

Examples 3.11. 1. The series

∑

sin(

nπ

) diverges.

2. The n

-term test tells us that the geometric series

∑

diverges whenever

≥ 1. We still need

our earlier analysis (Theorem 3.4) for when

< 1.

3. The converse of the n

-term test is false! Example 3.9 provides the canonical example: the

divergent harmonic series

∑

also satisﬁes lim

= 0.

Theorem 3.12 (Comparison test). 1. Let

∑

be a convergent series of non-negative terms and as-

sume

≤ b

for all (large) n. Then both

∑

and

∑

are convergent.

2. If

∑

= ∞ and a

≤ b

for all (large) n, then

∑

= ∞.

Proof. Suppose “large n” means n > M for some ﬁxed M.

1. Let ϵ > 0 be given. Since

∑

converges, ∃N ≥ M such that

m > n > N =⇒



∑

k=n+1



△

≤

∑

k=n+1

≤

∑

k=n+1

< ϵ

2. The n

partial sum (post M) of

∑

k=M+1

≥

∑

k=M+1

→ ∞

Corollary 3.13. 1. (Absolute convergence implies convergence) Take

= b

in part 1 to see that

∑

convergent =⇒

∑

convergent.

2. (Estimation of series) Suppose

∑

is a convergent series of non-negative terms and that

≤ b

for all n. Then

∑

≤

∑

≤

∑

Examples 3.14. 1. Since the geometric series

∑

converges,

2n+1

(n+2)3

≤

, we see that

∞

∑

n=0

2n + 1

(n + 2)3

≤ 2

∞

∑

n=0

−n

1 −

= 3

That is, the ﬁrst series converges (absolutely) to some value ≤ 3.

2. One can sometimes ﬁnd a sensible comparison series by considering how a

behaves for large

n. For instance, when n is large, a

+1)

1/2

(1+

√

behaves like

. Indeed, when n ≥ 2,

(1 +

√

16n

Comparison with the divergent series

∑

shows that

∑

also diverges to ∞.

3. Since ln n < n =⇒

ln n

, we see that

∑

ln n

diverges to ∞ by comparison with

∑

sin n

converges absolutely by comparison to

∑

(Example 3.7). Corollary 3.13 estimates its

value (≤

∞

∑

n=1

sin n

≤

∞

∑

n=1

sin n

≤

∞

∑

n=1

(approximately 1.014 ≤ 1.280 ≤ 1.645)

5. The alternating harmonic series s =

∑

∞

n=1

(−1)

n+1

converges via a sneaky comparison.

The series t =

∑

∞

n=1

2n(2n−1)

converges by comparison with

∑

4(n−1)

. Its n

partial sum

∑

k=1

2k(2k −1)

∑

k=1



2k −1

−



is precisely the even partial sum of the alternating harmonic series s

∑

k=1

(−1)

k+ 1

Plainly lim s

= t. Moreover s

2n+1

= s

2n+1

=⇒ lim s

2n+1

= t =⇒ lim s

= t. Since the

harmonic series

∑

diverges (Example 3.9), we conclude that the alternating harmonic series

converges conditionally. We’ll revisit this discussion in the next section.

∑



n+1



converges by comparison with

∑

−n

. To see this, recall Exercise 2.10.10:



n + 1



n + 1



1 −

n + 1



n+1

−−−→

n→∞

−1

Plainly e

−1

, whence for large n,



n + 1



≤

=⇒



n + 1



≤ 2

−n

In fact



n+1



is monotone-down, whence e

−1

≤



n+1



≤

for all n, and so

0.58198 ≈

−1

1 − e

−1

∞

∑

n=1

−n

≤

∞

∑

n=1



n + 1



≤

∞

∑

n=1

−n

1/2

1 −1/2

= 1

A computer estimate yields

∞

∑

n=1



n+1



≈ 0.8174.

Our last two tests in this section are less powerful but often easier to use.

Theorem 3.15 (Root test). Suppose lim sup

1/n

= L.

1. If L < 1, then

∑

converges absolutely.

2. If L > 1, then

∑

diverges.

If L = 1, then no conclusion can be drawn.

We defer the proof until after some examples. By combining with the inequalities of Theorem 2.51

lim inf



n+1



≤ lim inf

1/n

≤ lim sup

1/n

≤ lim sup



n+1



we obtain a second familiar test.

Corollary 3.16 (Ratio test). Suppose (a

) is a sequence of non-zero terms.

1. If lim sup



n+1



< 1, then

∑

converges absolutely.

2. If lim inf



n+1



> 1, then

∑

diverges.

In elementary calculus you likely saw the special cases when

L = lim

1/n

= lim



n+1



Our versions are more general since these limits might not exist.

Examples 3.17. 1. The ratio test is particularly useful for series involving factorials and exponentials.

(a)

∑

converges, since lim



n+1



= lim

(n+1)

< 1.

(b)

∑

diverges, since lim



n+1



= lim

(n+1)!

2n!

= lim

n+1

= ∞.

2. Both tests are inconclusive for rational sequences: if a

where b

, c

are polynomials, then

lim



n+1



= 1 = lim

1/n

For example, attempting to apply the ratio test to

∑

n+5

results in

lim



n+1



= lim

(n + 6)n

(n + 5)( n + 1)

= 1

This series is divergent by comparison with the harmonic series

∑

3. In Example 3.14.6, our use of the comparison test was really the root test in disguise:



n + 1



=⇒ lim

1/n

= lim



n + 1



= e

−1

< 1 =⇒

∑

converges

In this case the root test was much easier to apply.

4. The ratio test is the weakest test thus far; certainly it does not apply if any of the terms a

are

zero! For a more subtle example of its failure, consider

(

−n

if n is even

−n

if n is odd

=⇒

n+1

(





if n is even





if n is odd

=⇒ lim inf



n+1



= 0, lim sup



n+1



= ∞

The ratio test is therefore inconclusive. However, applying the root test it almost trivial!

1/n

(

if n is even

if n is odd

=⇒ lim sup

1/n

< 1 =⇒

∑

converges

We need not even have used the root test:

∑

plainly converges by comparison with

∑

−n

For a precise value, note that the sequence of n

partial sums converges monotone up to the

sum of two geometric series,

∞

∑

n=0

∞

∑

k=0

−2k

+ 3

−2k−1

1 −1/4

1/3

1 −1/9

Proof of the Root Test. 1. Suppose L < 1. Choose any ϵ > 0 such that L + ϵ < 1 (say ϵ =

1−L

). Since

= sup



1/n

: n ≥ N



deﬁnes a monotone-down sequence converging to L, we see that

∃N such that v

− L < ϵ

But then

n ≥ N =⇒

1/n

− L < ϵ =⇒

< (L + ϵ)

∑

therefore converges by comparison with the geometric series

∑

(L + ϵ)

2. If L > 1 then there exists some subsequence (a

) such that

1/n

→ L > 1. In particular,

inﬁnitely many terms of this subsequence must be greater than 1 and so (a

) does not converge

to zero.

∑

thus diverges by the n

-term test.

Summary The logical ﬂow of the tests in this section is as follows:

(divergence testing) Comparison n

-term

Root

Ratio

Deﬁnition of

∑

ks +3

Cauchy criterion



(convergence testing) Comparison

Root

Ratio

The ratio test is typically the easiest to use, but the least powerful. Every series which converges by

the ratio test can be seen to converge by the root and comparison tests, and the Cauchy criterion. If

you ﬁnd that a series diverges by the ratio test, you could have just used the n

-term test!

Exercises 3.14. Key concepts: Inﬁnite series, Cauchy criterion, Comparison/root/ratio tests

1. Determine which of the following series converge. Justify your answers.

(a)

∑

n−1

(b)

∑

(−1)

(c)

∑

(d)

∑

(e)

∑

(f)

∑

(g)

∑

(h)

∑

(i)

∞

∑

n=2



n + (−1)



−2

(j)

∑



√

n + 1 −

√



2. Let

∑

and

∑

be convergent series of non-negative terms. Prove that

∑

√

converges.

(Hint: start by showing that

√

≤ a

+ b

)

3. (a) If

∑

converges absolutely, prove that

∑

converges.

(b) More generally, if

∑

converges and (b

) is a bounded sequence, prove that

∑

converges absolutely.

4. Find a series

∑

which diverges by the root test but for which the ratio test is inconclusive.

5. Suppose lim inf

= 0. Prove that there is a subsequence (a

) such that

∑

converges.

(Hint: Try to construct a subsequence which converges to zero faster than

6. Prove that the harmonic series

∑

diverges by comparing with the series

∑

, where

) =



, . . .



7. Suppose b

≤ a

for all n and that

∑

and

∑

converge. Prove that

∑

≤

∑

(This also proves part 2 of Corollary 3.13)

8. Given

∑

∞

n=1

, ﬁnd the values of

∑

(2n)

∑

(2n+1)

and

∑

(−1)

n+1

9. The limit comparison test states:

Suppose

∑

are series of positive terms and that L = lim

∈ (0, ∞). Then the

series have the same convergence status (both converge or both diverge to ∞).

(a) Use the limit comparison test with b

to show that the series

∑



1 +



converges.

(Hint: Recall that e = lim



1 +



)

(b) Prove the limit comparison test.

(Hint: ﬁrst show that

for large n)

∑

and

∑

if L = 0 or L = ∞? Explain.

10. Euler asserted that the sine function, written as an inﬁnite polynomial in the form of a Maclau-

rin series, could also be expressed as an inﬁnite product,

sin x =

∞

∑

n=0

(−1)

(2n + 1)!

2n+1

= x



1 −



1 −

4π



1 −

9π



···

By considering the solutions to sin x = 0, give some weight to Euler’s claim. By comparing

coefﬁcients in these expressions, deduce the fact

∑

(As presented, this argument is non-rigorous!)

3.15 The Integral and Alternating Series Tests

In this section we develop two further standalone series tests, both with narrower applications than

our previous tests.

The ﬁrst is a little out of place given that it requires (improper) integration.

Theorem 3.18 (Integral test). Let a

= f (n) , where f is non-

negative, non-increasing, and integrable on [1, ∞). Then

∞

∑

n=1

converges ⇐⇒

∞

f (x) dx converges

In such a situation,

∞

f (x) dx ≤

∞

∑

n=1

≤ a

∞

f (x) dx

The statement is easily modiﬁed if the initial term is not a

f (x)

0 1 2 3 4 5 6

Proof. We need only interpret the picture as describing upper and lower Riemann sums:

n+1

f (x) dx ≤

∑

k=1

= s

= a

∑

k=2

≤ a

f (x) dx (∗)

Take limits as n → ∞ for the result.

Even for divergent sums, (∗) allows us to estimate the growth of (s

). For greater accuracy, we may

evaluate the ﬁrst few terms explicitly and modify the integral test to estimate the remainder.

An important application of the integral test is to provide a complete description of the convergence

status of p-series: a useful family of series to which others may be compared.

Corollary 3.19 (p-series). Let p > 0. The series

∑

converges if and only if p > 1.

Examples 3.20. 1.

∑

converges (it is a p-series with p > 1). Moreover,

∞

dx = lim

b→∞



−

−2



=⇒

≤

∞

∑

n=1

≤

This is a poor estimate, particularly the lower bound. For a quick improvement, evaluate the

ﬁrst term and re-run the test starting at n = 2:

1 +

∞

dx ≤

∞

∑

n=1

≤ 1 +

∞

dx =⇒ 1 +

≤

∞

∑

n=1

≤ 1 +

If greater accuracy is required, more terms can be explicitly evaluated.

Which in turn requires limits of functions:

∞

f (x) dx := lim

b→∞

f (x) dx. While we haven’t rigorously developed

these concepts, the relevant computations should be familiar from elementary calculus.

2. In Example 3.9, we used the Cauchy criterion to show that the harmonic series diverges to ∞.

The integral test makes this much easier and allows us to estimate how many terms are required

for the partial sum to s

to reach a certain threshold: 10 say. Since

ln(n + 1) =

n+1

dx ≤ s

∑

k=1

≤ 1 +

dx = 1 + ln n

we see that s

≈ 10 requires

ln(n + 1) ≤ 10 ≤ 1 + ln n =⇒ e

≤ n ≤ e

−1 =⇒ 8104 ≤ n ≤ 22025

The harmonic series diverges to inﬁnity, but it does so very slowly.

3. The integral test shows that

∑

∞

n=2

n ln n

= ∞. To exceed 10, somewhere between 10

3223

and 10

6631

terms are required. To exceed 100 requires ‘roughly’ 10

6×10

terms: 1 followed by 1000 zeros

for each water molecule in Lake Tahoe puts you at least in the right ballpark. . .

4. The series

∑

2n+1

√

−1

diverges to ∞ by comparison with the p-series

∑

√

Alternating Series and Conditional Convergence

Our ﬁnal test is unique in that it can detect conditional conver-

gence. The canonical example is the alternating harmonic series

(Example 3.14.5). With an eye on generalization, we re-index

so that the ﬁrst term is a

= 1:

s =

∞

∑

n=0

(−1)

n + 1

= 1 −

−

+ ···

∞

∑

n=0

(−1)

= a

− a

+ a

− a

+ ···

10 20

(−1)

−

The alternating ±-signs give the series its name. Consider the behavior of the sequence of partial

sums (s

), in particular two subsequences ( s

) = (s

) and ( s

−

) = (s

2n−1

∑

k=0

(−1)

= 1 −



−

|{z }

−a



−



−

|{z }

−a



−··· −



−

2n + 1

| {z }

2n−1

−a



(n ≥ 0)

−

2n−1

∑

k=0

(−1)



1 −

|{z}

−a





−

|{z }

−a



+ ··· +



2n −1

−

| {z }

2n−2

−a

2n−1



(n ≥ 1)

Each bracketed term is non-negative, so (s

) is monotone-down and ( s

−

) monotone-up. Moreover,

= s

−

≤ s

−

≤ s

−

+ a

= s

≤ s

= 1 (†)

from which both subsequences are bounded and thus convergent. Not only this, but

lim



−s

−



= lim a

= 0

shows that the limits of both subsequences are identical (of course both are s).

The above discussion depends only on two simple properties of the sequence (a

). We’ve therefore

proved a general statement.

Theorem 3.21 (Alternating series test). Suppose (a

) is monotone-down and that lim a

= 0. Then:

1. The alternating series s =

∑

(−1)

converges.

2. If (s

) is the sequence of partial sums, then

s −s

≤ a

n+1

Think about where the hypotheses regarding (a

) are used in the proof.

It can be shown that the alternating harmonic series converges to ln 2, though the estimates provided

by the alternating series test are very poor: to guarantee accuracy to two decimal place requires us to

sum 100 terms of the series!

Examples 3.22. 1. Since a

converges monotone-down to zero, the alternating series

∑

(−1)

converges. By evaluating s

and s

explicitly, we see that

0.3678791887 . . . ≤

∞

∑

n=0

(−1)

≤ 0.3678819444 . . .

yielding an estimate of 0.36788 to 5 decimal places (the exact value is in fact e

−1

). The alternating

series test is only needed for the estimate, since the series converges absolutely.

2. The series

∑

∞

n=2

sin

ln n

can be viewed as an alternating series since every even term is zero. Writ-

ing m = 2n + 1, we obtain

∞

∑

n=2

sin

ln n

∞

∑

m=1

sin(πm +

)

ln( 2m + 1)

∞

∑

m=1

(−1)

ln( 2m + 1)

Since

ln(2m+1)

decreases to zero, the alternating series test demonstrates convergence.

Rearranging Inﬁnite Series

A rearrangement of an inﬁnite series

∑

is a series that results from changing the order of the terms

of the sequence (a

) before computing the partial sums. The new series must still use every term of

the original. Since the new sequence of partials sums is likely different, we shouldn’t assume that the

rearranged series has the same convergence properties as the old.

Example 3.23. We rearrange the alternating harmonic series by summing two positive terms before

each negative term:

1 +

−

+ ··· +

4n −3

4n −1

−

+ ···

Every term of the original sequence is used here, so this is a genuine rearrangement. It is perhaps

surprising to discover that the new series converges, though its limit is not the same as the original

alternating harmonic series! We leave the details to Exercise 11. This behavior is quite different to

that of ﬁnite sums, where the order of summation makes no difference at all.

The general situation is summarized in a famous result of Riemann. The ﬁrst part says that absolutely

convergent series behave just like ﬁnite sums. Conditionally convergent series are much stranger.

Theorem 3.24 (Riemann rearrangement). 1. If a series converges absolutely, then all rearrange-

ments converge to the same limit.

2. If a series converges conditionally and s ∈ R ∪{±∞} is given, then there exists a rearrangement

which tends to s.

We omit the proofs since they are prohibitively lengthy. Instead we illustrate the rough idea of part 2

via an example.

Example 3.25. We show how to construct a rearrangement of the alternating harmonic series which

converges to s =

√

2 = 1.41421 . . .

First we convince ourselves that the sum of the positive terms

∑

diverges to inﬁnity. The compar-

ison test makes this easy:

2n −1

=⇒

∑

2n −1

∑

= ∞

The negative terms also diverge:

∑

−

= −∞. Construction of the rearrangement is inductive.

1. Sum just enough positive terms S

= a

+ a

+ ··· + a

in order until the partial sum exceeds

s: plainly S

= 1 +

≈ 1.53333 will do here.

2. Add negative terms starting at the beginning of the sequence until the sum is less than s:

= S

+ (a

−

+ a

−

+ ··· + a

−

) = 1 +

−

= 1.0333 ··· < s

3. Repeat: add positive terms until the sum just exceeds s, then add negative terms, etc.,

= S

= 1.4551 . . . > s, S

= S

−

= 1.2051 . . . < s

Continuing the process ad inﬁnitum, we claim that

s =

√

2 = 1 +

−

+ ···

To see why, observe:

• Since

∑

= ∞ and

∑

−

= −∞, at each stage we need only add/subtract ﬁnitely many terms.

• All terms of the original sequence (a

) are eventually used since we add the positive (negative)

terms in order. E.g., a

495

appears, at the latest, during the 495

positive-addition phase.

•

−s

≤

, where a

is the ﬁnal term used at the n

stage. The right hand side converges

to zero (n

-term test!), whence lim S

= s.

Riemann’s second result is in fact even stronger. Conditionally convergent series also have rearrangements whose

sequence of partial sums diverges by oscillation to any given lim inf s

< lim sup s

Exercises 3.15. Key concepts: Integral test and approximation, Alternating series and approximation

1. Use the integral test to determine whether the series

∑

∞

n=1

converges or diverges.

2. Prove Corollary 3.19 regarding the convergence/divergence of p-series.

3. Let s

∑

k=1

√

. Estimate how many terms are required before s

≥ 100.

4. (Example 3.20.3) Verify the claim that

∑

∞

n=2

n ln n

= ∞ and the claim regarding the estimate.

5. (a) Use calculus to show that a

ln n

is monotone-down whenever n ≥ 2.

(b) Show that lim a

= 0, and that the hypotheses of the integral test are therefore satisﬁed.

∑

∞

n=2

ln n

converges or diverges.

6. (a) Give an example of a series

∑

which converges, but for which

∑

diverges.

(Exercise 3.14.3 really requires that

∑

be absolutely convergent!)

(b) Give an example of a divergent series

∑

for which

∑

converges.

7. Suppose (a

) satisﬁes the hypotheses of the alternating series test except that lim a

= a is

strictly positive. What can you say about the sequences (s

) and (s

−

) and the series

∑

(−1)

8. Let a

have partial sum s

∑

k=1

, and deﬁne a new sequence (t

) by

= s

−ln n = 1 +

+ ··· +

−ln n

Prove that (t

) is a positive, monotone-down sequence, which therefore converges.

(Hint: You’ll need the mean value theorem from elementary calculus)

9. Suppose

∑

is conditionally convergent and let

∑

be the series obtained by summing, in

order, the positive terms of the sequence (a

). Prove that

∑

= ∞.

10. (a) Show that the series

∑

∞

n=1

(−1)

is conditionally convergent to some real number s.

(b) How many terms are required for the partial sum s

to approximate s to within 0.01.

of the series in part (a) which converges to 0.

11. Recall the rearrangement of the alternating harmonic series in Example 3.23.

(a) Verify that the subsequence of partial sums (s

) is monotone-up, by checking that

4n −3

4n −1

−

> 0, for all n ∈ N

(b) Use the comparison test to show that

∑

converges.

(Thus s > ln 2 ≈ 0.69, the limit of the original alternating harmonic series)

The limit γ := lim t

≈ 0.5772 is the Euler–Mascheroni constant. It appears in several mathematical identities, and yet

very little about it is understood; it is not even known whether γ is irrational!

4 Continuity

In this chapter we discuss continuous functions. Functions themselves should be familiar. For refer-

ence, we begin with a review of some basic concepts and conventions.

We are concerned with functions f : U → V where both U, V are subsets of the real numbers R and

f is some rule assigning to each real number x ∈ U a real number f (x) ∈ V. For instance

f (x) =

(x −7)

(x −2)(x

−9)

assigns to x = 1 the value f (1) =

1(−6)

(−1)(−8)

Domain dom f = U is the set of inputs to f . When f is deﬁned by a formula, its implied domain

is the largest set on which the formula is deﬁned: the above example has implied domain

dom f = R \ {2, 3, −3}. In examples, the domain is most often a union of intervals of positive

length.

Codomain codom f = V is the set of possible outputs. In real analysis, we often take V = R by default.

Range range f = f (U) =



f (x) : x ∈ U



is the set of realized outputs, and is a subset of V = codom f .

Injectivity f is injective/one-to-one if distinct inputs produce distinct outputs. This is usually stated in

the contrapositive: f (x) = f (u) =⇒ x = u.

Surjectivity f is surjective/onto if every possible output is realized: that is f (U) = V.

Inverses f is bijective/invertible if it is both injective and surjective. Equivalently, f has an inverse

function f

−1

: V → U deﬁned as follows:

• Given y ∈ V, f surjective =⇒ ∃x ∈ U such that f (x) = y.

• Since f is injective, f (x) = f (u) =⇒ x = u, so x is unique. We deﬁne f

−1

(y) = x.

Example 4.1. The function deﬁned by f (x) =

x(x−2)

has implied

dom f = R \{0, 2} = (−∞, 0) ∪ (0, 2) ∪ (2, ∞)

range f = (−∞, −1] ∪ (0, ∞)

The function is neither injective (e.g., f (3) = f (−1)) nor surjective

(e.g., 0 ∈ range f ).

We can remedy both issues by restricting the domain and codomain.

For instance, the same rule/formula but with

dom

f = [1, 2) ∪ (2, ∞)

codom

f = (−∞, −1] ∪ (0, ∞)

deﬁnes a bijection with inverse function

−1

(y) =

(

1 + y

−1

y + 1 if y > 0

1 − y

−1

y + 1 if y ≤ −1

Observe that dom

−1

= codom

f and codom

−1

= dom

f .

−2

−1

f (x)

−1 1 2 3

−2 −1 0 1 2

−1

(y)

4.17 Continuous Functions

To introduce continuity, consider two common na

ıve notions.

The graph of f can be drawn without removing one’s pen from the page This is intuitive but un-

usable: drawn is poorly deﬁned, so how might we calculate or prove anything with this concept?

It moreover cannot reasonably be extended to other situations or higher dimensions where

drawing a graph is meaningless.

If x is close to a, then f (x) is close to f (a) This is better and admits generalization. The major is-

sue is the unclear meaning of close. Our formal deﬁnition of continuity addresses this using

sequences and limits.

Deﬁnition 4.2 (Sequential continuity). A real-valued function f : U → V is continuous at a ∈ U if,

∀(x

) ⊆ U, lim x

= a =⇒ lim f (x

) = f (a)

f is continuous (on U) if it is continuous at every point a ∈ U.

We say that f is discontinuous at a ∈ U if,

∃(x

) ⊆ U, such that lim x

= a and



f (x

)



does not converge to f (a)

f (x

)

f (x

)

(a, f (a))

···

lim f (x

)

Continuity at a: every sequence

with lim x

= a has lim f (x

) = f (a)

(a, f (a))

f (x

)

f (x

)

···

lim f (x

)

f (a)

Discontinuous at a: at least one sequence

with lim x

= a has lim f (x

) = f (a)

Examples 4.3. 1. f : R → R : x 7→ x

is continuous (at every a ∈ R). To see this, suppose (x

)

converges to a, then, by the limit laws,

lim f (x

) = lim x

= (lim x

)

= a

= f (a)

2. The function with g(x) = 1 +

is continuous. Choose any a ∈ dom g = R \ {0} and any

) ⊆ dom g with lim x

= a. Again, by the limit laws,

lim g(x

) = lim



1 +



= 1 +

(lim x

)

= 1 +

= f (a)

This example (with a = 1 and x

= 1 +

) is the ﬁrst picture in the above deﬁnition.

3. h : [0, ∞) → R : x 7→ 3x

1/4

is continuous. Again, everything follows from the limit laws. If

→ a where x

≥ 0 and a ≥ 0, then

lim h(x

) = lim 3x

1/4

= 3(lim x

)

1/4

= 3a

1/4

= h(a)

4. The function deﬁned by

k(x) =

(

1 + 2

√

x if x < 1

2 − x if x ≥ 1

is discontinuous at a = 1. This seems obvious from the picture, but

we need to use the deﬁnition. The sequence with x

= (1 −

)

converges to 1 from below, however the limit laws tells us that

lim k(x

) = lim



1 + 2



1 −



= 3 = 1 = k(1)

k(x)

0 2

k(1)

→

Basic Examples and Combinations of Continuous Functions

By appealing to the limit laws for sequences (Theorem 2.15), continuous functions may be combined

in natural ways. For instance, if f , g are continuous at a, then

lim x

= a =⇒ lim f (x

) + g(x

) = lim f (x

) + lim g(x

) = f (a) + g(a)

whence f + g is continuous at a. Here is a general summary.

Theorem 4.4. 1. Suppose f , and g are continuous and that k is constant. Then the following functions

are continuous (on their domains):

k f ,

, f + g, f − g, f g,

, max( f , g), min( f , g)

2. If n ∈ N then f : x 7→ x

1/n

is continuous on its domain.

3. Compositions of continuous functions are continuous: if g is continuous at a and f is continu-

ous at g(a), then f ◦ g is continuous at a.

4. Algebraic functions are continuous (includes all polynomials and rational functions).

Proof. Parts 1, 2 are the limit laws; for the maximum and minimum, see Exercise 2. For part 3:

lim x

= a

g cont

=⇒ lim g(x

) = g(a)

f cont

=⇒ lim f



g(x

)



= f



g(a)



Part 4 follows by combining parts 1, 2 and 3.

Example 4.5. The following algebraic function is continuous on its domain

f : (7, ∞) → R : x 7→

5/2

+ 7x

+ 4

(x −7)

1/3

Theorem 4.6 (Squeeze theorem). Suppose f (x) ≤ g(x) ≤ h(x) for all x = a, that f , h are continuous

at a, and that f (a) = g(a) = h(a). Then g is continuous at a.

Proof. This is simply the squeeze theorem (2.12) for sequences: if lim x

= a, then

f (x

) ≤ g(x

) ≤ h(x

) =⇒ lim g(x

) = g(a)

To provide more interesting examples, we state the following without proof.

Theorem 4.7. The common trigonometric, exponential and logarithmic functions are continuous.

It is possible, though slow and ugly, to address some of this now. We won’t do this since it is

cleaner to deﬁne these functions using power series,

which makes their continuity (and differ-

entiability/integrability!) come for free.

Examples 4.8. 1. f (x) =

√

sin e

is continuous on its domain R \



ln(nπ) : n ∈ N



2. If g(x) = x sin

when x = 0, and g(0) = 0, then g is continuous

on R. When x = 0, this follows from Theorems 4.4 and 4.7, while

at a = 0 we rely on the squeeze theorem:

x = 0 =⇒ −x ≤ x sin

≤ x

g(x)

The ϵ-δ Deﬁnition of Continuity

The sequential deﬁnition of continuity uses limits twice. By stating each of these using the ϵ-deﬁnition

of limit, we can reformulate continuity without mentioning sequences!

To motivate this, consider f (x) = x

at a = 2. By continuity, if (x

) is a sequence with lim x

= 2,

then lim f (x

) = 4. We restate each of these using the deﬁnition of limit:

(a) (lim x

= 2) ∀δ > 0, ∃M such that n > M =⇒

−2

< δ

(b) (lim x

= 4) ∀ϵ > 0, ∃N such that n > N =⇒



−4



< ϵ

Here is a short argument that shows how (a) ⇒(b) (we’ll revisit this formally in a moment).

Assume (a) and suppose ϵ > 0 is given. Deﬁne δ = min(1,

). Since lim x

= 2, ∃M such that

n > M =⇒



−4



−2

+ 2

< δ

−2) + 4

(by (a))

≤ δ



−2

+ 4



(△-inequality)

< δ(δ + 4) ≤ 5δ ≤ ϵ ((a) again)

Let N = M to conclude (b).

It turns out not to be very important that (x

) be a sequence. In fact we can dispense with it entirely. . .

For instance via Maclaurin series: e

∞

∑

n=0

, sin x =

∞

∑

n=0

(−1)

(2n+1)!

2n+1

and cos x =

∞

∑

n=0

(−1)

(2n)!

Deﬁnition 4.9 (ϵ-δ continuity). A real-valued function f : U → V is continuous at a ∈ U if

∀ϵ > 0, ∃δ > 0 such that ( ∀x ∈ U)

x −a

< δ =⇒

f (x) − f (a)

< ϵ (∗)

We say that f is discontinuous at a ∈ U if,

∃ϵ > 0 such that ∀δ > 0, ∃x ∈ U with

x −a

< δ and

f (x) − f (a)

≥ ϵ (†)

f (a) + ϵ

f (a) − ϵ

a − δ a + δa

f (x)

To force f (x)

to live here. . .

. . . it is enough for x

to live here

Continuity at a

a + δa − δ

f (a) + ϵ

f (a) − ϵ

f (a)

f (x)

Regardless of δ, there is

always some x here. . .

. . . so that f (x)

does not live here

Discontinuous at a

This ﬁts with the intuitive interpretation of continuity: if x is close to a, then f (x) is close to f (a); ϵ

and δ are our measures of closeness. Many mathematicians consider the ϵ-δ version to be the deﬁnition

of continuity. Thankfully, it doesn’t matter which you prefer. . .

Theorem 4.10. The sequential and ϵ-δ deﬁnitions of continuity (4.2 & 4.9) are equivalent.

Examples (4.3, cont). Before seeing a proof, we repeat our earlier examples using the ϵ-δ deﬁnition.

As with ϵ-N arguments for limits, it is often useful to do some scratch work ﬁrst.

1. Suppose f (x) = x

and a ∈ R. Our goal is to control the size of



− a



whenever

x −a

small. To keep things simple, assume

x −a

< 1, then,



− a



x −a

x + a

x −a



(x − a) + 2a



△

≤

x −a



x −a

+ 2



x −a



1 + 2



Now let ϵ > 0 be given and deﬁne δ = min(1,

1+2

). Then

x −a

< δ =⇒

f (x) − f (a)



− a



< δ



1 + 2



≤ ϵ

Thus f is continuous at a. This is simply a general version of the argument on page 63 with all

mention of sequences removed!

The bracketed ∀x ∈ U is often omitted in (∗) since the implication requires that x be universally quantiﬁed. It is

important that x ∈ U = dom f rather than merely x ∈ R! By contrast, the expression ∃x ∈ U in (†) is always written.

2. Let g(x) = 1 +

and a = 0. The ﬁrst challenge is to keep away from zero so that

behaves.

To do this, we insist that δ ≤

, so that

x −a

< δ =⇒

=⇒

(∗)

Now consider the required difference. If

x −a

< δ, then

g(x) − g(a)



1 +

−1 −



− x



a + x

x −a

a + x

△

≤ 4





(∗)

< 4

δ =

Given ϵ > 0, it sufﬁces to let δ = min(

ϵ). Then

x −a

< δ =⇒

g(x) − g(a)

< ϵ.

3. For h(x) = 3x

1/4

there are two cases. Suppose ϵ > 0 is given.

• If a = 0, let δ =





, then

x −a

< δ =⇒ 0 ≤ x < δ =⇒

h(x) − h(a)

= 3x

1/4

< 3δ

1/4

= ϵ

• If a > 0, let δ =

3/4

ϵ. Then, if

x −a

< δ,

h(x) − h(a)

= 3



1/4

− a

1/4



x −a

+ a

≤

x −a

3/4

3δ

3/4

= ϵ

4. We could establish the discontinuity statement (†) directly, but it is

typically easier to argue by contradiction.

Suppose k is continuous at 1 and let ϵ = 1. Then ∃δ > 0 for which

x −1

< δ =⇒

k(x) − k( 1)

k(x) −1

< 1

=⇒ 0 < k(x) < 2

However, x = max(

, 1 −

) satisﬁes

x −1

≤

< δ and k(x) ≥

) = 1 +

= 2. Contradiction. Think this last bit through!

0 1 2x

k(x)

2δ

2ϵ

The basic rules for combining continuous functions may also be proved using ϵ-δ arguments. E.g.,

ϵ-δ proof of the squeeze theorem. Given ϵ > 0, we know there exist δ

, δ

> 0 for which

x −a

< δ

=⇒

f (x) − f (a)

< ϵ and

x −a

< δ

=⇒

h(x) − h(a)

< ϵ

Let δ = min(δ

, δ

), then

x −a

< δ =⇒

g(x) − g(a)

≤ max



f (x) − f (a)

h(x) − h(a)



< ϵ

whence g is continuous at 0.

Remember the hidden quantiﬁer:

x −a

< δ for all x ∈ dom f = [0, ∞), thus x ≥ 0 for the duration of this example.

Several other arguments are in the exercises. Finally, here is the promised proof of equivalence.

Proof of Theorem 4.10. (sequential ⇒ ϵ–δ) We prove the contrapositive. Suppose a is an ϵ-δ disconti-

nuity (†) and let δ =

. Then there exists x

∈ U such that

− a

and

f (x

) − f (a)

≥ ϵ

Repeating for all n ∈ N plainly produces a sequence (x

) for which lim x

= a and lim f (x

) =

f (a): otherwise said, a is a sequential discontinuity.

(ϵ-δ ⇒ sequential) Assume (∗), let (x

) ⊆ U and suppose lim x

= a; we must prove that lim f (x

) =

f (a). Let ϵ > 0 be given so that a suitable δ satisfying (∗) exists. Since lim x

= a,

∃N such that n > N =⇒

− a

< δ (since x

→ a and δ > 0 is given)

=⇒

f (x

) − f (a)

< ϵ (by (∗))

We conclude that lim f (x

) = f (a), as required.

Examples 4.11. We ﬁnish with a couple of esoteric examples on the same theme.

1. Let f : R → R be the indicator function for the rational numbers:

f (x) =

(

1 x ∈ Q

0 x /∈ Q

Suppose f is continuous at a and let ϵ = 1. Then ∃δ such that

x −a

< δ =⇒

f (x) − f (a)

< 1 (‡)

There are two cases; both rely on the fact that any interval contains both rational and irrational

numbers (Corollary 1.23, etc.).

(a) If a ∈ Q, then f (a) = 1. There exists an irrational number x ∈ (a − δ, a + δ), whence

f (x) − f (a)

0 −1

= 1 < 1.

(b) If a /∈ Q, then f (a) = 0. There exists a rational number x ∈ (a − δ, a + δ), whence

f (x) − f (a)

1 −0

= 1 < 1.

Either way, we have contradicted (‡). We conclude that f is nowhere continuous.

2. Let g : R → R be deﬁned by

g : R → R : x 7→

(

x x ∈ Q

0 x /∈ Q

Since 0 ≤

g(x)

≤

, the squeeze theorem tells us that g is continuous at x = 0.

Now suppose g is continuous at a = 0 and let ϵ =

. Then ∃δ such that

x −a

< δ =⇒

f (x) − f (a)

The same two cases as in the previous example provide contradictions. We conclude that g is

continuous at precisely one point!

Exercises 4.17. Key concepts: Sequential and ϵ-δ continuity deﬁnitions/equivalence, ϵ-δ examples

1. Consider the function with f (x) =

√

+2x−3

(a) The implied domain of f has the form dom f = (−∞, a) ∪ (b, ∞). Find a and b.

(b) What is the range of f ?

(d) Find the inverse function when we instead restrict the domain to (−∞, a).

(e) Brieﬂy explain why f is continuous on its domain.

2. Let f and g be continuous functions at a.

(a) Show that max( f , g) =

( f + g) +

f − g

and deduce that max( f , g) is continuous at a.

(b) How might you show continuity of min( f , g)?

3. Use ϵ-δ arguments to prove the following.

(a) f (x) = x

−3x is continuous at x = 1. (b) g(x) = x

is continuous at x = a.

√

x is continuous. (d) j(x) = 3x

−1

is continuous on R \ {0}.

4. Rephrase Example 4.3.4’s ϵ-δ argument by directly justifying the discontinuity deﬁnition (†).

5. Prove that each function is discontinuous at x = 0; use both sequential and ϵ-δ formulations.

(a) f (x) = 1 for x < 0 and f (x) = 0 for x ≥ 0.

(b) g(x) = sin

for x = 0 and g(0) = 0.

6. Suppose f and g are continuous at a. Prove the following using ϵ-δ arguments.

(a) f − g is continuous at a.

(b) If h is continuous at f (a), then h ◦ f is continuous at a.

7. Suppose f : U → V ⊆ R is a function whose domain U contains an isolated point a: i.e. ∃r > 0

such that (a −r, a + r) ∩U = {a}. Prove that f is continuous at a.

8. In Example 4.11.2, provide the details of the required contradiction.

9. (a) Suppose f : R → R is a continuous function for which f (x) = 0 whenever x ∈ Q. Prove

that f (x) = 0 for all x ∈ R.

(b) Suppose f , g : R → R are continuous functions such that f (x) = g(x) for all rational x.

Prove that f = g.

10. (Hard) Consider f : R → R where

f (x) =

(

whenever x =

∈ Q with q > 0 and gcd(p, q) = 1

0 if x ∈ Q

For instance, f (1) = f (2) = f (−7) = 1, and f (

) = f (−

) = f (

) = ··· =

, etc.

(a) Prove that f is discontinuous at each rational number r.

(b) Prove that f is continuous at each irrational number i.

(Hint: given ϵ > 0, let q = ⌈

⌉, A =



r ∈ Q : f (r) ≥



and let δ = min

r∈A

i −r

. . . )

4.18 Properties of Continuous Functions

In this section we consider how continuous functions transform intervals.

Example 4.12. f (x) = x

maps [−3, 2] onto [0, 9]. In particular:

• f transforms an interval into another.

• f transforms a closed bounded set into another.

Our goal is to see that these are general properties exhibited by any

continuous function.

f (x)

−3 −2 −1 0 1 2

First recall a couple of deﬁnitions.

Deﬁnition 4.13. Suppose f : U → V where U, V ⊆ R.

1. (a) U is bounded if ∃M such that ∀x ∈ U,

≤ M.

(b) f is bounded if its range is a bounded set: ∃M such that ∀x ∈ U,

f (x)

≤ M.

2. (Deﬁnition 2.46) U is closed if every convergent sequence in U has its limit in U:

∀(x

) ⊆ U, lim x

= s (∈ R) =⇒ s ∈ U

Theorem 4.14 (Extreme Value Theorem). Suppose f : U → V is continuous where U is closed and

bounded. Then f (U) is closed and bounded. In particular, f is bounded and attains its bounds:

∃s, i ∈ U such that f (s) = sup f (U) and f (i) = inf f (U)

Examples 4.15. 1. (Example 4.12) If f (x) = x

on U = [−3, 2], then f (U) = [0, 9] is closed and

bounded. Moreover, sup f (U) = f (−3) and inf f (U) = f (0) (i.e., s = −3 and i = 0).

2. Before seeing the proof, here are three examples where we weaken one of the hypotheses of the

extreme value theorem and see that the conclusion fails.

0 1 2

(a) f discontinuous (b) U not closed (c) U not bounded

(a) If U = [0, 2], f (x) = 3x when x < 1 and f (x) = 1 when x ≥ 1, then f (U) = [0, 3). In

particular, sup f (U) = 3 is not attained.

(b) If f (x) =

√

2−x

and U = [0, 2), then f (U) = [

√

, ∞) is unbounded.

and U = [0, ∞), then f (U) = [0, ∞) is unbounded.

The strategy of the proof is to show that every limit point of f (U) = range f lies in f (U). We break

things into simple steps; observe where each hypothesis is used.

Proof. 1. Suppose M is a limit point of f (U): that is, M = lim f (x

) for some sequence (x

) ⊆ U.

A priori, M need not be ﬁnite, but M = sup f (U) or inf f (U) are certainly possible.

2. Since (x

) ⊆ U is bounded, Bolzano–Weierstraß (Theorem 2.41) says it has a convergent sub-

sequence, lim

k→∞

= x.

3. Since U is closed, we have x ∈ U. This means f (x) can be evaluated (it is ﬁnite).

4. Since f is continuous, lim f (x

) = f (x).

5. Finally, M = f (x) since all subsequences of a convergent (or divergent to ±∞) sequence tend

to the same limit (Lemma 2.37). It follows that all limit points M are ﬁnite and lie in f (U):

otherwise said, f (U) is closed and bounded.

Choosing M = sup f (U) yields x = s ∈ U (similarly inf f (U) leads to i ∈ U).

Example 4.16. It is worth considering why we needed a subsequence in the proof. The reason is that

the bounds of f might be attained multiple times. For example, suppose

f : [0, 4π] → R : x 7→ sin x

This satisﬁes the hypotheses of the extreme value theorem: U = [0, 4π] is closed and bounded and f

is continuous. Indeed max f ( U) = 1 is attained at both x =

and

5π

. The sequence deﬁned by

(

if n is odd

5π

if n is even

has f (x

) = sin





−−−→

n→∞

1 = sup f (U)

and therefore satisﬁes step 1 of the proof. However, (x

) itself is divergent by oscillation. Bolzano–

Weierstraß is used to force the existence of a convergent subsequence; in this case the subsequence of

odd terms (x

) = (x

2k−1

) satisﬁes the remaining steps.

−1

f (x)

f (x

)

3π

2π

5π

3π

7π

4π

If M = sup f (U), then a suitable (x

) might be constructed as follows:

• If M ∈ R, then for each n ∈ N, ∃x

∈ U such that M −

< f (x

) ≤ M (Lemma 1.20).

• If M = ∞, then for each n ∈ N, ∃x

∈ U such that f (x

) ≥ n.

The Intermediate Value Theorem and its Consequences

This result should be familiar from elementary calculus, even if its proof is not. It should also be

intuitive: like the Grand Old Duke of York, if you march up a hill, then at some point you must be

half-way up. . .

Theorem 4.17 (Intermediate Value Theorem (IVT)). Suppose f is continuous on [a, b] and that y

lies strictly between f (a) and f (b). Then ∃ξ ∈ (a, b) such that f (ξ) = y.

Being an existence result, it should be no surprise that completeness is used in the proof.

Proof. WLOG assume f (a) < y < f (b). Let S =



x ∈ [a, b] : f (x) < y



and deﬁne ξ := sup S.

Since S is non-empty (a ∈ S) and bounded above

(by b), we see that ξ exists and is ﬁnite. It remains to

prove that f (ξ) = y and ξ = a, b.

First choose any (s

) ⊆ S such that lim s

= ξ. Con-

tinuity forces lim f (s

) = f (ξ). Moreover

f (s

) < y =⇒ f (ξ) ≤ y

Since f (b) > y, this also shows that ξ = b.

a b

f (a)

f (b)

→ ← x

We now play a similar game from the other side: deﬁne x

:= min(ξ +

, b), then lim x

= ξ and

> ξ = sup S =⇒ x

∈ S =⇒ f (x

) ≥ y

=⇒ f (ξ) = lim f (x

) ≥ y

again via the continuity of f and the convergence properties of bounded sequences. Since y > f (a),

we also conclude that ξ = a.

Putting it all together, f (ξ) = y and ξ ∈ (a, b).

Note how the value of ξ in the proof is always the largest of potentially several choices.

Examples 4.18. In elementary calculus, the intermediate value theorem is typically applied to

demonstrate the existence of solutions to equations.

1. We show that the equation x

+ 3x = 1 + 4 cos(πx) has a solution.

The trick is to express the equation in the form f (x) = y where f is continuous, then choose

suitable a, b to ﬁt the theorem. In this case,

f (x) = x

+ 3x −4 cos(πx) and y = 1

are suitable choices. Now observe

f (0) = −4 < y and f (1) = 1 + 3 + 4 = 8 > y (i.e., a = 0 and b = 1)

whence ∃ξ ∈ (0, 1) such that f (ξ) = y = 1. Otherwise said, ξ is a solution to the original

equation.

The function f is plainly continuous on R, a much larger interval than [a, b], but no matter.

2. The existence of a root ξ of the (continuous) polynomial

f (x) = x

−5x

+ 150

follows from the intermediate value theorem by observing that

f (0) = 150 > 0 and f (4) = −256 + 150 = −106 < 0

We conclude that such a root ξ exists satisfying ξ ∈ (0, 4).

As the graph suggests, there are other roots (η, ζ), the existence of

which may be shown by observing, say,

f (−3) = −798 < 0 and f (5) = 150 > 0

−100

100

200

−2 2 4

With an eye on generalizing, here is an alternative approach. Deﬁne sequences (s

), ( t

) via

f (−n)

= −1 −

150

f (n)

= 1 −

150

Since lim s

= −1 and lim t

= 1, we see that

∃a such that s

< −

=⇒ f (−a) = a

< −

< 0

∃b such that t

=⇒ f (b) = b

> 0

Applying the intermediate value theorem on [−a, b] shows the existence of a root.

The second approach in Example 4.16.2 may be applied to prove a general result.

Corollary 4.19. A polynomial function of odd degree has at least one real root.

The proof is an exercise. An even simpler exercise shows the existence of a ﬁxed point for a particular

type of continuous function.

Corollary 4.20 (Fixed Point Theorem). Suppose a and b are ﬁnite and that f : [a, b] → [a, b] is

continuous. Then f has a ﬁxed point:

∃ξ ∈ [a, b] such that f (ξ) = ξ

As the picture shows, a function could have several ﬁxed points.

This is the most basic ﬁxed-point theorem in analysis: if you continue

your studies you’ll meet several more. Many important consequences

ﬂow from such results, including a common fractal construction and

the standard existence/uniqueness result for differential equations.

a b

For a ﬁnal corollary, ﬁrst note a straightforward characterization that helps us consider all types of

interval simultaneously: U ⊆ R is an interval precisely when

a, b ∈ U and a < y < b =⇒ y ∈ U (∗)

Corollary 4.21 (Preservation of Intervals). Suppose U is an interval of positive length, and that

f : U → V is continuous and surjective (V = f (U)).

1. V is an interval or a point.

2. If f is strictly increasing (decreasing), then:

(a) V is an interval of positive length, f is injective, and therefore bijective.

(b) The inverse function f

−1

: V → U is also continuous and strictly increasing (decreasing).

Example 4.22. The interval V need not be of the same type as U. For

instance, if f (x) = 10x − x

, then f maps the open interval U = (2, 9)

to the half-open interval V = (9, 25] .

The extreme value theorem, however, guarantees that if U is closed

and bounded, then V is also. For instance,



[2, 9]



= [9, 25]

0 2 4 6 8

Proof. 1. If V is not a point, then ∃a, b ∈ U such that f (a) < f (b). If y lies between these, IVT says

∃ξ between a and b such that y = f (ξ). That is, y ∈ f (U). By (∗), V = f (U) is an interval.

2. (a) If f is strictly increasing, then ∀a, b ∈ U, a < b =⇒ f (a) < f (b). Plainly f is injective and

V contains at least two points; by part 1 it is an interval of positive length.

(b) Let y

< y

where both lie in V, and deﬁne x

= f

−1

) for i = 1, 2. Since f is increasing,

≤ x

=⇒ y

= f (x

) ≤ f (x

) = y

is a contradiction. Thus x

< x

and f

−1

is also strictly increasing.

If a ∈ U, it remains to show that f

−1

is continuous at b = f (a). Assume ﬁrst that a is not

an endpoint of U and let ϵ > 0 be given such that [a − ϵ, a + ϵ] ⊆ U,. Now deﬁne

δ := min



b − f (a −ϵ), f (a + ϵ) − b



This is positive since f is strictly increasing. But now

y −b

< δ =⇒ f (a −ϵ) − b < y − b < f (a + ϵ) −b

=⇒ f (a −ϵ) < y < f (a + ϵ)

=⇒ a − ϵ < f

−1

(y) < a + ϵ

=⇒



−1

(y) − f

−1

(b)



−1

(y) − a



< ϵ

where (=⇒) used the fact that f is strictly increasing.

a a + ϵa −ϵ

f (a + ϵ)

b + δ

b −δ

∥

f (a −ϵ)

If a is an endpoint of U, instead use [a −ϵ, a] ⊆ U or [a, a + ϵ] ⊆ U and only the corre-

sponding half of the expression deﬁning δ.

Example 4.23. The function f : [0, 2] → [0, 4] deﬁned by

f (x) =

(

√

x if 0 ≤ x ≤ 1

if 1 < x ≤ 2

is continuous, surjective and strictly increasing. It therefore has a continu-

ous inverse f

−1

: [0, 4] → [0, 2]. Compare this with the familiar statement

from elementary calculus: f

′

> 0 =⇒ f injective. We cannot apply this

here since f is not differentiable!

f (x)

0 1 2

Exercises 4.18. Key concepts: Extreme/Intermediate Value Theorems, Cont functions preserve intervals

1. (a) Give an example of a discontinuous function f : [0, 1] → R which is not bounded.

(b) State a continuous function with domain ( 1, ∞) whose range is bounded but not closed.

2. Let a < b be given. Give examples of continuous functions g, h : (a, b) → R such that:

(a) g is not bounded.

(b) h is bounded but does not attain its bounds.

3. Compute the inverse of the function f in Example 4.23.

4. Let S ⊆ R and suppose there exists a sequence (x

) in S converging to some x

∈ S. Show that

there exists an unbounded continuous function on S.

5. Prove that x = cos x for some x ∈ (0,

6. Suppose that f is a real-valued continuous function on R and that f (a) f (b) < 0 for some

a, b ∈ R. Prove that there exists some x between a, b such that f (x) = 0.

7. Suppose f is continuous on [0, 2] and that f (0) = f (2). Prove that there exist x, y ∈ [0, 2] such

that

y − x

= 1 and f (x) = f (y).

(Hint: consider g(x) = f (x + 1) − f (x) on [0, 1])

8. (a) Prove the ﬁxed point theorem (Corollary 4.20).

(Hint: If neither a nor b are ﬁxed points, consider g(x) = f (x) − x)

(b) Prove Corollary 4.19 for a general odd-degree monic polynomial f (x) = x

2m+1

∑

k=0

9. Consider f : R → R where f (x) = x sin

if x = 0 and f (0) = 0.

(a) Explain why f is continuous on any interval U.

(b) Suppose a < 0 < b and that f (a), f (b) have opposite signs. If y = 0, show that the

intermediate value theorem is satisﬁed by inﬁnitely many distinct values ξ.

10. (a) Suppose f : U → R is continuous and that U =

k=1

is the union of a ﬁnite sequence (I

)

of closed bounded intervals. Prove that f is bounded and attains its bounds.

(b) Let U =

∞

n=1

, where I

= [

2n−1

] for each n ∈ N. Give an example of a continuous

function f : U → R which is either unbounded or does not attain its bounds. Explain.

(This relates to the idea that ﬁnite unions of closed sets are closed, but inﬁnite unions need not be)

4.19 Uniform Continuity

Suppose f : U → V is continuous. By the ϵ-δ deﬁnition (4.9),

∀a ∈ U, ∀ϵ > 0, ∃δ(a, ϵ) > 0 such that (∀x ∈ U)

x −a

< δ =⇒

f (x) − f (a)

< ϵ (∗)

We write δ(a, ϵ) to stress that δ can depend both on the location a and the distance ϵ. The goal of this

section is to understand if/when it is possible to choose δ independently of the location a.

Example 4.24. We start with an example where our desire cannot be satisﬁed.

Consider f (x) = x

with domain U = [0, ∞). Since f is continu-

ous, given ϵ > 0 and a

∈ U, there exists δ such that

x −a

< δ =⇒

f (x) − f (a

)



− a



< ϵ

On page 64 we saw that δ = min(1,

1+2a

) was suitable, but this

depends on the location a

. Of course other expressions for δ will

also work. . .

Visualize what happens if we attempt to use the same constant δ

for different a

: imagine sliding the ﬁxed-width δ-interval along

the x-axis while simultaneously sliding the ϵ-interval vertically.

As a

increases, the image of the δ-interval eventually becomes

too large for the ϵ-interval to contain: if δ is constant, then

length



f (a

−δ, a

+ δ)



= (a

+ δ)

−(a

−δ)

= 4a

increases unboundedly with a

. For ﬁxed ϵ, as a increases, the increasing gradient of f means that we

need to choose a smaller δ.

By contrast, if f (x) = x

on a ﬁnite domain [0, b], then any δ that demonstrates continuity at x = b

will also do so everywhere else on [0, b]. We’ll check this explicitly in a moment.

To obtain a formal deﬁnition, we rewrite (∗) with the extra assumption that δ may be chosen inde-

pendently of the location a; this amounts to moving the quantiﬁer ∀a ∈ U after δ.

Deﬁnition 4.25. A function f : U → V ⊆ R is uniformly continuous if

∀ϵ > 0, ∃δ > 0 such that ( ∀x, y ∈ U)

x −y

< δ =⇒

f (x) − f (y)

< ϵ (†)

We use y instead of a for symmetry. Observe how δ, being quantiﬁed before x, y, now depends only

on ϵ. As before, the quantiﬁers for x, y are usually hidden. Note also how uniform continuity is only

relevant on the entire domain U; it makes no sense to speak of uniform continuity at a single point.

For the sake of tidiness, we make one more observation before seeing some examples.

Lemma 4.26. If f is uniformly continuous on U, then it is continuous on U.

This is trivial: (†) is the ϵ-δ continuity of f at y ∈ U, for all y simultaneously! The special feature of the

deﬁnition is that the same δ works for all y.

Examples 4.27. 1. We re-analyze f (x) = x

in view of the deﬁnition. Recall ﬁrst that

f (x) − f (y)



−y



x −y

x + y

where

x −y

is easily controlled by δ. We consider the behavior of

x + y

in two cases.

Bounded domain If U = dom f ⊆ [−T, T] for some T > 0, we show that f is uniformly

continuous. This will follow because

x + y

≤ 2T is also easily controlled.

Let ϵ > 0 be given and deﬁne δ =

, then

x −y

< δ =⇒

f (x) − f (y)

< δ ·2T = ϵ

Compare with Example 4.24. Our approach works for this function because the gradient

(and therefore the discrepancy between x

− y

and x −y) is greatest at the endpoints of

the interval. The same approach may not work for other functions!

Unbounded domain We show that f is not uniformly continuous when dom f = [0, ∞).

For contradiction, assume f is uniformly continuous; let ϵ = 1 and suppose δ > 0 satisﬁes

the deﬁnition. Taking x −y =

, we see that

x + y

= 2y +

=⇒

f (x) − f (y)



2y +



= δ



y +



> δy

Let y =

for the contradiction

f (x) − f (y)

> 1 = ϵ (large y are the problem!).

2. Let g(x) =

; we again consider two domains.

Uniform continuity on [a, b) whenever 0 < a < b ≤ ∞.

Let ϵ > 0 be given and let δ = a

ϵ. Then,

x −y

< δ =⇒

g(x) − g(y)



y − x



≤

= ϵ

where the last inequality follows because x, y ≥ a.

Non-uniform continuity on ( 0, b) whenever 0 < b ≤ ∞.

As before, let ϵ = 1 and suppose δ > 0 is given. Let

x = min



δ, 1,



and y =

Certainly x, y ∈ (0, b) and

x −y

≤

< δ. However,

f (x) − f (y)

≥ 1 = ϵ

g(x)

a b

2δ

2ϵ

Think about how ϵ and δ must relate as one slides the intervals in the picture up/down and

left/right. In this case, large values of x, y are not the problem, it’s the vertical asymptote at

zero that causes trouble.

General Conditions for Uniform Continuity

For the remainder of this section, we develop a few general ideas related to uniform continuity. The

ﬁrst is a little out of order since it depends on differentiation and the mean value theorem.

Theorem 4.28. Suppose f is continuous on an interval U (ﬁnite or inﬁnite) and differentiable except

perhaps at its endpoints. If f

′

is bounded, then f is uniformly continuous on U.

Proof. Suppose

′

(x)

≤ M. Let ϵ > 0 and δ =

. Then

x −y

< δ =⇒

f (x) − f (y)



′

(ξ)



x −y

< Mδ = ϵ

for some ξ between x and y. The existence of ξ follows from the mean value theorem.

Examples 4.29. 1. Compare the arguments in the previous exercise. For instance, if dom f ⊆ [−T, T],

f (x) = x

=⇒ f

′

(x) = 2x =⇒



′

(x)



≤ 2T

The derivative is bounded, whence f is uniformly continuous on [−T, T].

2. Any polynomial is uniformly continuous on any bounded interval.

3. The function f (x) = sin x is uniformly continuous on R since f

′

(x) = cos x is bounded (by 1).

4. Consider f (x) =

−

on ( 1, ∞). We have

′

(x) = −

=⇒



′

(x)



≤ 11

We conclude that f is uniformly continuous on (1, ∞).

The approach is often useful when you are asked to show using the deﬁnition that a function is

uniformly continuous; provided f

′

is bounded by M, you may always choose δ =

to obtain

an argument. For instance, with our function:

Given ϵ > 0, let δ =

. If x, y ∈ (1, ∞) and

x −y

< δ, then

f (x) − f (y)



−



x −y



5(x + y)

−



x −y



−



< 11

x −y

(△-inequality, since x, y > 1)

< 11δ = ϵ

As we’ll see very shortly, the above result isn’t a biconditional: non-differentiable functions and

functions with unbounded derivatives can be uniformly continuous.

If x < y then ∃ξ ∈ (x, y) such that f

′

(ξ) =

f (x) − f (y)

x −y

Our remaining conditions are variations on a theme: uniform continuity on a bounded interval U is

roughly the same thing as continuity on its closure U (Deﬁnition 2.46).

Theorem 4.30. A continuous function on a closed bounded domain is uniformly continuous.

Proof. Assume f is continuous, but not uniformly so, on a closed bounded domain U. Then

∃ϵ > 0 such that ∀δ > 0, ∃x, y ∈ U with

x −y

< δ and

f (x) − f (y)

≥ ϵ ( ∗)

Let δ =

for each n ∈ N to obtain sequences (x

), (y

) ⊆ U satisfying (∗).

Since (x

) ⊆ U is bounded, Bolzano–Weierstraß says there exists a convergent subsequence (x

)

which, since U is closed, converges to some x

∈ U.

Since

−y

≤

, we see that lim

k→∞

= x

. Finally, the continuity of f contradicts (∗):

ϵ ≤ lim

f (x

) − f (y

)

f (x

) − f (x

)

= 0

Both hypotheses on the domain are crucial: Examples 4.27 provide counter-examples if either is

weakened.

Example 4.31. f (x) =

√

x is uniformly continuous on [0, 1] since it is already continuous! This

cannot be concluded from Theorem 4.28, since the derivative f

′

(x) =

−1/2

is unbounded on (0, 1).

We now develop a partial converse, for which we ﬁrst need a lemma.

Lemma 4.32. If f is uniformly continuous on U and (x

) ⊆ U is Cauchy, then



f (x

)



is also Cauchy.

To apply the result, consider a convergent (Cauchy) sequence in U whose limit is not itself in U.

Example (4.27.2, just easier!). Let f (x) =

have U = dom f = (0, ∞) and consider the Cauchy

sequence deﬁned by x

; note crucially that its limit 0 does not lie in U. Moreover,

lim f (x

) = lim n = ∞

Plainly



f (x

)



is not Cauchy, whence f is not uniformly continuous.

Proof. Let ϵ > 0 be given. Since f is uniformly continuous,

∃δ > 0 such that ∀x, y ∈ U,

x −y

< δ =⇒

f (x) − f (y)

< ϵ

Now use this δ in the deﬁnition of (x

) being Cauchy:

∃N such that m, n > N =⇒

− x

< δ =⇒

f (x

) − f (x

)

< ϵ

Otherwise said,



f (x

)



is Cauchy.

The Cauchy condition is critical: we cannot apply uniform continuity directly to a convergent se-

quence (

− x

< δ . . .) if we do not already know that its limit (x) lies in U!

These arguments should feel familiar: compare this line to the proof of Theorem 4.10 and the rest to Theorem 4.14.

We apply the Lemma to show that a continuous function on a bounded interval is uniformly continu-

ous if and only if has a continuous extension.

Theorem 4.33. Suppose f is continuous on a bounded interval (a, b). Deﬁne g : [a, b] → R via

g(x) :=

(

f (x) if x ∈ (a, b)

lim f (x

) whenever (x

) ⊆ (a, b) and lim x

= a or b

Then f is uniformly continuous if and only if g is well-deﬁned; in such a case g is automatically

continuous.

Examples 4.34. 1. f (x) = x

− 3x + 4 is uniformly continuous on

(−2, 4) since it has a continuous extension

g : [−2, 4] → R : x 7→ x

−3x + 4

It should be obvious what is happening from the picture: to cre-

ate the extension g, we simply ﬁll in the holes at the endpoints of

the graph.

2. The function f (x) =

5−x

is continuous, but not uniformly, on the

interval ( 0, 5). This follows since

lim f



5 −



= lim n = ∞

means we cannot deﬁne g(5) unambiguously. Again the picture

is helpful; while we can ﬁll in the hole at the left endpoint (a =

0) , the vertical asymptote at b = 5 means that there is no hole to

ﬁll in and thereby extend the function.

−2 0 2 4

0 1 2 3 4 5

Proof. (⇐) Suppose g is well-deﬁned; we leave the claim that it is continuous as an exercise, but by

Theorem 4.30 it is uniformly so. Since f = g on a subset (a, b) ⊆ dom g, the same choice of δ

will work for f as it does for g: f is therefore uniformly continuous.

(⇒) Suppose f is uniformly continuous on (a, b). Let (x

), (y

) ⊆ (a, b) be sequences converging to

a. To show that g(a) is unambiguously deﬁned, we must prove that



f (x

)



and



f (y

)



are

convergent, and to the same limit.

Deﬁne a sequence

) = (x

, y

, x

, y

, x

, y

, . . .)

Plainly lim u

= a since (x

) and (y

) have the same limit. But then (u

) is Cauchy; by Lemma

4.32,



f (u

)



is also Cauchy and thus convergent. Since



f (x

)



and



f (y

)



are subsequences

of a convergent sequence, they also converge to the same (ﬁnite!) limit.

The case for g(b) is similar.

Examples 4.35. We ﬁnish with three related examples of continuous functions f : R \ {0} → R;

these will appear repeatedly as you continue to study analysis.

1. f (x) = sin

is continuous but not uniformly so. To see

this, note that x

(n+

)π

deﬁnes a Cauchy sequence

(lim x

= 0), and yet

f (x

) = sin



n +



π = (−1)

is not Cauchy (it diverges by oscillation). Consequently,

we cannot extend f to a continuous function on any in-

terval containing x = 0.

−1

−

2. f (x) = x sin

is uniformly continuous. One way to see

this is to extend the function to the origin by deﬁning

g(x) =

(

x sin

if x = 0

0 if x = 0

By the squeeze theorem, lim x

= 0 =⇒ lim f (x

) = 0,

so g is well-deﬁned and continuous on R. By Theorem

4.33, f is uniformly continuous on any bounded interval.

Moreover, the derivative

−

′

(x) = sin

−

cos

is bounded whenever x is large; together with Exercise 6 we conclude that f (x) is uniformly

continuous on R \ {0}. We cannot use the derivative argument on the whole domain R \ {0},

since f

′

(x) is unbounded when x is small (lim f

′



2πn



= lim(−2πn) = −∞ ).

3. f (x) = x

sin

is also uniformly continuous: again extend

by g(0) = 0. This time however, we could argue that the

derivative is bounded



′

(x)



2x sin

−cos



≤ 3

since

sin y

≤

and

cos y

≤ 1 for all y.

Something stranger is going on. As you may verify (see

Exercise 3), the extended function g is everywhere differen-

tiable with g

′

(0) = 0, and yet the derivative g

′

(x) itself is

discontinuous at x = 0!

−1

f (x)

−

′

(x)

Exercises 4.19. Key concepts: Order of quantiﬁers! Bounded derivative ⇒ Unif cont ⇔ Cont extension

1. Decide whether each f is uniformly continuous. Explain your answers.

(a) f (x) = x

on [−2, 4] (b) f (x) = x

on (−2, 4)

−3

on ( 0, 4] (d) f (x) = x

−3

on ( 1, 4]

(e) f (x) = e

on (−∞ , 100) (f) f (x) = e

on R

2. Prove that each f is uniformly continuous by verifying the ϵ-δ property.

(a) f (x) = 3x + 11 on R (b) f (x) = x

on [0, 3]

on [

, ∞) (d) f (x) =

x+2

x+1

on [0, 1]

3. Verify the claim in Example 4.35.3 that the function g(x) is differentiable at zero

but that the

derivative g

′

(x) is discontinuous there.

4. (a) If f is uniformly continuous on a bounded set U, prove that f is bounded on U.

(Hint: for contradiction, assume ∃(x

) ⊆ U for which

f (x

)

→ ∞ . . .)

(b) Use (a) to give another proof that

is not uniformly continuous on ( 0, 1).

could be unbounded.

5. Suppose g is deﬁned on U and a ∈ U. Give very brief (one line!) arguments for the following.

(a) Prove that g is continuous at a provided

∀ϵ > 0, ∃δ > 0 such that 0 <

x −a

< δ =⇒

g(x) − g(a)

< ϵ

(b) Prove that g is continuous at a provided

∀(x

) ⊆ U \ {a}, lim x

= a =⇒ lim g(x

) = g(a)

well-deﬁned.

6. (a) Suppose f is uniformly continuous on intervals U

, U

for which U

∩ U

is non-empty.

Prove that f is uniformly continuous on U

∪U

(Hint: if x, y do not lie in the same interval U

, choose some a ∈ U

∩U

between x and y)

(b) Prove that f (x) =

√

x is uniformly continuous on [0, ∞).

1/n

(n ∈ N) is uniformly continuous

on its domain (R if n is odd and [0, ∞) if n is even).

(d) (Hard) Given f (x) = x

1/n

, show that δ = ϵ

demonstrates uniform continuity when n is

even and δ =





when n is odd.

(Hint: use the binomial theorem to prove that 0 ≤ y < x + δ =⇒ y

1/n

< x

1/n

+ δ

1/n

)

Use the deﬁnition g

′

(0) = lim

x→0

g(x)−g(0)

x−0

. Limits of functions are covered formally in the next section (course!), but you

should be familiar with the idea from elementary calculus.