Math 140A - Notes

Neil Donaldson

Fall 2023

Introduction

Analysis is one of the major sub-disciplines of mathematics, being concerned with continuous func-

tions, limits, calculus and accurate approximations.

Analytic ideas date back over 2000 years. For instance, Archimedes (c. 287–212 BC) used limit-type

approaches to approximate the circumference of a circle and compute the area under a parabola.

The philosophical objections to such calculations are just as old: how can it make sense to sum up

inﬁnitely many inﬁnitesimally small quantities? This was part of a deeper debate among the ancient

Greeks: is the matter comprising the natural world atomic (consisting of minute, discrete, indivisible

pieces) or continuous (arbitrarily and inﬁnitesimally divisible). Several of Zeno’s famous paradoxes

C. BC) grapple with these difﬁculties: Achilles and the Tortoise essentially argues that the inﬁnite

series

∞

∑

n=1

+ ··· is meaningless.

···

0 1

···

As we’ll see, with modern deﬁnitions it makes sense for this sum to evaluate to 1.

The work of Newton and Leibniz in the late 1600s allowed the easy application of calculus to many

important problems in the sciences, but without properly addressing the ancient philosophical chal-

lenges. The logical development of calculus necessitated by this became the triumph of 18

–19

century mathematics. The critical notions of limit and continuity only became rigorous in the early

1800s, courtesy of Bolzano, Cauchy and Weierstrass (amongst others), with another 50 years before

Riemann provided a thorough description of the deﬁnite integral.

The Math 140A/B sequence introduces analysis by focusing on these ideas. In this course we pri-

marily consider sequences, limits, continuity and inﬁnite series. Power series, differentiation and

integration are the focus of 140B. We start, however, with something even more basic: to numeri-

cally measure continuous quantities, we need to familiarize ourselves with the real numbers. Since a

concrete description is quite difﬁcult, we build up to it using ﬁrst the natural numbers and then the

rationals. . .

Archimedes’ circle calculation is reminiscent of the Riemann sum approach to integration, whereas his parabolic area

method required the evaluation of the inﬁnite series

∑

∞

n=0

1 The Set N of Natural Numbers

You’ve been using the natural numbers N = {1, 2, 3, 4, 5, . . .} since you learned to count. In mathe-

matics, these must be axiomatically described. Here is one approach, known as Peano’s Axioms.

Axioms 1.1. 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 function is usually denoted ‘+1’

so that we may write,

n ∈ N =⇒ n + 1 ∈ N

3. (Initial element) f is not surjective. Otherwise said, there exists an element 1 ∈ range f which is

not the successor of any element.

4. (Unique predecessor/order) f is injective. If m and n have the same successor, then 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 are relatively straightforward, the natural numbers are deﬁned by repeatedly adding 1

to the initial element; for instance

3 := f



f (1)



= (1 + 1) + 1

To see why Axiom 5 is so-named, compare an easy example with a standard induction argument.

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.

The two arguments are precisely the familiar base case and induction step.

It is purely convention to denote the ﬁrst natural number by 1; we could use 0, x, α , or any symbol you wish!

What about the integers? It should be clear that the integers satisfy axioms 1, 2 and 4, but not 3 and

5. For instance:

3. The function f : Z → Z : n 7→ n + 1 is surjective (indeed bijective/invertible). The number 1 is

the successor of 0.

We can reverse this observation to provide an explicit construction of the integers from the natural

numbers. Simply extend the 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. Most of these exercises are to refresh your memory of mathematical induction. You can

use either the language of Peano’s axiom 5, or the (likely) 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. 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).

9. 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?

2 The Set Q of Rational Numbers

There are several ways to deﬁne the rational numbers from the integers. 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 are more familiar once we write

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

λp

λq

for

any non-zero λ ∈ Z. It is easy to deﬁne the usual operations (+, ·, etc.) consistently with those for

the integers (Exercise 7).

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 26x + 54 = 0 also corresponds to the same rational number!

Extending this process, we might consider higher degree polynomials.

Deﬁnition 2.1. 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 2.2. 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 2.3 (Rational Roots). Suppose that a

, . . . , a

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

lowest terms, then p | a

and q | a

Proof. Since x satisﬁes the polynomial, we see that





+ ··· + a





+ a

= 0 =⇒ a

+ a

n−1

q + ··· + a

n−1

+ a

= 0

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.

You should be alarmed by this! We have given up on constructing new numbers and instead are simply describing their

properties. No matter, a construction of the real numbers will come later.

Examples 2.4. 1. We show 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 numbers ±1, ±2 satisfy x

−2 = 0, we have a contradiction.

2. (

√

3 −1)

1/3

is irrational. It satisﬁes (x

+ 1)

= 3, from which

+ 2x

−2 = 0

By the theorem, if x =

were rational then p | 2 and q | 1, whence x = ±1, ±2, none of which

satisﬁes (x

+ 1)

= 3.



√



1/2

is irrational. It satisﬁes 5x

−4 =

√

3, from which

25x

−40x

+ 13 = 0

If x =

were rational, then p | 13 and q | 25. There are twelve possibilities in all:

x = ±1, ±13, ±

, ±

It is tedious to check all cases, but none satisfy the required polynomial.

With this example it is easier to bypass the theorem entirely: if x ∈ Q then

√

3 = 5x

−4 would

also be rational!

4. We factorize the polynomial 3x

+ x

+ x −2 = 0. By the rational roots theorem, if x =

is a

rational root, then p | 2 and q | 3 which gives several possibilities:

x ∈



±1, ±2, ±

, ±



It doesn’t take long to try these and observe that x =

is the only rational solution. The

polynomial has a factor of 3x −2 which we can extract 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 there exist non-algebraic (or transcendental) numbers, of which e and π are

the most famous examples. These satisfy no polynomial equation with integer coefﬁcients, though

demonstrating this is tricky.

Compare this to the standard proof of the irrationality of

√

2 as seen in a previous course. Note how easy it is to extend

our approach to

√

29,

√

8, etc.

Exercises 2. 1. Describe all the linear equations which correspond to the rational number

101

2. Show that

√

5 and

√

24 are not rational numbers.

(Hint: 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

is not rational.

5. Show that (3 +

√

2/3

is not rational.

6. Explain why 4 −7b

must be rational if b is rational.

7. Given rational numbers (p, q) and (r, s) as ordered pairs, what are the rational numbers (p, q) +

(r, s) and (p, q) · (r, s)?

(Hint: what is

8. Let n ∈ N. Use the rational roots theorem to prove that

√

n is rational if and only if it is an

integer.

3 Ordered Fields

Thus far, we have formally constructed the natural numbers and used them to (loosely) 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 3.1. A ﬁeld F is a set together with two binary operations + and ·which satisfy the following

(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 way: 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 haven’t yet been deﬁned!

Example 3.2. It is worth considering the rational numbers in a little more detail. Recall (Section 2)

how Q may be deﬁned as a set of ordered pairs

↭ (p, q) ∈ Z ×N. It moreover inherits a natural

ordering from Z and N:

≤

⇐⇒ ps ≤ qr (remember that q, s > 0)

We write multiplication · as juxtaposition unless it is helpful for clarity. We also use the common shorthand a

= a · a.

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.

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,

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

=⇒ a =

≤

= c

Basic Results about ordered ﬁelds

As with the axioms of an ordered ﬁeld, it is not worth memorizing these.

Theorem 3.3. 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 of these statements should be intuitive for the ﬁelds Q and R. Try proving a few using only the

axioms; they are easiest 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 one ﬁnal useful ingredient.

Deﬁnition 3.4. If F is an ordered ﬁeld, then the absolute value of a ∈ F is

(

a if a ≥ 0

−a if a < 0

Theorem 3.5. In any ordered ﬁeld:

≥ 0

a + b

≤

(△-inequality)

All three parts are immediate if you consider the ±-cases separately for a, b.

Exercises 3. 1. Which of the axioms of an ordered ﬁeld fail for N? For Z?

2. Prove parts 11 and 13 of Theorem 3.3.

(Remember 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) Use induction to prove

+ a

+ ··· + a

≤

+ ··· +

for any a

, . . . , a

∈ R.

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. Following Example 3.2, prove that Q satisﬁes axiom O5.

(Hint: if a =

, etc., what is meant by ac ≤ bc?)

7. (Hard!) The complex numbers C = {x + iy : x, y ∈ R} form a ﬁeld. Consider the lexicographic

ordering of C 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?

(Don’t prove your claims if an axiom is satisﬁed, but provide a counter-example if not)

4 The Completeness Axiom

While we still haven’t provided an explicit deﬁnition of the real numbers, you should be comfortable

with the fact that both Q and R are ordered ﬁelds. The question remains of how to distinguish them?

Perhaps surprisingly, only one additional axiom is required: the completeness axiom or least upper

bound principle. To explain this we ﬁrst need some terminology.

Deﬁnition 4.1 (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 above and below. We say that S is bounded by M if

∀s ∈ S,

≤ M

Examples 4.2. 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 interval S = [0, 3) = {x ∈ R : 0 ≤ x < 3} is bounded, for example by M = 5, it has

minimum 0 and no maximum. While this last is likely intuitive, it worth giving an explicit

argument, in this case by contradiction.

Suppose x = max S exists. It is helpful to draw a picture to get the lay of the land. Since x ∈ S,

we’ve placed x inside the interval, away from 3.

0 3x

s =

x+3

The crux of the proof is to observe that there exists s ∈ S which is larger than x. The natural

choice is the average s :=

(x + 3). Now observe that

3 −s = s − x =

(3 − x) > 0

In particular,

• s ∈ S since it is non-negative and s < 3.

• s > x.

Since S contains an element larger than x, it follows that x cannot be the maximum of S. In

conclusion, S has no maximum.

The following should be immediate: try proving them yourself.

Lemma 4.3. 1. If M is an upper bound for S, so is M + ε for any ε ≥ 0.

2. If max S exists, then it is unique.

3. A set is bounded if and only if it is bounded above and below. In particular, if m, M are

lower/upper bounds, then S is bounded by

∀s ∈ S,

≤ max(

)

Example 4.4. Before introducing the key axiom, we consider a variation on the previous example.

We show that the following set has no maximum:

S = Q ∩ [0,

√

2) = {x ∈ Q : 0 ≤ x <

√

The approach is similar to before: given a hypothetical maximum x, ﬁnd an element s ∈ S between

x and

√

2. The challenge is that we can’t simply use the average

(x +

√

2) : this isn’t rational (why?)

and so doesn’t lie in S!

To ﬁx this, we informally invoke sequences: this might seem quite hard at the moment, but will be

made rigorous later. The rough idea is to construct a sequence (s

) of elements of S which increases

√

2. Eventually one of these must be larger than x.

Deﬁne a sequence of rational numbers (s

) by s

⌊10

√

2⌋, where ⌊ ⌋denotes the ﬂoor function.

The sequence simply recovers the ﬁrst n decimal places of

√

= 1, s

= 1.4 =

, s

= 1.41 =

141

100

, s

= 1.414 =

1414

1000

, . . .

and has the following properties:

• s

∈ S since any truncating decimal is rational and certainly 0 ≤ s

√

•

√

2 −s

< 10

−n

follows since 10

√

2 −⌊10

√

2⌋ < 1.

Now suppose x = max S exists. Since x ∈ S, we have x <

√

2. Choose N ∈ N large enough so that

−N

√

2 − x. Then s

∈ S and

√

2 −s

< 10

−N

√

2 − x =⇒ x < s

The hypothetical maximum x is not an upper bound for S: contradiction.

≈

0 1 x

√

N−1

−N

⌊y⌋ is the greatest integer less than or equal to y; informally round down. For example ⌊π⌋ = 3. This approach is a

something of a hack: it can be sped up enormously using the upcoming density of Q in R (Corollary 4.12); indeed the

Archimedean property on which it depends is necessary for ⌊y⌋ to be well-deﬁned.

Suprema and Inﬁma

We now generalize the idea of maximum and minimum values for bounded sets.

Example 4.5. The interval [2, 5) has least upper bound 5: among all upper bounds, 5 is the smallest.

Deﬁnition 4.6. 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. If S is bounded below, its inﬁmum inf S is its greatest lower bound. Equivalently,

(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 (4.5 cont). We verify the supremum and inﬁmum for S = [2, 5); parts (a), (b) are the

properties in the above deﬁnition.

(a) Since s ∈ S ⇐⇒ 2 ≤ s < 5, we see that 5 is an upper bound and 2 a lower bound.

(b) Given x < 5, deﬁne

s := max{

(x + 5), 4}. Observe that x < s < 5 from which s ∈ S is larger

than x. It follows that x is not an upper bound for S, and that 5 is the least such.

Similarly, if y > 2, deﬁne t := min{

( y + 2), 4} to see that t ∈ S is smaller than y, which cannot

therefore be a lower bound for S.

We conclude that sup S = 5 and inf S = 2.

2 3 4 5

sup S

inf S

t s

We are assuming something quite important here!

Axiom 4.7 (Completeness of R). If S ⊆ R is non-empty and bounded above, then sup S exists (and

is a real number!).

It is this property that distinguishes the real numbers from the rationals.

Note that every bounded

set S of rational numbers has a supremum; the issue is that sup S might not be rational!

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

If you’ve studied abstract algebra, then a more rigorous statement should make sense: every ordered ﬁeld with 0 = 1

and which satisﬁes the completeness axiom is isomorphic to the real numbers.

Example (4.4 cont). The set S = Q ∩ [0,

√

2) has sup S =

√

2. We check conditions (a), (b) in

Deﬁnition 4.6.

(a) Certainly

√

2 is an upper bound for S, since every element is less than

√

(b) If x <

√

2 is given, then our previous argument says there exists some s

∈ S for which s

> x.

Plainly x isn’t an upper bound.

In conclusion,

√

2 is the smallest upper bound for S.

Consider the contrapositive of part (b) of Deﬁnition 4.6 after replacing M with x.

If x < sup S, then x is not an upper bound for S.

If we unpack this further, we recover a useful existence result. Indeed this is precisely what we did

in both previous examples.

Lemma 4.8. 1. If x < sup S, then ∃s ∈ S such that s > x.

2. If y > inf S, then ∃t ∈ S such that t < y.

sup S

inf S

This observation will be used repeatedly, so make sure it is well understood.

Examples 4.9. We state the following without proof or calculation. You should be able to justify all

these statements using the deﬁnition, or by mirroring the above examples.

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 min S = inf S if a minimum exists.

3. S = Q ∩(π, 4) has sup S = 4 and inf S = π.

4. S = {

: n ∈ N} = {. . . ,

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

5. S =

∞

n=1

[n, n +

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

6. S =

∞

n=1

[

, 1 +

) has inf S = 1 = sup S since S = {1}.

The completeness axiom only asserts the existence of the supremum of a bounded set. By reﬂecting

across zero (see Exercise 9), we obtain the same thing for the inﬁmum.

Theorem 4.10 (Existence of Inﬁma). If S ⊆ R non-empty and bounded below, then inf S ∈ R exists.

The Archimedean Property and the Density of the Rationals

We ﬁnish this section by discussing the distribution of the rational numbers among the real numbers.

Theorem 4.11 (Archimedean Property). If b > 0 is a real number, then ∃n ∈ N such that n > b.

Equivalently:

a, b > 0 =⇒ ∃n ∈ N such that an > b.

In this result we assume nothing about R except that is an ordered ﬁeld satisfying the completeness

axiom and 0 = 1. The natural numbers in this context are deﬁned as the subset

N = {1, 1 + 1, 1 + 1 + 1, . . .} ⊆ R

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 4.8, ∃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.

The use of completeness is necessary: there exist non-Archimedean ordered ﬁelds!

Corollary 4.12 (Density of Q in R). Between any two real numbers, there exists a rational number.

The idea is simple: 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. The Archimedean property shows the existence of m, n.

Proof. WLOG suppose 0 ≤ a < b. The Archimedean property applied to

b−a

> 0 says

∃n ∈ N such that n >

b−a

A second application says ∃k ∈ N such that k > an. Now consider

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

It is immediate that any interval (a, b) now contains inﬁnitely many rational numbers.

Just replace b with

This part of the argument is needed because, in this context, we haven’t established the well-ordering property of N

(equivalent to Peano’s ﬁfth axiom).

Exercises 4. 1. Decide if each set is bounded above and/or below. If it is, 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 4.4, 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?

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 4.10 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 4.7), sup T exists. Prove that inf S = −sup T by verifying Deﬁni-

tion 4.6 parts 2(a) and (b).

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 5.1. Let S ⊆ R be any subset. If S is bounded above/below, then sup S/inf S are as in

Deﬁnition 4.6. 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 4.8 are precisely statements 1 & 2 in the above deﬁnition!

Examples 5.2. 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 5. 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?

6 A Development of R (non-examinable)

The comment in footnote 8 essentially constitutes a synthetic deﬁnition of the real numbers: there

is essentially just one set with the required properties. It is nice, however, to be able to provide an

explicit construction. The following approach uses Dedekind cuts.

First one deﬁnes N, Z and Q. Use Peano’s axioms and proceed as in sections 1 and 2. The operations

+, · and ≤ are deﬁned, ﬁrst on N and then for Z and Q building on the concepts for the integers.

Deﬁnition 6.1. A Dedekind cut α

∗

is a non-empty proper subset of Q with the following properties:

1. If r ∈ α

∗

and s ∈ Q with s < r, then s ∈ α

∗

2. α

∗

has no maximum.

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. While this is the idea, it doesn’t stand up as a deﬁnition due to circular logic: α cannot be

deﬁned in terms of itself!

Examples 6.2. 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 Dedekind cuts corresponding to irrational numbers,

though some are relatively straightforward. For instance the real number

√

2 would be the

Dedekind cut

√

∗

= {x ∈ Q : x < 0 or x

< 2}

It remains to prove that the set of Dedekind cuts satisﬁes all the axioms of a complete ordered ﬁeld.

The full details are too much for us, 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 can then prove

the multiplication axioms, the ﬁnal order axiom O5, and the distributive axiom.

• The completeness axiom must also be veriﬁed, though it comes almost for free! If A ⊆ R (so

that A is a set of Dedekind cuts), then the supremum of A is

sup A =

[

∗

∈A

∗

An alternative approach to R using sequences of rational numbers will be given later in the course.

Exercises 6. 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 deﬁnition of ‘product’ is unsatisfactory for deﬁning multiplication in R.

3. We verify the Archimedean property (Theorem 4.11) 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 6.1 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

∗

7 Limits of Sequences

Sequences are the fundamental tool in our approach to analysis.

Deﬁnition 7.1. A sequence of real numbers is a list indexed by the natural numbers

( s

) = (s

, s

, . . .)

We call s

the initial term/element.

This is strictly the deﬁnition of an inﬁnite sequence; ﬁnite sequences don’t appear in this course. Other

letters may be used (a

, b

, etc.), though s

is most common in the abstract. It is also common to have

sequences which start with a different initial term (n = 0 is particularly common). If you need to be

explicit, describe the range of indices with sub/superscripts, e.g. (s

)

∞

n=0

Examples 7.2. 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 t

= 1 and t

n+1

= 3t

− 1 together deﬁne

the sequence

( t

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

3. u

−4

deﬁnes a sequence with initial term u

( u

)

∞

n=3



, . . .



Limits In analysis we are typically interested in what happens to the terms of a sequence (s

) when

n gets large (as such, it is common to be non-explicit as to the initial term). In elementary calculus,

you should have become used to writing expressions 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, we have

( s

) = (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. In the next section we will do so

by developing the formal deﬁnition of limit. Before seeing this, we quickly refresh a few simple

examples from elementary calculus. At the moment, all these rely on your intuition and experience.

This is a good thing to practice: in analysis it is often essential to have a good idea of the correct

answer before you try to prove it!

If there are multiple letters in your expression, then for clarity it can be helpful to write lim

n→∞

with a subscript.

Examples 7.3. 1. lim

= 0. Our instinct is 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). Indeed

( s

)

∞

n=0

= (1, −1, 1, −1, 1, −1, . . .)

isn’t getting closer to any real number.

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

( s

) =



2, 4, 5,

, . . .



This appears to have limit lim s

= 6. Indeed it is not hard to spot the pattern s

= 6 −

which

is easily veriﬁed by induction: for the induction step, simply observe that

+ 3 =



6 −



+ 3 = 6 −

n+1

Exercises 7. 1. Decide whether each sequence converges; if it does, give 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 7.2.2 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?

8 The Formal Deﬁnition of Limit

Deﬁnition 8.1. A sequence (s

) converges to a limit s ∈ R, if

∀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.

This isn’t as hard as it looks! The best way to understand it is to work a lot of examples. . .

Example 8.2. We show that the sequence with n

term s

= 2 −

√

converges to s = 2.

If we plot the sequence like a function, we see how ϵ controls the distance from s

is to the limit s; the

deﬁnition requires us to show that no matter how small we make ϵ, there is some tail of the sequence

(all s

with n > N) whose terms are less than a distance ϵ from the limit.

0 20 40 60 80 100

s + ϵ

s −ϵ

s = 2

. . . guarantees that s

lives here

n being larger than this. . .



To verify a ‘for all, there exists’ statement requires an argument with a speciﬁc structure:

• Suppose ϵ > 0 has been provided and describe N, dependent on 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.

We want

−s





2 −

√



−2



√



< ϵ (equivalently n >

) whenever n > N.

Choosing N =

should be enough to complete the proof!

Warning! We do not yet have a proof: “N =

” is not the correct conclusion! We ﬁnish by rear-

ranging our scratch work to make it clear that the deﬁnition is satisﬁed.

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

. Then

n > N =⇒

−s



2 −

√

−2



√

= ϵ

Thus s

→ 2, as required.

N can be quantiﬁed as either a real or a natural number, the deﬁnitions being equivalent by the Archimedean property:

if N ∈ R satisﬁes the deﬁnition, then ∃

N ∈ N such that

N ≥ N; but then n >

N =⇒ n > N . . . It tends to be easier to use

R for convergence and N when directly proving divergence (see Deﬁnition 8.5).

The last three lines are all we need—think of them as the concert performance after much rehearsal!

With practice, you might be able to do simple ϵ–N arguments like these without scratch work, though

even experts usually require some.

Before seeing more examples, we prove a hopefully intuitive result.

Lemma 8.3 (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 deﬁni-

tion we obtain a tail of the sequence (all terms s

coming after some N) 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 8.1 twice: ∃N

, N

such that

n > N

=⇒

−s

s −t

and n > N

=⇒

−t

s −t

Deﬁne N := max{N

, N

}. Then,

n > N =⇒

s −t

s −s

+ s

−t

≤

−s

−t

(△-inequality)

s −t

Contradiction.

Examples 8.4. We give several more examples of using the limit deﬁnition. Remember that only the

formal arguments needs to be presented; some scratch work is included to show the thought process.

1. For any k ∈ R

, we prove that lim

= 0.

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

n > N =⇒

< ϵ

This amounts to having n >

1/k

, so it is enough to choose N to be the right hand side.

Formal argument. Let ϵ > 0 be given, and let N =

1/k

. Then

n > N =⇒



−0



= ϵ

We conclude that

→ 0, as required.

2. We prove that lim



√

n + 4 −

√



= 0.

Scratch work. Everything follows from a (hopefully) familiar algebraic trick for manipulating

surd expressions:

√

n + 4 −

√

n =

√

n + 4 +

√

Formal argument. Let ϵ > 0 be given, and let N =

. Then

n > N =⇒



√

n + 4 −

√



√

n + 4 +

√

= ϵ

Thus lim



√

n + 4 −

√



= 0, as required.

3. 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 for us to have

n −7 >

or equivalently n > 7 +

Formal argument 1. Let ϵ > 0 be 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 (cont). An alternative approach is available if we play with (∗) a little. By insisting

that n ≥ 14, we may simplify the denominator

n −7 ≥

n =⇒

n −7

≤

Formal argument 2. Let ϵ > 0 be 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. Observe how the second is always

larger than the ﬁrst: if N = N

( ϵ) works in a proof, then any larger choice N

( ϵ) will also,

n > N

≥ N

=⇒ n > N =⇒

−s

< ϵ

Use this to your advantage to produce simpler arguments.

100

0 1 2 3 4 5

= 7 +

= max



14,



4. Given s

−3n+1

, we prove that lim s

Scratch work. We want to conclude that

n > N =⇒



−3n + 1

+ n

+ 4

−



−2n

−9n −5

3( 3n

+ n

+ 4)



< ϵ

Attempting to solve for n (as in the ﬁrst method previously) is crazy! Instead we simplify the

fraction by observing that since n ≥ 1, we have



−2n

−9n −5

3( 3n

+ n

+ 4)



≤

16n

3( 3n

+ n

+ 4)

(1 ≤ n ≤ n

and the △-inequality)

16n

+ 4 > 0 =⇒ 3n

+ n

+ 4 > 3n

)

The ﬁnal simpliﬁcation is merely for additional tidying.

Formal argument. Let ϵ > 0 be given, and let N =

. Then

n > N =⇒



−



−2n

−9n −5

3( 3n

+ n

+ 4)



16n

(since n ≥ 1)

= ϵ

Other choices of N are feasible (see e.g. Exercise 5); everything depends on how you want to

simplify things in your scratch work.

Divergent sequences

By negating Deﬁnition 8.1, we obtain a new deﬁnition.

Deﬁnition 8.5. 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 8.6. 1. We prove that the sequence with s

does not converge to s = 2.

Visualization. We intuitively know that s

→ 0. If ϵ is anything smaller than 2, then the terms

will eventually be further than this from s = 2.

0 5 10 15

s = 2

s + ϵ

s −ϵ

Every tail of the sequence contains

elements s

that do not lie here

Direct argument. Let ϵ = 1. Since we are only concerned with large values of n, we see that

−s



−2



= 2 −

≥ ϵ = 1 ⇐⇒

≤ 1 ⇐⇒ n ≥ 7

Given N ∈ N, let

n = max{7, N + 1}. But then

−s



−2



≥ ϵ, from which we

conclude that s

↛ 2.

Contradiction argument. For an alternative approach, we suppose s

→ 2 and let ϵ = 1 in

Deﬁnition 8.1. Then ∃N such that

n > N =⇒



−2



< 1 =⇒ 1 <

< 3 =⇒

< n < 7

Regardless of the value of N, this cannot hold for all n > N: in particular n := max{7, N + 1}.

Contradiction.

The two arguments are very similar, though consider that a signiﬁcant advantage of the contra-

diction approach is that you only have to remember one deﬁnition!

This is why we prefer to let N be a natural number when proving divergence. If N ∈ R, then we’d have to use the

ceiling function (n = max{14, ⌈N⌉ + 1}), or resort to the Archimedean property on which it depends (∃n > max{14, N}).

Either way is ugly and potentially confusing, so better avoided.

2. We prove that the sequence deﬁned by s

= (−1)

is divergent.

Suppose, for contradiction, that s

→ s. The picture shows the case s ≥ 0 and strongly suggests

that ϵ = 1 will lead to a contradiction (why?).

−1

10 20

s + ϵ

s −ϵ

Regardless of N, every tail

of the sequence after here. . .

. . . contains some elements

that do not lie here











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

n > N =⇒

( −1)

−s

< 1

One each of the values {n

, n

} = {N, N + 1} is even and the other odd. There are two cases:

• If s ≥ 0 then

( −1)

−s

−1 −s

= s + 1 ≥ 1 = ϵ.

• If s < 0 then

( −1)

−s

1 −s

= 1 −s ≥ 1 = ϵ.

Either way we have a contradiction. We conclude that (s

) is divergent.

3. We show that the sequence deﬁned by s

= ln n is divergent.

Our intuition from calculus is that logarithms increase unboundedly. For any s ∈ R, letting

ϵ = 1 should be enough, for eventually ln n ≥ s + 1. This time we prove directly using the

deﬁnition of divergence (8.5).

Suppose s ∈ R, let ϵ = 1, and suppose that N ∈ N is given. Deﬁne n = max{N + 1, e

s+1

} and

observe that. Then

n > N and ln n ≥ ln(e

s+1

) = s + 1 (ln is increasing, and so respects inequalities!)

In particular,

−s

= ln n − s ≥ 1 = ϵ

We conclude that (s

) is divergent.

In the next section we’ll have a deﬁnition of what it means for a sequence to diverge to ∞: this is what’s happening for

= ln n, but it’s not (yet) what we’re trying to demonstrate.

A Little Abstraction

Working explicitly with the limit deﬁnition is tedious. In the next section we’ll develop and summa-

rize the limit laws so we can combine limits of sequences without providing new ϵ-N proofs. To start

working towards this, here are three general results.

Lemma 8.7. If lim s

= s, then lim s

= s

The challenge is that we want to bound



−s



−s

+ s

, which means we need some

control over

+ s

. One way uses the triangle-inequality,

+ s

−s + 2s

≤

−s

+ 2

Assuming

−s

≤ 1 gives us a ﬁxed bound

s + s

≤ 1 + 2

. We may now begin a proof.

Proof. Suppose s

→ s. Let ϵ > 0 be given, and let δ = min{1,

1+2

}. Since s

→ s, ∃N such that

n > N =⇒

−s

< δ

But then

n > N =⇒



−s



−s

+ s

≤

−s

(

−s

+ 2

)

(△-inequality)

< δ(1 + 2

) (since

−s

< δ ≤ 1)

≤ ϵ

Theorem 8.8. Suppose lim s

= s.

1. If s

≥ m for all except ﬁnitely many n, then s ≥ m.

2. If s

≤ M for all except ﬁnitely many n, then s ≤ M.

Proof. We prove part 1 by contradiction—part 2 is similar.

Suppose s

→ 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 (add s −m to both sides)

By assumption, s

< m holds for at most ﬁnitely many n. Contradiction.

The expression for all but ﬁnitely many n can be added to many abstract limit theorems; other com-

mon variants are for all large n, and for some tail of the sequence. To avoid cumbersome language, the

expression is often omitted. Remember that convergence/divergence is concerned with what hap-

pens when n is large: we can change or delete the ﬁrst million terms of (s

) without altering it’s

convergence status!

Theorem 8.9 (Squeeze Theorem). Suppose three sequences satisfy a

≤ s

≤ b

(for all large n) and

that (a

) and (b

) both converge to s. Then lim s

= s.

Proof. By subtracting s from our assumed inequality, we see that

−s ≤ s

−s ≤ b

−s =⇒

−s

≤ max{

−s

}

It remains to bound the right hand side by ϵ. Let ϵ > 0 be given, then there exist N

, N

such that

n > N

=⇒

−s

< ϵ and n > N

=⇒

−s

< ϵ

Finally let N = max{N

, N

} to see that

n > N =⇒

−s

≤ max{

−s

} < ϵ

Example 8.10. If s

1+sin n

, then 0 ≤ s

≤

. The squeeze theorem quickly forces lim s

= 0.

Exercises 8. 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. Let (t

) be a bounded sequence (there exists M such that

≤ M for all n), and let (s

) be a

sequence such that lim s

= 0. Prove that lim(s

) = 0.

(Hint: given ϵ, note that

is also a small number. . . )

3. Prove the following

(a) lim[

√

+ 1 − n] = 0 (b) lim[

√

+ n −n] =

√

+ n −2n] =

4. Let (s

) be a convergent sequence, and suppose lim s

> a. Prove that there exists N such that

n > N =⇒ s

> a.

5. (a) Show that n ≥ 2 =⇒ 2n

+ 9n + 5 ≤ 9n

(b) (Recall Example 8.4.4) Provide another argument that lim

−3n+1

by choosing N =

max{2,

√

6. (a) Prove that the sequence with n

term s

does not converge to −1.

(b) Prove that ( s

) does not converge to 1.

7. Prove that the sequence deﬁned by t

= n

diverges.

8. Provide a contradiction argument to justify Example 8.6.3: (ln n) diverges.

9. (Recall Theorem 8.8) Suppose lim s

= s where every s

> m. Can we conclude that s > m?

Explain your answer.

10. (a) If

−s

< 1, explain why



+ ss

+ s



< 1 + 3

+ 3

(b) Suppose s

→ s. Prove that s

→ s

9 Limit Theorems for Sequences

We’d like to develop some rules for working with limits so that we don’t have to resort to an ϵ–N

proof every time. The rough idea is that limits respect the basic rules of algebra. For instance. . .

Example 9.1. If lim s

= s, it seems natural that a new sequence (5s

) obtained by multiplying the

original terms by 5 should have limit lim 5s

= 5s. Consider what we have to prove to conﬁrm this:

∀ϵ > 0, ∃N such that n > N =⇒

−5s

< ϵ

This last amounts to observing that

−s

. The challenge here is to see that we’re essentially

done: this is just the statement lim s

= s in disguise! Here is a more formal argument.

Let ϵ > 0 be given. Since lim s

= s, we know that

∃N such that n > N =⇒

−s

=⇒

−5s

< ϵ

The trick in the example will be used repeatedly in the proofs that follow. What’s critical is that

you read the limit deﬁnition correctly: given any small number (ϵ,

, etc.) there is some tail of the

sequence which remains closer to the limit than this.

Theorem 9.2 (Limit laws). Limit calculations respect algebraic operations: ±, ×, ÷ and roots.

More speciﬁcally, if (s

) converges to s and (t

) to t, then,

1. lim(s

±t

) = s ±t

2. lim(s

) = st; as a special case, 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)

Our ﬁrst example was the special case of part 2 with k = 5. Note also how parts 2 and 4 extend

Lemma 8.7: by induction we now have s

→ s

for any q ∈ Q.

Proving the limit laws takes a little work, including a small lemma. To advertise their beneﬁt, we

repeated apply them to a limit calculation as you might have seen it in elementary calculus.

Examples 9.3. 1. We evaluate lim

√

n−1

−2

using the limit laws.

lim

+ 2

√

n −1

−2

= lim

3 +

3/2

−

5 −

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 lim

= 0 (Example 8.4.1))

This calculation involves some generally accepted sleight of hand; formally we’re working from

the bottom up since lim

√

n−1

−2

shouldn’t really be written until you know it exists!

2. Suppose (s

) is deﬁned inductively via s

= 2 and s

n+1

( s

) =



577

408

, . . .



This sequence in fact converges, though we’ll need to wait until the next section to see why.

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. The strategy for the ﬁrst is simple: control both

sequences so that both

−s

−t

, then add. The only challenge is writing it formally.

Proof of Theorem 9.2, part 1. Let ϵ > 0 be given. Since s

→ s and t

→ t, we see that ∃N

, N

such that

∃N

such that n > N

=⇒

−s

and,

∃N

such that 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.

The multiplicative limit law requires a preparatory result.

Lemma 9.4. (s

) convergent =⇒ (s

) bounded (∃M such that ∀n,

≤ M).

The converse to this statement is false: why?

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.

Proof. Suppose lim s

= s and 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

The approach to part 2 is similar to part 1, we just need to be a bit cleverer to break up

−st

Proof of Theorem 9.2, part 2. Exercise 8.2 deals with (and extends) the case when s = 0. Instead sup-

pose s = 0, and let ϵ > 0 be given. Since s

→ s and t

→ t,

( t

) is bounded (Lemma) : ∃M such that ∀n,

≤ M

∃N

, N

such that n > N

=⇒

−s

and n > N

=⇒

−t

Again let N = max{N

, N

}, then

−st

+ st

−st

△

≤

−s

−t

M +

= ϵ

The proofs of parts 3 and 4 are in Exercise 6.

More basic examples With a few simple general examples, the limit laws allow us to rapidly com-

pute the limits of a great variety of sequences.

Theorem 9.5. 1. If k > 0 then lim

= 0

2. If

< 1 then lim a

= 0

3. If a > 0 then lim a

1/n

= 1

4. lim n

1/n

= 1

Examples 9.6. 1. lim(3n)

2/n

= (lim 3

1/n

)

(lim n

1/n

)

= 1.

2. 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 −

√

Note that lim

sin n

= 0 follows from the squeeze theorem:



sin n



≤

→ 0.

Proof. 1. This is Example 8.4.1.

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

ϵ and observe that

n > N =⇒



= ϵ

3. 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 (8.9) shows that s

→ 0, whence lim a

1/n

= 1.

We leave the a < 1 case and part 4 to Exercise 7.

(1 + x)

∑

k=0

(

)

= 1 + nx +

n(n−1)

n(n−1)(n−2)

2·3

+ ··· + x

Divergence to ±∞ and the ‘divergence laws’

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

Deﬁnition 9.7. We say that (s

) diverges to ∞ if,

∀M > 0, ∃N such that n > N =⇒ s

> M

We write s

→ ∞ or lim s

= ∞. The deﬁnition for s

→ −∞ is similar.

If (s

) neither converges nor diverges to ±∞, we say that it diverges by oscillation.

Consider how M is trying to describe “closeness” to inﬁnity similarly to how ϵ measures closeness

to s in the original deﬁnition of limit (8.1).

Examples 9.8. As with convergence proofs, it is a good idea to try some scratch work ﬁrst!

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. Prove that s

= n

−n

−2n + 1 → ∞.

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. Prove that the sequence deﬁned by 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

In such cases lim s

is meaningless; you likely wrote lim s

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

The notion that s

→ −∞ can be phrased in multiple ways: some prefer

∀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 9.9. 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 s

→ −∞ should be clear.

Proof. We prove two of the results: 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 ﬁnd the limit of any rational sequence:

where (p

), (q

) are

polynomials in n.

Example 9.10. By applying Theorem 9.9 (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

= ∞

Indeed, you should be able to conﬁrm the familiar result from elementary calculus:

Corollary 9.11. 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 9. 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. Consider s

= (100n)

100

. Describe s

(1 followed by how many zeros?). Repeat for s

. Now

compute the limit lim s

3. 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) .

4. Prove the following:

(a) lim(n

−98n) = ∞ (b) lim



√

n −n +



= −∞

5. 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.

6. We prove parts 3 and 4 of the limit laws (Theorem 9.2). 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 t

→ t, ∃N

such that n > N

=⇒

−t

ϵ. Combine

and N

to provide a proof that lim

(d) Use the following inequality (valid when s

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



1/k

−s

1/k



−s

k−1

+ s

k−2

+ ··· + s

k−1

≤

−s

k−1

7. We ﬁnish the proof of Theorem 9.5.

(a) Suppose 0 < a < 1. Prove that lim a

1/n

= 1.

(Hint: consider b =

. . . )

(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.

8. Prove the remaining parts of Theorem 9.9.

9. 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

)

n→∞

depend on the value of a?

10 Monotone and Cauchy Sequences

The deﬁnition of limit (Deﬁnition 8.1) exhibits 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 10.1. A sequence (s

) is:

• 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 10.2. 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 10.3 (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 ∞ (see Exercise 5). The statement is lim s

= inf{s

} for monotone-down sequences.

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 4.8, there exists some s

> s −ϵ. Since (s

) is monotone-up, we have

n > N =⇒ s

≥ s

> s −ϵ =⇒ 0 ≤ s −s

< ϵ =⇒

s −s

< ϵ

The monotone-down case is similar.

This gets at the typical role of sequences in analysis: to demonstrate the existence of and deﬁne a new object (the limit)

and, more broadly, to transfer useful properties from the sequence to the limit. For instance, if ( f

) is a sequence of differ-

entiable functions, we’d like to know if lim f

(x) exists and is itself differentiable with derivative lim f

′

(x): discussions of

this ilk will dominate Math 140B.

Some authors describe such as sequence as either non-decreasing or increasing. We prefer monotone-up/down since this

directly describes the direction of any potential movement in the sequence and prevents confusion over whether the in-

equality is strict. A sequence with s

n+1

> s

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

Examples 10.4. 1. Deﬁne (s

) via s

= 1 and s

n+1

( s

+ 8):

( s

) =



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



The sequence certainly appears to be monotone-up and converging to 2. We prove this:

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 9.3.2, 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 9.3.2 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 0.d

. . . may be viewed as the limit of a monotone-up sequence of rational numbers:

0.d

. . . = lim

n→∞

∑

k=0

This is bounded above by 1 and so converges. Compare this with Example 4.4.

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 +



In case you’ve seen it before, 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 remains useful to be able to describe its long-term behavior.

Deﬁnition 10.5. 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

n/N

The original sequence (s

) is almost wedged between

( v

) and (u

) in a situation reminiscent of

the squeeze theorem (except 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 10.6. 1. (v

) is monotone-down, (u

) is monotone-up, and u

≤ s

N+1

≤ v

2. lim sup s

and lim inf s

exist for any sequence (they might be inﬁnite).

3. lim inf s

≤ lim sup s

Examples 10.7. 1. The picture shows 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

= 4

= sup{s

: n > 7} = s

= 7.625

0 10

n/N

3 7

= inf{s

: n > 3}

2. If s

, then v

= s

N+1

and u

= 0 for all N, whence lim sup s

= lim inf s

= 0.

A minor redeﬁnition would remove the ‘almost,’ but at the cost of making some subsequent arguments a little messier.

It is still reasonable to think of (u

) and (v

) as providing a long-term envelope for the original sequence.

3. Let s

= (−1)

. This time the calculation is easy:

= inf{s

: n > N} = −1 and v

= sup{s

: n > N} = 1

Therefore lim sup s

= 1 and lim inf s

= −1.

Lemma 10.8. For any sequence (s

lim inf s

= lim sup s

=⇒ lim s

exists

(the limit can be inﬁnite!).

In such a case all three values are equal.

lim s

In fact the converse to this is also true: we could prove it now, but it will come for free a little later. . .

Proof. (s ﬁnite) Since u

n−1

≤ s

≤ v

n−1

for all n, the squeeze theorem tells us that lim s

= s.

(s = ∞) Since u

n−1

≤ s

for all n and lim u

n−1

= ∞, it follows (Theorem 9.9.1) that lim s

= ∞.

(s = −∞) This time s

≤ v

n−1

→ −∞ =⇒ lim s

= −∞.

Cauchy Sequences

We now come to a class of sequences whose analogues will dominate your study of analysis.

Deﬁnition 10.9. 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 10.10. 1. Let s

. Let ϵ > 0 be given and let N =

. Then

m > n > N =⇒

−s

−

= ϵ

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

( s

) =



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 picture in the last example illustrates the essential point regarding Cauchy sequences: (s

) ap-

pears very much to converge. . .

Theorem 10.11 (Cauchy Completeness). A sequence of real numbers is convergent if and only if it

is Cauchy.

Proof. ( ⇒) Suppose lim s

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

m, n > N =⇒

−s

and

−s

=⇒

−s

−s + s −s

△

≤

−s

s −s

< ϵ

whence (s

) is Cauchy.

( ⇐) To discuss the convergence of (s

) we ﬁrst need a potential limit! In view of Lemma 10.8, 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 10.9:

∃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 ﬁnite.

Since m, n are arbitrary, WLOG we may assume m > n; equality is never interesting in these situations. This assump-

tion is very common and we’ll use it repeatedly without comment.

(lim sup s

= lim inf s

) Since (s

) is Cauchy, given ϵ > 0,

∃N such that m, n > N =⇒

−s

< ϵ =⇒ s

< s

+ ϵ

Taking v

= sup{s

: n > N}, we see that

m > N =⇒ v

≤ s

+ ϵ

Taking the inﬁmum of the right hand side yields

≤ u

+ ϵ (since u

= inf{s

: m > N})

Since (v

) is monotone-down and (u

) monotone-up, we see that

lim sup s

≤ v

≤ u

+ ϵ ≤ lim inf s

+ ϵ =⇒ lim sup s

≤ lim inf s

+ ϵ

This last holds for all ϵ > 0, whence lim sup s

≤ lim inf s

. By Lemma 10.6 we have

equality.

By Lemma 10.8, we conclude that (s

) converges to lim sup s

= lim inf s

In view of the Theorem, the previous examples 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 (10.10.2 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 therefore convergent; good luck explicitly ﬁnding its limit though!

The main point is easy to miss: Cauchy Completeness provides a powerful tool for determining

whether a sequence converges without ﬁrst guessing a limit. While the result depends on monotone

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

sequences. We ﬁnish with an application of this idea.

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

does not rely on Dedekind cuts (Section 6).

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

( s

) ∼ (t

) ⇐⇒ lim(s

−t

) = 0

We then deﬁne R :=



∼

. All this is done without reference to Cauchy Completeness, though it

certainly informs our intuition that (s

) and (t

) have the same limit. Some work is still required to

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 10. 1. Use the deﬁnition to show that the sequence with n

term 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

( s

+ 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 10.10.3. 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

+ ···+ 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.

) is monotone-up.

(Hint: consider

n+1

)

(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 limit is e).

(Hint: compute t

−s

)

11 Subsequences

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

might obtain a subsequence with interesting behavior.

Deﬁnition 11.1. Let (s

) be a sequence. A subsequence (s

) is a subset (s

) ⊆ (s

), where

< n

< ···

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

Example 11.2. Take s

= (−1)

(recall Example 8.6.2) 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 result illustrated in the example, that every bounded

sequence has a convergent subsequence (the famous Bolzano–Weierstraß theorem).

Lemma 11.3. If lim

n→∞

= s, then every subsequence (s

) also satisﬁes lim

k→∞

= s.

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

−s

< ϵ. Since

≥ k for all k, we see that

k > N =⇒ n

> N =⇒

−s

< ϵ

The case where s = ±∞ is an exercise.

Lemma 11.4. 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 11.5. Given a sequence (s

), there exist subsequences (s

) and (s

) such that

lim s

= lim sup s

and lim s

= lim inf s

Combining with the lemmas, we may assume these subsequences are monotonic.

Example 11.6. 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

, s

> k}

Choosing the minimum isn’t necessary here, but it at

least 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 10.8 says that lim s

= −∞, whence

( s

) itself is a suitable subsequence.

(lim sup s

= v ﬁnite) We follow an inductive construction: let n

= 1 and deﬁne s

for k ≥ 2 via,

• 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 4.8 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, n

k−1

Example (11.6 cont.). The example shows why the two-step construction is necessary. It may seem

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

) and (v

). Indeed,

( u

) =



−1, −1, −1, −1, −

, −

, . . .



contains a monotonic subsequence of (s

) converging to lim inf s

= 0. Unfortunately, the same isn’t

true for (v

) = (2, 1, 1, 1, 1 . . .), where v

= 1 for all k ≥ 2; taking n

= 2k −1 results in

= 1 −

2k −1

> 1 −

= v

−

The above discussion rapidly provides two results, the ﬁrst of which is Exercise 3.

Theorem 11.7 (Lemma 10.8 with converse). For any sequence,

lim sup s

= lim inf s

⇐⇒ lim s

exists

Theorem 11.8 (Bolzano–Weierstraß). Every bounded sequence has a convergent subsequence.

Proof 1. Lemma 11.4 says there exists a monotone subsequence. This is bounded and thus converges

by the monotone convergence theorem.

Proof 2. By Theorem 11.5, 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

generalizing to higher dimensions (rather than intervals, take boxes. . . ).

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!). 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

for which 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 11.9. (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

) come before s

. . .

Subsequential Limits, Divergence by Oscillation & Closed Sets

Recall Deﬁnition 9.7, where we stated that a sequence (s

) diverges by oscillation if it neither converges

nor diverges to ±∞. We can now give a positive statement of this idea.

( s

) diverges by oscillation

Thm 11.7

⇐⇒ lim inf s

= lim sup s

Thm 11.5

⇐⇒ ( 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 11.10. We call s ∈ R ∪ {±∞} a subsequential limit of a sequence (s

) if there exists a

subsequence (s

) such that lim

k→∞

= s.

Examples 11.11. 1. The sequence deﬁned by s

has only one subsequential limit, namely zero.

Recall Lemma 11.3: 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 full set of R ∪ {±∞}!

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

) inductively:

• By the density of Q in R (Corollary 4.12), 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 11.9, cont.) We won’t prove it, but the set of subsequential limits of

( s

) = (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 11.12. 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 11.5, 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 11.7, 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. Using sequences, we can make a formal deﬁnition.

Deﬁnition 11.13. 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.

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

Examples 11.14. 1. The interval [0, 1] is closed. If (s

) ⊆ [0, 1] has lim s

= s, then

0 ≤ s

≤ 1

Thm 8.8

=⇒ 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 interval (0, 1] is not closed since its closure is (0, 1] = [0, 1]. In particular, the sequence

lies in the original interval but has limit 0. Indeed this example shows that an inﬁnite

union of closed intervals need not be closed.

Theorem 11.15. If (s

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

We omit the proof since it is hard to read, involving unpleasantly many subscripts (subsequences of

subsequences. . . ).

As in the proof of Theorem 11.5, we could make this more explicit by choosing minimums, but there is no need: if

there are inﬁnitely many r

in S

, then only ﬁnitely many of them can come before r

k−1

Exercises 11. 1. Consider the sequences with the following n

terms:

= (−1)

= n

6n + 4

7n −3

(a) For each sequence, give an example of a monotone subsequence.

(b) For each sequence, state its set of subsequential limits.

(d) Which of the sequences converge? diverge to +∞? diverge to −∞?

(e) Which of the sequences are bounded?

2. Prove the case of Lemma 11.3 when lim s

= ∞

3. Suppose that lim s

= s (could be ±∞). Use Theorem 11.5 and Lemma 11.3 to prove that

lim sup s

= s = lim inf s

(This completes the proof of Theorem 11.7)

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 11.8) 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 11.13.

(b) Is there a sequence ( s

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

7. 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 11.11.5)

8. (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

···

12 Lim sup and Lim inf

In this 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 12.1. Let (s

), (t

) be bounded sequences.

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

Modiﬁcations can be made inﬁma and products of sequences (Exercise 3).

Example 12.2. To see that equality is unlikely, take s

= (−1)

= −t

, then

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}

Simply take limits as N → ∞ for the ﬁrst result.

2. By part 1, we already know that

lim sup(s

+ t

) ≤ s + lim sup t

For the other direction, let a

= s

+ t

and apply part 1 again:

lim sup t

= lim sup



( s

+ 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 12.3. 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 12.4. 1. Here is a quick proof that lim n

1/n

= 1 (recall Theorem 9.5): let s

= n, then

lim



n+1



= lim

n + 1

= 1 =⇒ lim n

1/n

= lim

1/n

= 1

2. Let s

= n! and apply the corollary to see that

lim(n!)

1/n

= lim



n+1



= lim(n + 1) = ∞

Proof. Assume lim sup



n+1



= L = ∞ (otherwise the third inequality is trivial) and let ϵ > 0. Then

lim

N→∞

sup





n+1



: n > N



< L + ϵ =⇒ ∃N such that sup





n+1



: n > N



< L + ϵ

For brevity, denote a = L + ϵ and b = a

−N−1

N+1

. For any n > N, we therefore have



n+1



< a =⇒

< a

n−N−1

N+1

=⇒

1/n

< a



−N−1

N+1



1/n

= ab

1/n

=⇒ lim sup

1/n

≤ a lim b

1/n

= a = L + ϵ

Since this holds for all ϵ > 0, we conclude the third inequality: lim sup

1/n

≤ L.

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

Exercises 12. 1. Compute lim

( n!)

1/n

(Hint: let s

in Theorem 12.3 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

( s

)

∞

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 12.3 and hence conclude that the converse of (†) is false.

14 Series

What should be meant by the following expression?

∞

∑

n=1

= 1 +

+ ···

The ∞-symbol in the summation should lead you to suspect a role for limits. . .

Deﬁnition 14.1. Deﬁne the n

partial sum s

of a sequence (a

)

∞

n=m

via

∑

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 ∞).

To return to our motivating example,

∞

∑

n=1

= lim s

where s

∑

k=1

= 1 +

+ ··· +

We don’t (yet) know whether the series converges or diverges to ∞. . .

Theorem 14.2 (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.

Note that series do not behave nicely with respect to multiplication (see also Exercise 3):

+ a

+ ··· =

∑

=



∑



∑





+ a

+ ···



+ b

+ ···



It is common to denote a series

∑

if the initial term is understood (typically a

or a

), or is irrelevant to the situation.

Series which may be evaluated exactly

Our major goal is to develop techniques for answering the binary question, “Does

∑

converge?”

Even when the answer is yes, a precise computation of the limit is usually beyond us. However,

our techniques (the upcoming series tests) will typically rely on comparing

∑

to a ‘standard’ series

whose properties are completely understood. You have met two such series in elementary calculus.

Deﬁnition 14.3 (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 14.4. If 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 14.5. 1.

∞

∑

n=−1



−



= 2



−



−1

1 +

= −

2. Consider the series

∑

∞

∑

n=3





+ 2

. If this were convergent, then

∑

−

∑





would converge (Theorem 14.2); a contradiction.

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

Example 14.6. 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 general series convergence uses Cauchy completeness.

Example 14.7. Consider again the series

∑

. We show that the sequence of partial sums (s

) is

Cauchy. Let ϵ > 0 be 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 we follow the telescoping series approach to cancel most terms. By Cauchy completeness

(Theorem 10.11), (s

) converges and we conclude

The series

∑

is convergent

Computing the value of this series rigorously is signiﬁcantly harder, though a sketch argument for

why

∑

is in Exercise 10.

Theorem 14.8 (Cauchy Criterion for Series). A series

∑

converges if and only if

∀ϵ > 0, ∃N such that m ≥ n > N =⇒

−s

n−1



∑

k=n



< ϵ

In the previous example we essentially veriﬁed the Cauchy criterion for the series

∑

Proof. Let (s

) be the sequence of partial sums. Then

∑

converges ⇐⇒ (s

) converges ⇐⇒ (s

) is Cauchy (Thm 10.11)

⇐⇒



∀ϵ > 0, ∃N such that m > n > N =⇒

−s

< ϵ



To ﬁnish, simply replace n with n −1.

Example 14.9. Assume, for contradiction, that the harmonic series

∑

converges. Now take ϵ =

the Cauchy criterion:

∃N such that m ≥ n > N =⇒



∑

k=n



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



∑

k=n



+ ··· +



≥

m −(n −1)

= 1 −

n −1

We conclude that the harmonic series diverges to ∞.

The Series Tests

For the remainder of this section we develop several standard tests for the convergence/divergence

of an inﬁnite series: the divergence, comparison, root and ratio tests. The ﬁrst of these follow quickly

from the Cauchy criterion.

Theorem 14.10 (Divergence/n

-term test). If lim a

= 0 then

∑

is divergent.

Proof. We prove the contrapositive. Suppose

∑

is convergent, let ϵ > 0 be given and take m = n in

the Cauchy criterion. Then

∃N such that n > N =⇒

< ϵ

Otherwise said, lim a

= 0.

Examples 14.11. 1. The series

∑

sin(

nπ

) diverges.

2. The divergence test tells us that the geometric series

∑

diverges whenever

≥ 1. We still

need our earlier analysis for when

< 1.

3. The converse of the n

-term test is false! For the canonical example, consider the divergent har-

monic series

∑

(Example 14.9, even though lim

= 0.

Theorem 14.12 (Comparison test). 1. Let

∑

be a convergent series of non-negative terms and

assume

≤ 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.

1. Let ϵ > 0 be given. Then ∃N ≥ M such that

m ≥ n > N =⇒



∑

k=n



△

≤

∑

k=n

≤

∑

k=n

< ϵ

2. The n

partial sum of

∑

k=M

≥

∑

k=M

→ +∞

Corollary 14.13. 1. Take

= b

in part 1 to see that

∑

converges =⇒

∑

converges. Thus

absolute convergence implies convergence.

2. If

∑

is a convergent series of non-negative terms and

≤ b

for all n, then

∑

≤

∑

≤

∑

Examples 14.14. 1. Since

2n+1

(n+2)3

≤ 2 · 3

−n

and the geometric series

∑

2 ·3

−n

converges, we see that

the resulting series converges (absolutely), to some value

∞

∑

n=0

2n + 1

( n + 2)3

≤ 2

∞

∑

n=0

−n

1 −

= 3

2. One can usually ﬁnd a sensible series to compare with just by thinking about how a

behaves

when n is very large. For instance, a

+1)

1/2

(1+

√

behaves like

and we see that

(1 +

√

16n

Comparison with

∑

shows that

∑

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

∑

. Corollary 14.13 gives an estimation for the

value of the series, though it is not accurate! The n

partial sums satisfy

∞

∑

n=1

sin n

≤

∞

∑

n=1

sin n

≤

∞

∑

n=1

(approximately 1.014 ≤ 1.280 ≤ 1.645)

5. The alternating harmonic series

∞

∑

n=1

(−1)

n+1

converges via a sneaky comparison.

Consider the series t =

∞

∑

n=1

2n(2n−1)

which converges by comparison with

∑

4(n−1)

. Its n

par-

tial sum is

∑

k=1

2k(2k −1)

∑

k=1



2k −1

−



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

∑

m=1

(−1)

m+1

Take limits of s

2n+1

= s

2n+1

, we see that lim s

2n+1

= t from which lim s

= t. Since the

harmonic series

∑

diverges (Example 14.9), we conclude that the alternating harmonic series

converges conditionally. We will revisit this discussion in the next section.

∑



n+1



converges by comparison with the geometric series

∑

−n

. To see this, note that



n + 1



n + 1



1 −

n + 1



n+1

−−−→

n→∞

−1

from which we see that, for all large n,



n + 1



=⇒



n + 1



< 2

−n

In fact (compare Exercise 10.10),



n+1



is monotone-down, whence e

−1

≤



n+1



≤

and

0.58198 ≈

e −1

−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 ﬁnal two tests in this section are less powerful, but have the advantage of being easier to use.

Theorem 14.15 (Root test). Let 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 seeing some examples.

Corollary 14.16 (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

Proof. This follows directly from the root test and Theorem 12.3:

lim inf



n+1



≤ lim inf

1/n

≤ lim sup

1/n

≤ lim sup



n+1



The versions of these tests familiar from elementary calculus are the special cases when

L = lim

1/n

= lim



n+1



Our versions are more general since these limits are not guaranteed to exist.

Examples 14.17. 1. The ratio test is particularly useful for series involving factorials and exponentials.

(a)

∑

converges: just observe that lim



n+1



= lim

(n+1)

< 1.

(b)

∑

diverges: in this case 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,

∑

n + 5

⇝ lim



n+1



= lim

( n + 6)n

( n + 5)(n + 1)

= 1

In fact this example is divergent by comparison with the harmonic series

∑

3. Recall Example 14.14.6: 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, consider:

(

−n

if n is even

−n

if n is odd

First we try applying the ratio test:

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

1/n

(

if n is even

if n is odd

=⇒ lim sup

1/n

< 1

By the root test, the series

∑

converges. We need not even have used the root test:

∑

converges by comparison with

∑

−n

Proof of the Root Test. 1. Suppose lim sup

1/n

= L < 1. Recall that v

= sup{

1/n

: n > N}

deﬁnes a monotone-down sequence converging to L. Choose any ϵ > 0 such that L + ϵ < 1 (say

ϵ =

1−L

) to 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. Clearly a

does not converge

to zero whence the series diverges by the n

term test.

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

(divergence tests) n

term

Root

Ratio

(testing both) Deﬁnition of

∑

ks +3

Cauchy Criterion

Comparison



(convergence tests) 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, etc. If a series diverges by the

ratio test, then it in fact diverges by the n

term test.

Exercises 14. 1. Determine which of the following sequences converge. Justify your answers.

(a)

∑

n−1

(b)

∑

( −1)

(c)

∑

(d)

∑

(e)

∑

(f)

∑

(g)

∑

(h)

∑

(i)

∞

∑

n=2



n+(−1)



(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. (Hard) Let (a

) be a sequence such that 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 14.13)

8. Use the basic series laws to 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 we’ve presented it, this argument is non-rigorous!)

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

Theorem 15.1 (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 which case

∞

f (x) dx ≤

∞

∑

n=1

≤ a

∞

f (x) dx

The statement is easily modiﬁed if the initial term is a

f (x)

0 1 2 3 4 5 6

Proof. We need only interpret the picture:

n+1

f (x) dx ≤

∑

k=1

= s

= a

∑

k=2

≤ a

f (x) dx (∗)

Taking limits gives the result.

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

and analyze its rate of growth. For greater accu-

racy, explicitly evaluate the ﬁrst few terms and use a modiﬁed integral test to estimate the remainder.

The big application of the integral test is a complete description of the convergence status of p-series,

another useful collection of series to which others may be compared.

Corollary 15.2 (p-series). Let p > 0. The series

∑

converges if and only if p > 1.

Examples 15.3. 1.

∑

converges (it is a p-series with p > 1). For a simple estimate, observe that

∞

dx = lim

b→∞



−

−2



=⇒

≤

∞

∑

n=1

≤

This is a poor estimate, especially the lower bound. For a quick improvement, we could explic-

itly 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. Even though we haven’t developed these

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

2. In Example 14.9, we used the Cauchy criterion to show that the harmonic series diverges to ∞.

The integral test makes this much easier! The integral test also allows us to estimate how many

terms are required for the partial sum to s

to reach a certain threshold, say 10. Since

ln(n + 1) =

n+1

dx ≤

∑

k=1

≤ 1 +

dx = 1 + ln n

we see that

≈ 10 =⇒ ln(n + 1) ≤ 10 ≤ 1 + ln n =⇒ e

≤ n ≤ e

−1 =⇒ 8104 ≤ n ≤ 22025

Somewhere between 8 and 22 thousand terms are required! The harmonic series diverges to

inﬁnity, but it does so very slowly.

3. The integral test shows that

∑

∞

n=2

n ln n

= ∞ and moreover that, to exceed 10, somewhere be-

tween 10

3223

and 10

6631

terms are required!

4. The series

∑

2n+1

√

−1

diverges to ∞ by comparison with the p-series

∑

√

Alternating Series and Conditional Convergence

Our ﬁnal test is the only one capable of detecting conditional

convergence, the canonical example of which is the alternat-

ing harmonic series (recall Example 14.14.5). 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. For us, however, the behavior of the sequence (s

)

of partial sums is more interesting. Consider 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)

Since the brackets are non-negative, (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 depended only on two simple properties of the sequence (a

); we’ve therefore

proved a general statement.

Theorem 15.4 (Alternating series test). Let (a

) be monotone-down with lim a

= 0.

1. The series

∑

( −1)

converges.

2. If (s

) is the sequence of partial sums converging to s =

∑

( −1)

, then

s −s

≤ a

n+1

Think about where our assumptions on (a

) are used in the proof. It can, in fact, be shown that

the alternating harmonic series converges to ln 2, although the estimates provided by the alternating

series test make this a terrible method of approximation. Even summing the ﬁrst 100 terms only

results in 2 decimal places of accuracy!

Examples 15.5. 1. Since a

converges monotone-down to zero, the alternating series

∑

(−1)

n+1

converges. By taking the ﬁrst 9 and 10 terms of this series, we see that

0.9010498898 . . . ≤

∞

∑

n=1

( −1)

n+1

≤ 0.90105016538 . . .

which at least yields the estimate 0.9015 to 5 decimal places. The alternating series test is not

required for this example, since it in fact 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 shows convergence.

Rearranging Inﬁnite Series

A rearrangement of an inﬁnite series

∑

arises when we change the order of the terms of the sequence

) before computing the sequence of partial sums. We still have to use every term a

in the new

series. Since the resulting sequence of partials sums is likely completely different, we shouldn’t

assume that the new series has the same convergence properties as the old.

Example 15.6. We rearrange the alternating harmonic series by summing two positive terms before

each negative term:

1 +

−

+ ··· +

4n −3

4n −1

−

+ ···

Every term in 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 9.

This behavior is quite different to that of ﬁnite sums, where the order of summation makes no differ-

ence at all. The situation can be summarized in a famous result of Riemann.

Theorem 15.7 (Riemann rearrangement). 1. If

∑

is conditionally convergent and s ∈ R ∪ {±∞},

then there exists a rearrangement of the series which tends

to s.

2. If

∑

converges absolutely, then all rearrangements converge to the same limit.

The second part says that absolutely convergent series behave just like ﬁnite sums! We omit the

proofs since they are lengthy and require nasty notation. Instead we illustrate the rough idea of part

1 via an example.

Example 15.8. We show how to construct a rearrangement of the alternating harmonic series which

converges to s =

√

2 −1 = 0.41421 . . .

First we convince ourselves that the sum of the positive terms

∑

diverges to inﬁnity. In this case

the comparison test comes to our rescue:

2n −1

=⇒

∑

2n −1

∑

= ∞

The negative terms also diverge

∑

−

= −∞. Construction of the rearrangement is inductive.

1. Sum just enough positive terms until the partial sum exceeds s: plainly S

= 1 will do.

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

= 1 −

−

= 0.25 < s

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

= S

= 0.583 . . . > s, S

= S

−

= 0.291 . . . < s

Continuing the process ad inﬁnitum, we claim that

s = 1 −

−

−···

To see why, observe:

• Since

∑

= ∞ and

∑

−

= −∞, at each stage we 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 last term used at the n

stage. This plainly converges to zero,

whence lim S

= s.

Riemann’s result is in fact even stronger. Rearrangements also exist which diverge by oscillation between any given

lim inf and lim sup!

Exercises 15. 1. Use the integral test to determine whether the series

∞

∑

n=1

converges or diverges.

2. Prove Corollary 15.2 regarding the convergence/divergence of p-series.

3. Let s

∑

k=1

√

. Estimate how many terms are required before s

≥ 100.

4. (Example 15.3.3) Verify the claim that

∞

∑

n=2

n ln n

= ∞. If you want a challenge, verify the estimate

claim also.

5. (a) Give an example of a series

∑

which converges, but for which

∑

diverges.

(Exercise 14.3 really requires that

∑

be absolutely convergent!)

(b) Give an example of a divergent series

∑

for which

∑

converges.

6. Suppose (a

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

= a > 0.

What can you say about the sequences (s

) and (s

−

) and the series

∑

( −1)

7. We know that the harmonic series has a growth rate comparable to ln n. Let a

and deﬁne

a new sequence (t

) by

= s

−ln n = 1 +

+ ··· +

−ln n

where s

∑

k=1

is the n

partial sum. Prove that (t

) is a positive, monotone-down sequence,

which therefore converges.

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

8. (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.

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

9. In Example 15.6 we rearranged the terms of the alternating harmonic series by taking two pos-

itive terms before each negative term.

(a) Verify, for each n ∈ N, that

4n −3

4n −1

−

> 0

whence the subsequence of partial sums (s

) is monotone-up.

(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 many mathematical identities, and yet

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

17 Continuous Functions

For the rest of the course, we discuss continuous functions. Functions themselves should be familiar.

For reference, 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: for the above example, dom f = R \ {2, 3, −3}. In

examples, the domain is typically 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. It is a subset of V.

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: otherwise said, 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 17.1. 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)

To introduce continuity, we 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 could we calculate or prove anything with this concept?

Moreover, it cannot reasonably be extended to other situations (e.g., multivariable functions)

where drawing a graph is meaningless.

If x is close to a, then f (x) is close to f (a) This is better and can be generalized to other situations.

The major issue is the unclear meaning of close. Our formal deﬁnition of continuity addresses

this using sequences and limits.

Deﬁnition 17.2 (Sequential continuity). Let f : U → V ⊆ R be a function. We say that f is

continuous at u ∈ U if,

∀(x

) ⊆ U, we have lim x

= a =⇒ lim f (x

) = f (a)

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

A discontinuity of f is a value a ∈ U at which f is discontinuous (not continuous),

∃(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)

Discontinuity at a: at least one sequence

with lim x

= a has lim f (x

) = f (a)

Examples 17.3. 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 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 + 2x

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, whence

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 9.2), we can combine continuous functions 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 17.4. 1. Suppose f , 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 the function f : x 7→ x

1/n

is continuous on its domain.

3. Compositions of continuous functions are continuous. Speciﬁcally, if g is continuous at a and f

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

4. Algebraic functions are continuous (this 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 from parts 1, 2 and 3.

Example 17.5. Part 3 of the theorem says that the following algebraic function is continuous

f : (7, ∞) → R : x 7→

5/2

+ 7x

+ 4

(x −7)

1/3

Theorem 17.6 (Squeeze theorem). Suppose f (x) ≤ g(x) ≤ h(x) for all x = a, that f , h are continu-

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

Proof. This is simply the squeeze theorem (8.9) 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 17.7. The common trigonometric, exponential and logarithmic functions are continuous.

It is possible, though slow and ugly to address some of this now, though it is not very proﬁtable. It is

better to deﬁne these functions later using power series

when their continuity (and differentiabil-

ity/integrability!) come for free.

Examples 17.8. 1. f (x) =

√

sin e

is continuous on its domain R \{ln(nπ) : n ∈ N

2. The function deﬁned by g(x) = x sin

if x = 0 and g(0) = 0 is

continuous on R. When x = 0, this follows from Theorems 17.4

and 17.7, 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 ever 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), let ϵ > 0 be given, and 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 the Maclaurin series e

∞

∑

n=0

, sin x =

∞

∑

n=0

(−1)

(2n+1)!

2n+1

and cos x =

∞

∑

n=0

(−1)

(2n)!

Deﬁnition 17.9 (ϵ–δ continuity). A function f : U → V ⊆ R is continuous at a if

∀ϵ > 0, ∃δ > 0 such that (∀x ∈ U)

x −a

< δ =⇒

f (x) − f (a)

< ϵ (∗)

A discontinuity of f is a value a ∈ U for which,

∃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

Discontinuity at a

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

are merely measures of closeness. Most mathematicians consider the ϵ–δ version to be the deﬁnition

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

Theorem 17.10. The sequential and ϵ–δ deﬁnitions of continuity (17.2 & 17.9) are equivalent.

Examples (17.3, cont). Before seeing the proof, we repeat our earlier examples using the ϵ–δ deﬁni-

tion. 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 66 with all

mention of sequences removed!

The bracketed statement ∀x ∈ U is often omitted in (∗), since the implication requires x to 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 control

by staying away from zero: to

do this, we start by insisting that δ ≤

. But now,

x −a

< δ =⇒

=⇒

(∗)

Now consider the required difference; if

x −a

< δ, then

g(x) − g(a)



1 +

−1 −



− x



a + x

x −a

△

≤ 4





x −a

(∗)

< 4

x −a

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

ϵ). Then

x −a

< δ =⇒

f (x) − f (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, we have

h(x) − h(a)

= 3



1/4

− a

1/4



x −a

+ a

≤

x −a

3/4

3δ

3/4

= ϵ

4. 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, if we choose x = max(

√

, 1 −

), then

x −1

≤

< δ

and k(x) ≥ k(

√

) = 1 +

= 2. Contradiction.

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 promise proof of equivalence.

Proof of Theorem 17.10. (sequential ⇒ ϵ–δ) We prove the contrapositive. Suppose a is an ϵ–δ discon-

tinuity (†) and let δ =

. Then there exists x

∈ U such that

− a

and

f (x

) − f (a)

≥ ϵ

Repeating for all n ∈ N results in 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 17.11. 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; these rely on the fact that any interval contains both rational and irrational

numbers (Corollary 4.12, 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 17. 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. Prove that x = 0 is a discontinuity of each function: use both deﬁnitions of continuity.

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

(b) g(x) = sin(1/x) for x = 0 and g(0) = 0.

5. 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.

6. 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.

7. In Example 17.11.2, provide the details of the required contradiction.

8. (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.

9. (Hard) Consider f : R → R where

f (x) =

(

whenever x =

∈ Q with q > 0 and gcd(p, q) = 1

0 if x ∈ Q

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

) = f (−

) = f (

) = ··· =

, etc. Prove that f

is continuous at each irrational number, and discontinuous at each rational number.

18 Properties of Continuous Functions

In this section consider how continuous functions transform intervals.

Example 18.1. f (x) = x

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

• f transforms one interval into another.

• f transforms one closed bounded set into another.

The purpose of this section is to see that these are general properties

exhibited by any continuous function.

f (x)

−3 −2 −1 0 1 2

Before stating our ﬁrst result, recall a couple of deﬁnitions.

Deﬁnition 18.2. Let U, V ⊆ R and f : U → V.

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 11.13) U is closed if every convergent sequence in U has its limit in U:

∀(x

) ⊆ U, lim x

= s (∈ R) =⇒ s ∈ U

Theorem 18.3 (Extreme Value Theorem). Suppose f : U → V is continuous where U is closed and

bounded. Then f is bounded and attains its bounds:

∃s, i ∈ U such that f ( s) = sup



f (U)



and f (i) = inf



f (U)



In fact f (U) is also closed and bounded.

In Example 18.1, if U = [−3, 2], then s = −3 and i = 0.

Examples 18.4. Before seeing the proof, here are three examples where we weaken one of the hy-

potheses and see that the result fails.

0 1 2

1. f discontinuous 2. U not closed 3. U not bounded

1. sup(range f ) = 3 is not attained by f (x) =

(

3x if 0 ≤ x < 1

1 if 1 ≤ x ≤ 2

2. If f (x) =

√

2−x

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

√

, ∞ ) is unbounded.

3. If f (x) = x

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

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

broken 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 it is possible

that M = sup



f (U)



or inf



f (U)



2. Since (x

) ⊆ U is bounded, Bolzano–Weierstraß (Theorem 11.8) says it has a convergent subse-

quence, lim

k→∞

= x.

3. Since U is closed, we have x ∈ U. This means f (x) makes sense (it is a real number!).

4. Since f is continuous, we have 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 11.3). It follows that all limit points M are ﬁnite and lie in f (U):

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

In particular, choosing M = sup



f (U)



yields x = s ∈ U as in the Theorem.

Example 18.5. It is worth thinking about why we needed to use a subsequence in the proof. The

reason is that it is possible for the bounds of f to be attained multiple times. For example, consider

f : [0, 4π] → R : x 7→ sin x

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

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

and

5π

. The sequence deﬁned by

(

if n odd

5π

if n even

has f (x

) = sin





−−−→

n→∞

1 = sup(range f )

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 of the proof.

−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 4.8).

• 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 is also intuitive: if

you climb a hill, then at some point you must be half-way up the hill. . .

Theorem 18.6 (Intermediate Value Theorem (IVT)). Let f be continuous on the interval [a, b] and

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

Proof. WLOG assume f (a) < y < f (b). Now 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 a < ξ < b and f (ξ) = y.

First choose any (s

) ⊆ S such that

lim s

= ξ.

Continuity forces lim f (s

) = f (ξ). Moreover

f (s

) ≤ y =⇒ f (ξ) ≤ 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, whence

f (x

) ≥ y =⇒ f (ξ) = lim f (x

) ≥ y

This also shows that ξ > a. Putting it all together, we conclude that 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 18.7. The intermediate value theorem was typically used in elementary calculus to show

the existence of solutions to equations. Here are a couple of examples of this process.

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 that

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) = 01. Otherwise said, ξ is a solution to the original equation.

The function f is in fact continuous on R, a much larger interval that [a, b], but no matter!

Similarly to step 1 of the proof of the Extreme Value Theorem.

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 exists and that ξ ∈ (0, 4).

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

which may be shown by also evaluating, say,

f (−3) = −798 < 0 and f (5) = 150 > 0

−100

100

200

−2 2 4

With an eye on generalizing, consider a slightly different approach. Deﬁne two sequences (s

)

and (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 18.5.2 may be applied to prove the general result.

Corollary 18.8. 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 18.9 (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 ﬁrst of several ﬁxed-point theorems you’ll meet if your

studies of analysis continue. Many important consequences ﬂow from

these, including a common fractal construction and the standard exis-

tence/uniqueness result for differential equations.

a b

For our ﬁnal corollary, we ﬁrst note a straightforward characterization: a set I ⊆ R is an interval if

a, b ∈ I and a < y < b =⇒ y ∈ I (∗)

Corollary 18.10 (Preservation of Intervals). Suppose f : U → V is continuous where U = dom f is

an interval (of positive length) and V = range f .

1. V is an interval or a point.

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

(a) V is an interval (it has positive length).

(b) f is injective (and thus bijective).

−1

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

Example 18.11. In part 1, note that 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 a closed

bounded interval, 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). Let y ∈



f (a), f (b)



; IVT says

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

2. (a,b) If f is strictly increasing, then ∀a, b ∈ U, a < b =⇒ f (a) < f (b). It follows that f is

injective and that V contains at least 2 points; by part 1 it has positive length.

< y

where both lie in V, and deﬁne x

= f

−1

( y

) 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.

It remains to show that f

−1

is continuous at b = f

−1

(a). Assume ﬁrst that a is not an

endpoint of U. Given ϵ > 0 for which [a − ϵ, a + ϵ] ⊆ U, let

δ := min



b − f (a −ϵ), f (a + ϵ) − b



and observe that

y −b

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

=⇒ a −ϵ < y < a + ϵ ( f strictly increasing)

=⇒



−1

( y) − 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 δ.

Example 18.12. The function f : [0, 2] → [0, 4] deﬁned by

f (x) =

(

√

x if 0 ≤ x ≤ 1

if 1 < x ≤ 2

is continuous and strictly increasing. It therefore has a continuous in-

verse function f

−1

: [0, 4] → [0, 2].

Compare this with the result from elementary calculus: f

′

> 0 =⇒ f

injective. We cannot apply this here since f is not differentiable!

f (x)

0 1 2

Exercises 18. 1. Give an example of a discontinuous function f : [0, 1] → R which is not bounded.

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 18.12.

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

) in S that converges to a number x

∈ S.

Show that there exists an unbounded continuous function on S.

5. Prove that x = cos x for some x in (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 x between a, b such that f (x) = 0.

7. Suppose that 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 18.9).

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

(b) Prove Corollary 18.8 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 is related to the idea that ﬁnite unions of closed sets remain closed, but inﬁnite unions need not)

19 Uniform Continuity

Suppose f : U → V is continuous. By the ϵ -δ deﬁnition (17.9),

∀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 the fact 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 19.1. We start by considering how this desire might be impossible to satisfy.

Consider f (x) = x

with domain U = [0, ∞). Given ϵ > 0

and a

∈ U, we can certainly ﬁnd some

δ such that

x −a

< δ =⇒

f (x) − f (a

)



− a



< ϵ

Visualize what happens if we try to use the same δ for dif-

ferent a

: imagine sliding the ﬁxed-width δ-interval along

the x-axis while simultaneously sliding the ϵ-interval verti-

cally. As a

increases, the image of the δ-interval eventually

becomes too large for the ϵ-interval to contain:

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 we consider the same formula f (x) = x

but on a restricted ﬁnite domain [0, b], then

any δ that sufﬁces to demonstrate continuity at x = b will also do so everywhere else on [0, b]. We’ll

check this explicitly in a moment.

To formalize things, consider rewriting (∗), where we additionally assume that δ may be chosen

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

Deﬁnition 19.2. 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 (†)

For reasons of symmetry we use y instead of a. Note how δ now depends only on ϵ since it is

quantiﬁed before x and y; as previously, 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 point a.

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

Lemma 19.3. If f is uniformly continuous on U, then it is continuous on U.

This should be 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.

For instance δ = min(1,

1+2a

), as we saw on page 67.

Examples 19.4. 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 follows 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 19.1. Our approach works for this function because the gradient

(and therefore potential 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. Supposing x −y =

, we see that

x + y

= 2y +

=⇒

f (x) − f (y)



2y +



= δ



y +



> δx

Letting y =

(x =

) yields the contradiction

f (x) − f (y)

> 1 = ϵ.

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 19.5. 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

where the existence of ξ between x, y follows from the Mean Value Theorem.

Examples 19.6. 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 (recall Deﬁnition 11.13).

Theorem 19.7. 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 are crucial: Examples 19.4 provide counter-examples if either is weakened.

Example 19.8. f (x) =

√

x is uniformly continuous on [0, 1]. This cannot be concluded from Theorem

19.5, since its derivative f

′

(x) =

−1/2

is unbounded on ( 0, 1).

Our next goal is to develop a partial converse, for which we ﬁrst need a lemma.

Lemma 19.9. 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 19.10. Let f (x) =

be deﬁned on U = (0, ∞) and consider the sequence deﬁned by x

This is plainly Cauchy since it converges; note crucially that its limit 0 does not lie in U. Moreover,

lim f (x

) = lim n = ∞



f (x

)



is not Cauchy, whence f is not uniformly continuous. This is a far simpler argument than

that presented previously!

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.

These arguments should feel familiar: compare this line to the proof of Theorem 17.10 and the rest to Theorem 18.3.

The Cauchy condition is important here: we cannot apply the uniform continuity condition directly to a convergent

sequence (

− x

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

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 19.11. 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 19.12. 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 19.7 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

( u

) = (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

19.9,



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 19.13. 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 since it diverges by oscillation.

Consequently, there is no way to extend f to a continuous

function on any interval 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

19.11, f is uniformly continuous on any bounded inter-

val. Moreover, the derivative

−

′

(x) = sin

−

cos

is bounded whenever x is large; together with Exercise 6 we could use this to conclude uni-

form continuity of f (x). Note however that f

′

(x) is unbounded when x small (lim f

′



2πn



lim(−2πn) = −∞) so we can’t use Theorem 19.5 to conclude that f is uniformly continuous on

its entire domain.

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

≤

for all y.

In fact something stranger is going on. As you may ver-

ify (see Exercise 3), the extended function g is everywhere

differentiable with g

′

(0) = 0, and yet the derivative g

′

(x)

itself is discontinuous at x = 0!

−1

f (x)

−

′

(x)

Exercises 19. 1. Decide whether each function is uniformly continuous on the given interval.

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 function is uniformly continuous on the indicated domain 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 19.13.3 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.