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der™) with the properties that we expect. I have listed the properties in the
appendix on page 379 but only to reassure the reader. We are not interested
in the properties of general ordered ¬elds but only in that particular prop-
erty (whatever it may be) which enables us to avoid the problems outlined
in Example 1.1.3 and so permits us to do analysis.


1.2 Limits
Many ways have been tried to make calculus rigorous and several have been
successful. We choose the ¬rst and most widely used path via the notion of
a limit. In theory, my account of this notion is complete in itself. However,
my treatment is unsuitable for beginners and I expect my readers to have
substantial experience with the use and manipulation of limits.
Throughout this section F will be an ordered ¬eld. The reader will miss
nothing if she simply considers the two cases F = R and F = Q. She will,
however, miss something if she fails to check that everything we say applies
to both cases equally.

De¬nition 1.2.1. We work in an ordered ¬eld F. We say that a sequence
a1 , a2 , . . . tends to a limit a as n tends to in¬nity, or more brie¬‚y

an ’ a as n ’ ∞

if, given any > 0, we can ¬nd an integer n0 ( ) [read ˜n0 depending on ™]
such that

|an ’ a| < for all n ≥ n0 ( ).
4 A COMPANION TO ANALYSIS

The following properties of the limit are probably familiar to the reader.

Lemma 1.2.2. We work in an ordered ¬eld F.
(i) The limit is unique. That is, if an ’ a and an ’ b as n ’ ∞, then
a = b.
(ii) If an ’ a as n ’ ∞ and 1 ¤ n(1) < n(2) < n(3) . . . , then an(j) ’ a
as j ’ ∞.
(iii) If an = c for all n, then an ’ c as n ’ ∞.
(iv) If an ’ a and bn ’ b as n ’ ∞, then an + bn ’ a + b.
(v) If an ’ a and bn ’ b as n ’ ∞, then an bn ’ ab.
(vi) Suppose that an ’ a as n ’ ∞. If an = 0 for each n and a = 0,
then a’1 ’ a’1 .
n
(vii) If an ¤ A for each n and an ’ a as n ’ ∞, then a ¤ A. If bn ≥ B
for each n and bn ’ b, as n ’ ∞ then b ≥ B.

Proof. I shall give the proofs in detail but the reader is warned that similar
proofs will be left to her in the remainder of the book.
(i) By de¬nition:-
Given > 0 we can ¬nd an n1 ( ) such that |an ’ a| < for all n ≥ n1 ( ).
Given > 0 we can ¬nd an n2 ( ) such that |an ’ b| < for all n ≥ n2 ( ).
Suppose, if possible, that a = b. Then setting = |a ’ b|/3 we have > 0. If
N = max(n1 ( ), n2 ( )) then

|a ’ b| ¤ |aN ’ a| + |aN ’ b| < + = 2|b ’ a|/3

which is impossible. The result follows by reductio ad absurdum.
(ii) By de¬nition,

Given > 0 we can ¬nd an n1 ( ) such that |an ’ a| < for all n ≥ n1 ( ),
()

Let > 0. Since n(j) ≥ j (proof by induction, if the reader demands a proof)
we have |an(j) ’ a| < for all j ≥ n1 ( ). The result follows.
(iii) Let > 0. Taking n1 ( ) = 1 we have

|an ’ c| = 0 <

for all n ≥ n1 ( ). The result follows.
(iv) By de¬nition, holds as does

Given > 0 we can ¬nd an n2 ( ) such that |bn ’ b| < for all n ≥ n2 ( ).
( )
5
Please send corrections however trivial to twk@dpmms.cam.ac.uk

Observe that

|(an + bn ) ’ (a + b)| = |(an ’ a) + (bn ’ b)| ¤ |an ’ a| + |bn ’ b|.

Thus if > 0 and n3 ( ) = max(n1 ( /2), n2 ( /2)) we have

|(an + bn ) ’ (a + b)| ¤ |an ’ a| + |bn ’ b| < /2 + /2 =

for all n ≥ n3 ( ). The result follows.
(v) By de¬nition, and hold. Let > 0. The key observation is
that

|an bn ’ ab| ¤ |an bn ’ an b| + |an b ’ ab| = |an ||bn ’ b| + |b||an ’ a| (1)

If n ≥ n1 (1) then |an ’ a| < 1 so |an | < |a| + 1 and (1) gives

|an bn ’ ab| ¤ (|a| + 1)|bn ’ b| + |b||an ’ a|. (2)

Thus setting2 n3 ( ) = max(n1 (1), n1 ( /(2(|b| + 1)), n2 ( /(2(|a| + 1))) we see
from (2) that

|an bn ’ ab| < /2 + /2 =

for all n ≥ n3 ( ). The result follows.
(vi) By de¬nition, holds. Let > 0. We observe that

|a ’ an |
1 1
’ = . (3)
|a||an |
an a

Since a = 0 we have |a|/2 > 0. If n ≥ n1 (|a|/2) then |an ’ a| < |a|/2 so
|an | > |a|/2 and (3) gives

2|a ’ an |
1 1
’ ¤ . (4)
|a|2
an a

Thus setting n3 ( ) = max(n1 (|a|/2), n1 (a2 /2)) we see from (4) that

1 1
’ <
an a

for all n ≥ n3 ( ). The result follows.
2
The reader may ask why we use n1 ( /(2(|b| + 1)) rather than n1 ( /(2|b|)). Observe
¬rst that we have not excluded the possibility that b = 0. More importantly, observe that
all we are required to do is to ¬nd an n3 ( ) that works and is futile to seek a ˜best™ n3 ( )
in these or similar circumstances.
6 A COMPANION TO ANALYSIS

(vii) The proof of the ¬rst sentence in the statement is rather similar to
that of (i). By de¬nition, holds. Suppose, if possible, that a > A, that is,
a ’ A > 0. Setting N = n1 (a ’ A) we have

aN = (aN ’ a) + a ≥ a ’ |aN ’ a| > a ’ (a ’ A) = A,

contradicting our hypothesis. The result follows by reduction ad absurdum.
To prove the second sentence in the statement we can either give a similar
argument or set an = ’bn , a = ’b and A = ’B and use the ¬rst sentence.
[Your attention is drawn to part (ii) of Exercise 1.2.4.]

Exercise 1.2.3. Prove that the ¬rst few terms of a sequence do not a¬ect
convergence. Formally, show that if there exists an N such that an = bn for
n ≥ N then, an ’ a as n ’ ∞ implies bn ’ a as n ’ ∞.

Exercise 1.2.4. In this exercise we work within Q. (The reason for this will
appear in Section 1.5 which deals with the axiom of Archimedes.)
(i) Observe that if ∈ Q and > 0, then = m/N for some strictly posi-
tive integers m and N . Use this fact to show, directly from De¬nition 1.2.1,
that (if we work in Q) 1/n ’ 0 as n ’ ∞.
(ii) Show, by means of an example, that, if an ’ a and an > b for all
n, it does not follow that a > b. (In other words, taking limits may destroy
strict inequality.)
Does it follow that a ≥ b? Give reasons.

Exercise 1.2.5. A more natural way of proving Lemma 1.2.2 (i) is to split
the argument in two
(i) Show that if |a ’ b| < for all > 0, then a = b.
(ii) Show that if an ’ a and an ’ b as n ’ ∞, then |a ’ b| < for all
> 0.
(iii) Deduce Lemma 1.2.2 (i).
(iv) Give a similar ˜split proof ™ for Lemma 1.2.2 (vii).

Exercise 1.2.6. Here is another way of proving Lemma 1.2.2 (v). I do not
claim that it is any simpler, but it introduces a useful idea.
(i) Show from ¬rst principles that, if an ’ a, then can ’ ca.
(ii) Show from ¬rst principles that, if an ’ a as n ’ ∞, then a2 ’ a2 .
n
2 2
(iii) Use the relation xy = ((x + y) ’ (x ’ y) )/4 together with (ii), (i)
and Lemma 1.2.2 (iv) to prove Lemma 1.2.2 (v).

The next result is sometimes called the sandwich lemma or the squeeze
lemma.
7
Please send corrections however trivial to twk@dpmms.cam.ac.uk

Exercise 1.2.7. Suppose am ≥ cn ≥ bm for all m. Then, if an ’ c and
bn ’ c, it follows that cn ’ c as n ’ ∞.
Suppose |am | ≥ |cm | ≥ |bm | for all m and that an ’ c and bn ’ c as
n ’ ∞. Does it follow that cn ’ c? Give a proof or counterexample as
appropriate.


1.3 Continuity
Our de¬nition of continuity follows the same line of thought.
De¬nition 1.3.1. We work in an ordered ¬eld F. Suppose that E is a subset
of F and that f is some function from E to F. We say that f is continuous
at x ∈ E if given any > 0 we can ¬nd δ0 ( , x) [read ˜δ0 depending on and
x™] with δ0 ( , x) > 0 such that

|f (x) ’ f (y)| < for all y ∈ E such that |x ’ y| < δ0 ( , x).

If f is continuous at every point of E we say that f : E ’ F is a continuous
function.
The reader, who, I expect, has seen this de¬nition before, and is, in any
case, a mathematician, will be anxious to move on to see some theorems
and proofs. Non-mathematicians might object that our de¬nition does not
correspond to their idea of what continuous should mean. If we consult
the dictionary we ¬nd the de¬nition ˜connected, unbroken; uninterrupted in
time or sequence: not discrete™. A mathematician would object that this
merely de¬nes one vague concept in terms of other equally vague concepts.
However, if we rephrase our own de¬nition in words we see that it becomes
˜f is continuous if f (y) is close to f (x) whenever y is su¬ciently close to
x™ and this clearly belongs to a di¬erent circle of ideas from the dictionary
de¬nition.
This will not be a problem when we come to de¬ne di¬erentiability since
there is no ˜common sense™ notion of di¬erentiability. In the same way the
existence of a ˜common sense™ notion of continuity need not trouble us pro-
vided that whenever we use the word ˜continuous™ we add under our breath
˜in the mathematical sense™ and we make sure our arguments make no appeal
(open or disguised) to ˜common sense™ ideas of continuity.
Here are some simple properties of continuity.
Lemma 1.3.2. We work in an ordered ¬eld F. Suppose that E is a subset
of F, that x ∈ E, and that f and g are functions from E to F.
(i) If f (x) = c for all x ∈ E, then f is continuous on E.
8 A COMPANION TO ANALYSIS

(ii) If f and g are continuous at x, then so is f + g.
(iii) Let us de¬ne f — g : E ’ F by f — g(t) = f (t)g(t) for all t ∈ E.
Then if f and g are continuous at x, so is f — g.
(iv) Suppose that f (t) = 0 for all t ∈ E. If f is continuous at x so is
1/f .
Proof. Follow the proofs of parts (iii) to (vi) of Lemma 1.2.2.
By repeated use of parts (ii) and (iii) of Lemma 1.3.2 it is easy to show
that polynomials P (t) = n ar tr are continuous. The details are spelled
r=0
out in the next exercise.
Exercise 1.3.3. We work in an ordered ¬eld F. Prove the following results.
(i) Suppose that E is a subset of F and that f : E ’ F is continuous at
x ∈ E. If x ∈ E ‚ E then the restriction f |E of f to E is also continuous
at x.
(ii) If J : F ’ F is de¬ned by J(x) = x for all x ∈ F, then J is continuous
on F.
(iii) Every polynomial P is continuous on F.
(iv) Suppose that P and Q are polynomials and that Q is never zero on
some subset E of F. Then the rational function P/Q is continuous on E (or,
more precisely, the restriction of P/Q to E is continuous.)
The following result is little more than an observation but will be very
useful.
Lemma 1.3.4. We work in an ordered ¬eld F. Suppose that E is a subset
of F, that x ∈ E, and that f is continuous at x. If xn ∈ E for all n and
xn ’ x as n ’ ∞, then f (xn ) ’ f (x) as n ’ ∞.
Proof. Left to reader.
We have now done quite a lot of what is called , δ analysis but all we
have done is sharpened our proof of Example 1.1.3. The next exercise gives
the details.
Exercise 1.3.5. We work in Q. The function f is that de¬ned in Exam-
ple 1.1.3.
(i) Show that the equation x2 = 2 has no solution. (See any elementary
text on number theory or Exercise K.1.)
(ii) If |x| ¤ 2 and |·| ¤ 1 show that |(x + ·)2 ’ x2 | ¤ 5|·|.
(iii) If x2 < 2 and δ = (2 ’ x2 )/6 show that y 2 < 2 whenever |x ’ y| < δ.
Conclude that f is continuous at x.
(iv) Show that if x2 > 2 then f is continuous at x.
(v) Conclude that f is a continuous function.
9
Please send corrections however trivial to twk@dpmms.cam.ac.uk

Unless we can isolate the property that distinguishes the rationals from
the reals we can make no progress.


1.4 The fundamental axiom
The key property of the reals, the fundamental axiom which makes everything
work, can be stated as follows:

The fundamental axiom of analysis. If an ∈ R for each n ≥ 1, A ∈ R
and a1 ¤ a2 ¤ a3 ¤ . . . and an < A for each n, then there exists an a ∈ R
such that an ’ a as n ’ ∞.

Less ponderously, and just as rigorously, the fundamental axiom for the
real numbers says every increasing sequence bounded above tends to a limit.
Everything which depends on the fundamental axiom is analysis, every-
thing else is mere algebra.
I claim that all the theorems of classical analysis can be obtained from the
standard ˜algebraic™ properties of R together with the fundamental axiom. I
shall start by trying to prove the intermediate value theorem. (Here [a, b] is
the closed interval [a, b] = {x ∈ R : a ¤ x ¤ b}. )

Theorem 1.4.1. (The intermediate value theorem.) If f : [a, b] ’ R
is continuous and f (a) ≥ 0 ≥ f (b) then there exists a c ∈ [a, b] such that
f (c) = 0.

(The proof will be given as Theorem 1.6.1.)

Exercise 1.4.2. Assuming Theorem 1.4.1 prove the apparently more general
result:“
If f : [a, b] ’ R is continuous and f (a) ≥ t ≥ f (b) then there exists a
c ∈ [a, b] such that f (c) = t.

How might our programme of obtaining the intermediate value theorem
from the fundamental axiom fail?
(1) The reader has higher standards of rigour than I do but can ¬ll in
the gaps herself. For example, in the statement of Theorem 1.4.1, I do not
explicitly say that b ≥ a. Again, I talk about the ˜algebraic™ properties of R
when, strictly speaking, a set cannot have algebraic properties and I should
refer instead to the algebraic properties of (R, +, —, >). Such problems may
annoy the reader but do not cause the programme to fail.
(2) As is almost certainly the case, my proofs contain errors or have gaps
which do not occur in other accounts of the material and which can thus be
10 A COMPANION TO ANALYSIS

corrected. In such a case, I apologise but it is I who have failed and not the
programme.
(3) My proofs contain a serious error or have a serious gap which occurs
in all accounts of this material. If this can be corrected then the programme
survives but looks a great deal more shaky. If a serious error has survived
for a century who knows what other errors may lurk.
(4) All accounts contain an error which cannot be corrected or a gap that
cannot be ¬lled. The programme has failed.
We start our attempt with a simple consequence of the fundamental ax-
iom.
Lemma 1.4.3. In R every decreasing sequence bounded below tends to a
limit.
Proof. Observe that if a1 , a2 , a3 , . . . is a decreasing sequence bounded below
then ’a1 , ’a2 , ’a3 , . . . is an increasing sequence bounded above. We leave
the details to the reader as an exercise.
Exercise 1.4.4. (i) If m1 , m2 , . . . is an increasing sequence of integers
bounded above, show that there exists an N such that mj = mN for all j ≥ N .
(ii) Show that every non-empty set A ⊆ Z bounded above has a maximum.
More formally, show that if A ⊆ Z, A = … and there exists a K such that
K ≥ a whenever a ∈ A then there exists an a0 ∈ A with a0 ≥ a whenever
a ∈ A.


1.5 The axiom of Archimedes
Our ¬rst genuinely ˜analysis™ result may strike the reader as rather odd.
Theorem 1.5.1. (Axiom of Archimedes.)
1
’ 0 as n ’ ∞
n
Proof. Observe that the 1/n form a decreasing sequence bounded below.
Thus, by the fundamental axiom (in the form of Lemma 1.4.3), 1/n tends
to some limit l. To identify this limit we observe that since the limit of a
product is a product of the limits (Lemma 1.2.2 (v))
1 11 l
=—’
2n 2n 2
and since the limit of a subsequence is the limit of the sequence (Lemma 1.2.2 (ii))
1
’l
2n
11
Please send corrections however trivial to twk@dpmms.cam.ac.uk

as n ’ ∞. Thus, by the uniqueness of limits (Lemma 1.2.2 (i)), l = l/2 so
l = 0 and 1/n ’ 0 as required.

Exercise 1.5.2. [Exercise 1.2.4 (ii) concerned Q. We repeat that exercise
but this time we work in R.] Show, by means of an example, that, if an ’ a
and an > b for all n, it does not follow that a > b. (In other words, taking
limits may destroy strict inequality.)
Does it follow that a ≥ b? Give reasons.

Theorem 1.5.1 shows that there is no ˜exotic™ real number say (to choose an
exotic symbol) with the property that 1/n > for all integers n ≥ 1 and yet
> 0 (that is is smaller than all strictly positive rationals and yet strictly
positive). There exist number systems with such exotic numbers (the famous
˜non-standard analysis™ of Abraham Robinson and the ˜surreal numbers™ of
Conway constitute two such systems) but, just as the rationals are, in some
sense, too small a system for the standard theorems of analysis to hold so
these non-Archimedean systems are, in some sense, too big. Eudoxus and
Archimedes3 realised the need for an axiom to show that there is no exotic
number bigger than any integer (i.e. > n for all integers n ≥ 1; to see the
connection with our form of the axiom consider = 1/ ). However, in spite
of its name, what was an axiom for Eudoxus and Archimedes is a theorem
for us.

Exercise 1.5.3. (i) Show that there does not exist a K ∈ R with K > n for
all n ∈ Z by using Theorem 1.5.1.
(ii) Show the same result directly from the fundamental axiom.

Exercise 1.5.4. (i) Show that if a is real and 0 ¤ a < 1 then an ’ 0 as
n ’ ∞. Deduce that if a is real and |a| < 1 then an ’ 0.
(ii) Suppose that a is real and a = ’1. Discuss the behaviour of

1 ’ an
1 + an
as n ’ ∞ for the various possible values of a.
[ Hint (1 ’ an )/(1 + an ) = (a’n ’ 1)/(a’n + 1).]

Here is an important consequence of the axiom of Archimedes.

Exercise 1.5.5. (i) Use the fact that every non-empty set of integers bounded
above has a maximum (see Exercise 1.4.4) to show that, if x ∈ R, then there
exists an integer m such that m ¤ x < m + 1. Show that |x ’ m| < 1.
3
This is a simpli¬cation of a more complex story.
12 A COMPANION TO ANALYSIS

(ii) If x ∈ R and n is a strictly positive integer, show that there exists an
integer q such that |x ’ q/n| < 1/n.
(iii) Deduce Lemma 1.5.6 below, using the axiom of Archimedes explicitly.
Lemma 1.5.6. If x ∈ R, then, given any > 0, there exists a y ∈ Q such
that |x ’ y| < .
Thus the rationals form a kind of skeleton for the reals. (We say that the
rationals are dense in the reals.)
The reader will probably already be acquainted with the following de¬-
nition.
De¬nition 1.5.7. If a1 , a2 , . . . is a sequence of real numbers, we say that
an ’ ∞ as n ’ ∞ if, given any real K, we can ¬nd an n0 (K) such that
an ≥ K for all n ≥ n0 (K).
Exercise 1.5.8. Using Exercise 1.5.3 show that n ’ ∞ as n ’ ∞.
Exercise 1.5.8 shows that two uses of the words ˜n tends to in¬nity™ are
consistent. It is embarrassing to state the result but it would be still more
embarrassing if it were false. Here is a another simple exercise on De¬ni-
tion 1.5.7.
Exercise 1.5.9. Let a1 , a2 , . . . be a sequence of non-zero real numbers.
Show that, if an ’ ∞, then 1/an ’ 0. Is the converse true? Give a proof

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