Chapter 6
Mathematical analysis
Reading
Simon and Blume contains most of what is needed for this chapter, but for some
topics (such as the ‘sandwich theorem’ and the ‘intermediate value theorem’), you
should consult a specialist text in mathematical analysis (or ‘real analysis). Some
examples are listed below.
Simon, C.P. and Blume, L., Mathematics for Economists Appendix A1, sections
A1.1 and A1.2; Chapter 12; Chapter 13, section 13.4.
Bartle, R.G. and Sherbert, D.R., Introduction to Real Analysis. Chapter 2,
section 2.4; Chapter 3, sections 3.1–3.4 and 3.6; Chapter 4, sections 4.1 and
4.2; Chapter 5, sections 5.1–5.3; Chapter 10, sections 10.1–10.3.
Bryant, V., Yet Another Introduction to Analysis. Chapters 1,2, and 3.
Introduction
In this large chapter we introduce the topic of mathematical analysis, often referred to as real analysis or simply analysis. This is a formal approach to the
ideas underlying calculus. The emphasis in mathematical analysis is on formalising
notions such as the limit of a sequence, and on proving certain facts using these
formalisations. Students do not find analysis easy when they first meet it, but it is
probably the single most important part of this subject, so you should persevere!
Sets and functions: some terminology
For sets A, B, we define the set
A \ B = {a ∈ A : a 6∈ B} .
If B ⊆ A, then A \ B is the complement of B in A.
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Sets and functions,
supremum and infimum,
limits of sequences,
limits and continuity of
real functions, open and
closed sets, compact
sets, analysis in Rm
Suppose that S is some set and I is some non-empty (indexing) set such that for each
i ∈ I, we have a set Ai ⊆ S. Thus, {Ai : i ∈ I} is a set, or collection, of sets. Then
\
Ai = {x : x ∈ Ai for all i ∈ I}
i∈I
and
[
Ai = {x : x ∈ Ai for at least one i ∈ Ai } .
i∈I
We have the De Morgan laws of complementation:
\
[
[
\
S \ Ai = (S \ Ai ), S \ Ai = (S \ Ai ).
I
I
I
I
In what follows, let f : A → B be a function or mapping. (The terms function
and mapping will, for our purposes, mean the same thing.) The set A is called the
domain of f and B is called the codomain of f . For C ⊆ A, we define
f (C) = {f (c) : c ∈ C} .
The set f (A) is called the image of f .
The function f is surjective if f (A) = B. It is injective if
f (x) = f (y) =⇒ x = y.
It is bijective if it is both injective and surjective, in which case it has an inverse
function f −1 : B → A.
The cartesian product of A and B is the set
A × B = {(a, b) : a ∈ A, b ∈ B}
of all ordered pairs (a, b). Given sets A1 , A2 , . . . , An , we define
A1 × A2 × . . . × An = {(a1 , a2 , . . . , an ) : a1 ∈ A1 , a2 ∈ A2 , . . . , an ∈ An } .
Quantifiers
There are two very useful symbols often used in mathematical analysis.
First, we have the symbol ∀, called the universal quantifier, which means ‘for all’.
For example, consider the following mathematical statement:
∀x ∈ R, x2 ≥ 0.
This says, literally, that ‘for all x ∈ R, x2 is non-negative’; that is, the square of any
real number is non-negative, and this is something we know is true. On the other
hand, the statement
∀x ∈ [1, 2], x > 1
is clearly false, since if we take x = 1, we have x ∈ [1, 2] but x is not greater than 1.
This serves to demonstrate a useful observation. Suppose that P (x) is a statement
depending on x, which is either true or false for each particular x. (So, for example,
P (x) could be x > 1.) A statement of the form ∀x ∈ S, P (x) is called an existential
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statement, and it says that P (x) is true for every x ∈ S. Such a statement can
be shown to be false (or, equivalently, can be disproved) by exhibiting a particular
x0 ∈ S such that P (x0 ) is false.
Next, we have the existential quantifier ∃, meaning ‘there exists’ (or ‘there is’).
Consider the statement
∃x ∈ R such that x2 > 5.
This says there is an x ∈ R such that x2 > 5, and this is clearly true. (Often, one
abbreviates ‘such that’ to ‘s. t.’, or simply omits it.) On the other hand, the statement
∃x ∈ R such that x2 + 1 = 0
is false, since x2 + 1 = 0 has no real solutions; that is, for all x ∈ R, x2 + 1 6= 0. Again
we have a general principle. Suppose that P (x) is a statement depending on x, which
is either true or false for each particular x. A statement of the form ∃x ∈ S, P (x) is
called a universal statement, and it says that P (x) is true for some x ∈ S. Such
a statement can be shown to be false (or, equivalently, can be disproved) by proving
that for every x ∈ S, P (x) is false.
Statements can involve both types of quantifier. Consider
∀M ∈ R ∃x ∈ R such that x2 > M,
or, equivalently (omitting the ‘such that’)
∀M ∈ R, ∃x ∈ R, x2 > M.
This says ‘for every real number M there is some real number x such that x2 > M ’.
Well, this is true: for example (and there are other possibilities!), if M < 0 we may
take x = 0, if 0 < M < 1 we may take x = 1, and if M > 1 then (since M 2 > M ) we
may take x = M .
The order in which the quantifiers appear can be crucial. Suppose we were to reverse
the order of the universal and existential quantifiers in the statement just considered.
Then we obtain
∃x ∈ R, ∀M ∈ R, x2 > M.
Think about what this means: it asserts that there is some fixed real number x such
that for every real number M , x2 > M . But this is nonsense. For a fixed x, there is
always M such that x2 ≤ M . Indeed, take M = x2 to see this. So the statement we
have now is very different from the previous one.
Activity 6.1 Ensure you understand why the order of the quantifiers is vital in these
examples.
The triangle inequality
For any real number x, the absolute value of x, denoted by |x|, is x itself if x ≥ 0 and
−x if x < 0. For instance, |5| = 5 and | − 3.5| = 3.5.
The basic triangle inequality is very simple, but extremely useful. It states that for
any two real numbers x, y, we have
|x + y| ≤ |x| + |y|.
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Activity 6.2 Prove the triangle inequality. (Just consider each of four cases: x >
0, y < 0, etc.)
This has many useful consequences, which we shall use often. In particular, for any
three real numbers x, y, z, we have
|x − y| ≤ |x − z| + |y − z|
(simply because |x − y| = |(x − z) + (z − y)| ≤ |x − z| + |z − y|).
Sometimes useful is the fact that for any two real numbers x, y,
|x − y| ≥ ||x| − |y||.
Properties of real numbers: supremum and infimum
Recall that for real numbers a, b, we define different types of interval as follows:
[a, b] = {x ∈ R : a ≤ x ≤ b}
(a, b] = {x ∈ R : a < x ≤ b}
(a, b) = {x ∈ R : a < x < b}
[a, b) = {x ∈ R : a ≤ x < b}
[a, ∞) = {x ∈ R : x ≥ a}
(a, ∞) = {x ∈ R : x > a}
(−∞, b] = {x ∈ R : x ≤ b}
(−∞, b) = {x ∈ R : x < b} .
We shall sometimes use the symbol R+ to denote the set of all positive real numbers.
It is very important (especially in what follows) that you understand that ∞ is not
a number: it is merely a symbol. You cannot perform arithmetic with it. It is quite
simply nonsense to write 1/∞ = 0 and so on.
Some of these intervals are ‘bounded above’, some ‘bounded below’ and some ‘bounded
above and below’. The formal definitions of boundedness are:
Definition 6.1 S ⊆ R is bounded above if and only if there is M ∈ R such that
for all x ∈ S, x ≤ M.
In this case, M is called an upper bound of (or ‘on’ or ‘for’) S.
Definition 6.2 S ⊆ R is bounded below if and only if there is m ∈ R such that
for all x ∈ S, x ≥ m.
In this case, m is called a lower bound of (or ‘on’ or ‘for’) S.
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A set is said simply to be bounded if it is bounded above and bounded below.
Definition 6.3 An element x ∈ S is the maximum of S if it is an upper bound of
S (similarly, we can define the minimum of S).
A set that is bounded above (below) need not have a maximum (minimum); a set has
a maximum element if and only if it has an upper bound that belongs to the set; this
will not be true in general.
Every subset of the integers has a maximum and a minimum. The corresponding
notions for subsets of the real line are supremum and infimum, defined below.
We assume the following property of real numbers (as an axiom).
The Continuum property: Every non-empty set of real numbers that is bounded
above has a least upper bound and any non-empty set of real numbers that is bounded
below has a greatest lower bound.
Definition 6.4 For a non-empty subset S of R, if S is bounded above, the least upper
bound of S (which exists, by the continuum property) is called the supremum of S,
and is denoted sup S. If S is bounded below, the greatest lower bound is called the
infimum of S and is denoted inf S.
Note that the supremum and infimum of a set S (when they exist; i.e., when S is
bounded above/below) need not belong to S.
Example: Consider the real interval S = (1, 2). This is bounded above: for example,
2 is an upper bound, since every member of S is at most 2. In fact, any number
greater than or equal to 2 is an upper bound, so the set of upper bounds is [2, ∞).
It is clear that the least upper bound is therefore 2. That is, sup S = 2. But notice
that the set S has no maximum. For, suppose that M is a maximum. Then (two
things!): (i) M ∈ S and, (ii) for all x ∈ S, x ≤ M . Now, because M ∈ S, we must
have M < 2. But now consider the number (M + 2)/2, half-way between M and 2.
This is still less than 2, so it belongs to S. But it is not true that x ≤ M for all
x ∈ S, because this doesn’t hold when we take x = M . So S has a supremum but no
maximum.
Example: On the other hand, the interval S = (1, 2] has a maximum: its maximum
(and its supremum) is 2.
Example: Consider the set S = {1/n : n ∈ N} = {1, 1/2, 1/3, . . .}. This is clearly
bounded below, by 0 for example. In fact, inf S = 0. To see this, we observe (as we
already have) that 0 is a lower bound on S. To prove it is the greatest lower bound,
we need to show that no number τ > 0 is a lower bound on S. So let’s suppose that
τ is any positive number. To prove it is not a lower bound on S we have to show
that some member of S is less than τ . Well, this means we need to find a number of
the form 1/n such that 1/n < τ . But it’s clear that this inequality is equivalent to
n > 1/τ . So if we take any natural number n that is greater than 1/τ , then 1/n ∈ S
and 1/n < τ . So we are done: any τ > 0 can’t be a lower bound, and 0 is, so 0 is the
greatest lower bound, the infimum.
Activity 6.3 Make sure you understand the preceding three examples.
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An alternative description of the supremum is: σ = sup S if and only if σ is an upper
bound for S and if for any σ 0 < σ, there is some x ∈ S with x > σ 0 . (A similar
condition can be given for infimum.)
Activity 6.4 Write down the corresponding description of infimum.
A further characterisation (which follows from the one just given by taking σ 0 = σ −)
is as follows.
Theorem 6.1 For S ⊆ R, S non-empty, σ = sup S if and only if σ is an upper
bound for S and
∀ > 0, ∃x ∈ S with x > σ − .
Activity 6.5 Convince yourself that this characterisation of the supremum is equivalent to the one given in the definition.
There is also a similar characterisation of the infimum.
Theorem 6.2 For S ⊆ R, S non-empty, τ = inf S if and only if τ is a lower bound
for S and
∀ > 0, ∃x ∈ S with x < τ + .
Sequences of real numbers
Formally, a sequence is a function f from N to R. We call f (n) the nth term of the
sequence and we often denote the sequence by (f (n))∞
n=1 or simply (f (n)). Informally,
and more usually, a sequence is thought of as an infinite list of real numbers, one for
each positive integer: for example,
a1 , a2 , a3 , . . .
th
We denote it (an )∞
term
n=1 or (an ) (or indeed, (ar ), (ai ) etc.). Then we call an the n
of this sequence.
A sequence may be defined by giving an explicit formula for the nth term. For example
the formula an = 1/n defines the sequence whose value at the positive integer n is
1/n.
A sequence may also be defined inductively. For instance, we might have
a1 = 1, an+1 =
an
3
+
(n ≥ 1).
2
2an
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Limits of sequences
The definition of limit, and some basic results
As an example, let us calculate the first few terms of the sequence given by
a1 = 1, an+1 =
an
3
+
(n ≥ 1).
2
2an
We have
a1 = 1, a2 = 2, a3 = 1.75, a4 = 1.73214, a5 = 1.7320508, . . . .
Activity 6.6 Check these calculations of a2 , . . . , a5 .
Note that the terms get closer and closer to 1.73205
√ . . . as n increases. In fact, the
3√= 1.7320508057 . . .. We say that
number the terms approach
can
be
shown
to
be
√
the sequence has limit 3, or that it converges to 3.
What do we really mean by saying that a sequence (xn ) of numbers ‘tends towards’
some fixed ‘limit’ L? Well, one informal approach is to say that the terms of the
sequence are ‘eventually’ ‘approximately equal to’ L. But this will not do: we need
something more formal. First, the notion of ‘approximately equal’ is very vague.
What we really mean is that for any small given > 0, the terms of the sequence
should be between L − and L + ; that is,
L − < xn < L + ,
or, equivalently, |xn − L| < . And by ‘eventually’, we mean for large enough n: that
is, for all n > N , where N is some number. The smaller is, the larger N might have
to be, so N will generally depend on . The formal definition of a limit is as follows.
Definition 6.5 (Finite limit of a sequence) The sequence (xn ) is said to tend
to the (finite) limit L if for all > 0, there is N such that for all n > N we have
|xn − L| < . That is, (xn ) tends to L if
∀ > 0, ∃N s.t. for n > N, |xn − L| < .
We write
xn → L as n → ∞
or
lim xn = L,
n→∞
and say that xn tends to L as n tends to ∞.
You will need to know this definition thoroughly, but don’t just memorise it: understand it! Without the understanding, you won’t be able to use it. Be very aware that
the order of the quantifiers is important: ∀ comes before ∃N .
Any sequence that tends to a finite limit is said to be convergent.
In the definition, we can assume if we like that N is an integer, or we can replace
n > N by n ≥ N : neither of these changes makes any difference.
The following result is easy to prove, but very useful.
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Theorem 6.3 A sequence has no more than one limit.
Activity 6.7 Prove this. [Hint: Suppose that L1 and L2 are two limits of a sequence.
Show that in fact L1 must equal L2 by proving that for all > 0, |L1 − L2 | < 2, which
can only be true if |L1 − L2 | = 0.]
We can also talk of a sequence tending to plus or minus infinity, as in the following
definition.
Definition 6.6 (Infinite limit) The sequence (xn ) tends to infinity as n → ∞
if:
∀K, ∃N s.t. for n > N, xn > K.
We write xn → ∞ or
lim xn = ∞.
n→∞
Similarly we say that the sequence (xn ) tends to minus infinity as n → ∞ if:
∀K, ∃N s.t. n > N ⇒ xn < K.
Bounded sequences
A sequence is bounded above if its terms are bounded above (and similarly for bounded
below, and bounded). Formally:
Definition 6.7 (Bounded sequences) A sequence is (respectively) bounded above,
bounded, or bounded below if the set
S = {xn : n ∈ N} ⊆ R
is (respectively) bounded above, bounded, or bounded below.
Note that a sequence is bounded if and only if it is both bounded below and bounded
above.
The following result is often useful.
Theorem 6.4 Any convergent sequence is bounded.
Proof Suppose (xn ) is convergent, with limit L. Then for all > 0 there is N such
that if n > N then |xn − L| < . Take = 1. Then there is N (which we may assume
is a positive integer) so that for all n > N , |xn − L| < 1. Then, for such n,
|xn | = |(xn − L) + L| ≤ |xn − L| + |L| < 1 + L.
So all the terms xN +1 , xN +2 , . . . are bounded in absolute value by the constant 1+|L|.
The remaining terms x1 , x2 , . . . , xN are only finite in number (there are N of them)
and so they can be bounded in absolute value by
M = max{|x1 |, |x2 |, . . . , |xN |}.
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Then, for all n, we have either |xn | ≤ |L| + 1 or |xn | ≤ M , so certainly, |xn | <
|L| + 1 + M , and this shows that the sequence is bounded.
This result says that convergent sequences are bounded, but there’s no reason at
all why a bounded sequence must converge. For example, the sequence given by
an = (−1)n is certainly bounded, but does not converge.
Activity 6.8 Convince yourself that an = (−1)n is bounded, but does not converge.
Monotonic sequences
Definition 6.8 (Monotonic sequences) A sequence (an ) is increasing (respectively, decreasing) if for all n, an+1 ≥ an (respectively, an+1 ≤ an ). A sequence is
monotonic if it is either increasing or decreasing.
We noted above that bounded sequences need not converge. However, as the following
result shows, bounded monotonic sequences do converge1 .
Theorem 6.5 An increasing (decreasing) sequence that is bounded above (below) is
convergent. That is,
bounded + monotonic =⇒ convergent.
The Algebra of Limits
It is often possible to find the limit of a sequence using results on the limits of related
sequences. The following so-called ‘heredity’ results state several ways in which this
is possible.
Theorem 6.6 (‘Heredity’ results) Suppose (an ) and (bn ) are convergent sequences
with
an → a, bn → b, as n → ∞.
Then as n → ∞,
1. Can → Ca for any real number C
2. |an | → |a|
3. an + bn → a + b
4. an bn → ab
5. If bn 6= 0, ∀ n and if b 6= 0, then 1/bn → 1/b
6. akn → ak for any positive integer k.
Activity 6.9 Prove 3. by using the fact that
|(an + bn ) − (a + b)| = |(an − a) + (bn − b)| ≤ |an − a| + |bn − b|.
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1
For a proof, see, for
example, Simon and Blume,
Section 29.1.
It’s important to understand what the last of these results says. It says that for any
fixed positive integer k, akn → ak as n → ∞: the integer k cannot depend on n.
Example: We find the limit as n → ∞ of
an =
1 + n + n2
.
3n + 5n2
First, we note that if we divide both the numerator and the demoninator by n2 , we
have
1/n2 + 1/n + 1
.
an =
3/n + 5
Now, 1/n → 0 and hence (by the Heredity results) 1/n2 = (1/n)2 → 02 = 0. So
(again by the Heredity results) we have
1/n2 + 1/n + 1 → 0 + 0 + 1 = 1,
and
3/n + 5 = 3(1/n) + 5 → 3(0) + 5 = 5.
So (again using the Heredity results),
an →
1
5
as n → ∞.
The following result is useful.
Theorem 6.7 If |a| < 1, then limn→∞ an = 0.
Proof Suppose first that 0 < a < 1, and let xn = an . Then each term in the
sequence (xn ) is positive. Also, xn+1 = an+1 = axn < xn , so (xn ) is a decreasing
sequence. Since xn ≥ 0 for all 0, the sequence is bounded below. So the sequence is
bounded and monotonic and hence it must converge to some limit L. Now (convince
yourself!), if xn → L then xn+1 → L too. But xn+1 = aL, so by the Heredity results,
lim xn+1 = lim(axn ) = a lim xn = aL.
Hence L = aL, so L(1 − a) = 0. But since 1 − a 6= 0, this means L = 0. To sum
up, then, xn = an → 0 as n → ∞. If −1 < a < 0 then for any n, |an − 0| = |a|n .
But 0 < |a| < 1, so we know from what we’ve just shown that |a|n → 0 as n → ∞.
Thus |an − 0| → 0. Hence an → 0. (This last step can be seen as an application of
a general result, namely that if |xn | → 0 then xn → 0. It should not be thought,
however, that if |xn | → L for general L then xn → L; that simply isn’t true, as can
be seen by considering the sequence xn = (−1)n .)
In this proof we used the simple observation that if xn → L then xn+1 → L. This is
often useful in determining what the limit of a recursively defined sequence must be
(if it converges).
Example: Consider the example we met earlier:
a1 = 1, an+1 =
an
3
+
(n ≥ 1).
2
2an
91
If this sequence converges, to L, then an → L and an+1 → L too. But
an+1 =
an
3
L
3
+
→ +
,
2
2an
2
2L
so we must have
L=
3
L
+
,
2
2L
which may be written as
3
L2
+ , L2 = 3,
2
2
√
√
so L = 3 or − 3. Since (as can easily be shown) an ≥ 0 for all n, we have L ≥ 0
and therefore the limit (if it converges,
as in fact it does, though to show this requires
√
extra work), then the limit is 3.
L2 =
The Sandwich Theorem
A useful result is the Sandwich Theorem (also known as the Squeeze Theorem). You
should be able to prove it and use it.
Theorem 6.8 (Sandwich Theorem) Let (an ), (bn ), (cn ) be sequences such that an ≤
bn ≤ cn for all n ≥ 1 and
lim an = L = lim cn .
n→∞
n→∞
Then
lim bn = L.
n→∞
Proof Because an → L and cn → L, given any > 0, there are N1 , N2 such that for
n > N1 , |an − L| < and, for n > N2 , |cn − L| < . If we set N = max{N1 , N2 }, the
larger of N1 and N2 , then for n > N we have n > N1 and n > N2 , so both the above
conditions hold: |an − L| < and |cn − L| < . These may be written as
L − < an < L + , L − < cn < L + ,
so for n > N ,
L − < an ≤ bn ≤ cn < L + ,
which says |bn − L| < . Thus bn → L as n → ∞.
Example: 0 <
1
1
1
1
< and lim
= 0. So, by the Sandwich Theorem, lim n = 0.
n
n→∞
n→∞
2
n
n
2
Example: Consider the sequence (xn )∞
n=1 where
xn =
Clearly xn ≤ n ×
Thus, since
n2
n2
1
1
1
+ 2
+ ... + 2
.
+1 n +2
n +n
1
n
1
n
1
≤ . Also we have that xn ≥ n × 2
=
.
=
1
n
n +n
n+1
+1
n + n2
1
1
= 0 = lim ,
n→∞ n
n+1
we have by the Sandwich Theorem that
1
1
1
lim
+ 2
+ ... + 2
= 0.
n→∞ n2 + 1
n +2
n +n
lim
n→∞
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Limits of real functions
Definition of limit
6
6
L
L
u
-
a
a
We start by defining what we mean by the limit of a function f (x) as the variable x
approaches some value a ∈ R. For example, for the first function illustrated in the
figure above, we would say that the limit of the function as x approaches a is the
value L. In fact, we are not concerned about the value of the function at the point a
(indeed we don’t even care if f is defined at a); for example, for the second function
illustrated in the above figure, we would still say that the limit is L, even though f (a)
is defined to be some value strictly greater than L. Thus the notion of limit applies
only to the behaviour of the function f in a neighbourhood near to a. The formal
definition now follows.
Definition 6.9 (Limit of a function at a point) Let f : R → R be a function.
We say that L is the limit of f (x) as x approaches a, denoted by limx→a f (x) = L,
if for each > 0, there exists δ > 0 such that
0 < |x − a| < δ =⇒ |f (x) − L| < .
The definition states that if someone gives us any arbitrarily small , then there is
some neighbourhood of a, (a − δ, a + δ), such that any x in this neighbourhood —
other than a itself — will have f (x) in the -neighbourhood (L − , L + ) of L.
Examples of limit
Example: Suppose that f (x) = 3x−1. We can then show directly from the definition
of a limit that limx→1 f (x) = 2. To prove this, we must argue that for any we can
bound the value |f (x) − 2| = |(3x − 1) − 2| < in some neighbourhood of 1. But we
easily see that
|f (x) − 2| = 3|x − 1|.
In other words, the distance of f (x) from 2 is three times the distance of x from 1.
Thus if we choose x to be within distance /3 from 1, then f (x) is within distance of 2.
A formal proof would go as follows. Let > 0. Then, let δ = /3. For any x such
that 0 < |x − 1| < δ we have:
|f (x) − 2| = 3|x − 1| < 3
< .
3
93
Therefore f (x) → 2 as x → 1.
Example: As a second example, consider f (x) = x2 +x. We show that limx→2 f (x) =
6. To see this, suppose that > 0 is given. If x is such that 0 < |x − 2| < δ, then
|f (x) − 6| = |x2 + x − 6| = |x + 3||x − 2| < |x + 3|δ.
Now, if |x − 2| < δ then |x + 3| ≤ 5 + |x − 2| < 5 + δ, by the triangle inequality. So
to make |f (x) − 6| less than , it suffices to have (5 + δ)δ ≤ . There is no need to
solve a quadratic equation in δ. If we just put δ = min{1, /6}, the lesser of 1 and
/6, then we have 5 + δ ≤ 6 and δ ≤ /6, so
(5 + δ)δ ≤ 6(/6) = ,
and we’re done.
Algebra of limits
Let f, g : R → R be two functions and c any real number. Then a new function
(f + g) is obtained by defining for each x, (f + g)(x) = f (x) + g(x). We say, as
earlier, that (f + g) is derived pointwise since the value of (f + g) at x is defined
by the normal arithmetic sum of the two real numbers f (x) and g(x). Similarly, we
may define the functions |f |, (cf ), (f − g), (f + g), (f g) and (f /g), provided g(x) 6= 0.
(For example, (f g)(x) = f (x)g(x). This should not be confused with the composite
function f (g(x)).)
Theorem 6.9 Let f, g : R → R be two functions and c be any real number. Suppose
that limx→a f (x) = L and limx→a g(x) = M . Then
1. limx→a (cf )(x) = cL
2. limx→a (|f |)(x) = |L|
3. limx→a (f + g)(x) = L + M
4. limx→a (f − g)(x) = L − M
5. limx→a (f g)(x) = LM
6. limx→a (f /g)(x) = L/M provided g(x) 6= 0 for each x in some neighbourhood of
a.
One-sided limits
Sometimes, we may have a configuration as in the next figure, where the limit from
one side is different from that from the other. We adapt the definition as follows.
Definition 6.10 (One-sided limits) Let f : R → R be a function. We say that L
is the limit of f (x) as x approaches a from the left, denoted by limx→a− f (x) = L
if for each > 0, there exists δ > 0 such that
0 < a − x < δ =⇒ |f (x) − L| < .
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A similar definition applies to limits from the right, denoted limx→a+ f (x) = L. In
the following figure, the left-hand and right-hand limits are L, M respectively.
6
M
u
L
a
We can also define the limit of a function as x tends to ∞ (with a similar definition
for the limit as x tends to −∞). This definition is similar to the one we met for the
limit of a sequence.
Definition 6.11 (Limit as x approaches infinity) Let f : R → R be a function.
We say that L is the limit of f (x) as x approaches ∞, denoted by limx→∞ f (x) = L
if for each > 0, there exists N > 0 such that
x ≥ N =⇒ |f (x) − L| < .
Continuity of real functions
Loosely speaking, a function is continuous if its graph can be drawn without lifting
pen from paper.
Definition 6.12 (Continuity at a point) A function is continuous at the point
a if f (a) is defined and is equal to limx→a f (x).
Definition 6.13 (Continuity) A function is continuous if it is continuous at each
point a.
Definition 6.14 (Continuity on interval) A function is continuous on the interval [a, b] if it is continuous at each point in (a, b) and (i) f (a) = limx→a+ f (x) and
(ii) f (b) = limx→b− f (x).
A function is discontinuous at a if it is not continuous there.
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It follows from the results on the algebra of limits that there are ‘heredity’ results for
continuity.
Theorem 6.10 (Heredity results for continuity) Let f, g : R → R be functions
that are continuous at a ∈ R and c be any real number. Then |f |, (cf ), (f − g), (f +
g), (f g) are all continuous at a, and (f /g) is continuous provided g(x) 6= 0 for any x
in some neighbourhood of a.
Pk
As a corollary, we show that any polynomial p(x) = i=0 ai xi is continuous. Clearly
any constant function is continuous by the above results. Moreover, the function
f (x) = x is continuous. Thus by the product rule, so is f (x) = x2 , and by extension,
f (x) = xk is continuous for any finite k. Hence so is any function of the form
f (x) = ai xi . Finally, by repeated application of the summation rule (the fact that
f + g is continuous if f and g are), the polynomial p(x) is deduced to be continuous.
Recall that if f, g are functions, then we may define the composite function f (g(x)).
The following result shows that continuity of the composite function follows from
continuity of f and g.
Theorem 6.11 If g is a function that is continuous at a, and f is a function that is
continuous at g(a), then the composite function f (g(x)) is continuous at a.
Continuity and sequences
We now give a result linking continuity with the concept of limits for sequences.
Theorem 6.12 Suppose that the function f is continuous at a and that (an ) is a
sequence such that limn→∞ an = a. Then we have limn→∞ f (an ) = f (a).
We find this a very useful result for the sake of manipulation of limits since it states
that for any sequence an → a, the limit of the f (an )’s is simply f applied to the limit
of the an ’s:
lim f (an ) = f lim an .
n→∞
n→∞
Theorem 6.12 could in fact be used as the basis of a definition of continuity: a function
f : R → R is continuous at a ∈ R if and only if for every sequence xn converging
to a, f (xn ) → f (a). (The theorem as stated just states the ‘only if’ part, but the ‘if’
is also true.)
Continuous functions on closed intervals
Definition 6.15 (Bounded function) For a subset X of the domain of a function
f , we say that f is bounded on X if there exists M such that |f (x)| ≤ M for each
x ∈ M.
Definition 6.16 (Sup and max of a function) We define the supremum (or maximum) of f on X as sup{f (x) : x ∈ X} (or max{f (x) : x ∈ X} if it exists).
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The following result is very important. You will not need to know how to prove it,
but you must be able to state it accurately and use it.
Theorem 6.13 Let f be continuous on [a, b]. Then f is bounded on [a, b] and it
achieves its maximum and minimum; that is, explicitly, the set {f (x) : x ∈ [a, b]} is
bounded, and has a maximum and a minimum, i.e., there are x1 , x2 in [a, b] such that
f (x1 ) = max{f (x) : x ∈ [a, b]} and f (x2 ) = min{f (x) : x ∈ [a, b]}.
This theorem shows that it makes sense to try to optimise (that is, maximise or
minimise) a continuous function on a closed and bounded interval. It is necessary
for the interval to be both closed and bounded: for example, although the function
f (x) = 1/x is continuous on (0, 1), it has no maximum on the interval, and although
g(x) = x is continuous on the whole of the real numbers, it has no maximum. We
return later in this chapter to a discussion of what is so special about closed and
bounded intervals.
The Intermediate Value Theorem
In this section we prove one of the most fundamental (and obvious!) theorems in real
analysis. There are many proofs of this result, and if you don’t like the one given
here, you can find others in textbooks. Roughly speaking, the Intermediate Value
Theorem states that any continuous function cannot ‘hop over’ intermediate values
as it increases from one value to another.
Theorem 6.14 (Intermediate Value Theorem) Let f be a continuous function
on [a, b] and let K be such that f (a) < K < f (b). Then for some c ∈ (a, b), f (c) = K.
Fairly obviously, (just by considering −f in place of f ), if f (a) > K > f (b) then the
same conclusion holds.
We prove a special case of the result, from which the full theorem follows.
Theorem 6.15 Let f be a continuous function on [a, b] such that f (a) < 0 and
f (b) > 0. Then for some c ∈ (a, b), f (c) = 0.
Proof We construct a sequence of intervals [an , bn ] such that
1. f (an ) < 0, f (bn ) > 0 for each n
2. [an+1 , bn+1 ] ⊆ [an , bn ] for each n.
We start by letting [a1 , b1 ] = [a, b]. Then for each n ≥ 1, we define [an+1 , bn+1 ] as
follows. Let cn = (an + bn )/2 be the midpoint of the previous interval. If f (cn ) =
0, then the theorem is proved and so we need not continue constructing intervals!
Otherwise, if f (cn ) < 0, we define an+1 = cn and bn+1 = bn . And if f (cn ) > 0, we
define bn+1 = cn and an+1 = an . Note that condition 1. is satisfied by choosing our
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intervals in this manner. Moreover, note that the (n + 1)st interval is half the size of
the nth interval and so bn+1 − an+1 ≤ (b1 − a1 )/2n . It follows that
lim (bn − an ) = 0.
n→∞
(6.1)
Finally, note that (an ) is increasing and bounded above (by b1 ) and so it has a limit;
similarly (bn ) is decreasing and bounded below and so has a limit. Thus by (6.1)
(and algebra of limits) these limits are equal to, say, c. Thus by continuity (using
Theorem 6.12)
f (c) = lim f (bn ) ≥ 0
n→∞
where the last inequality follows from the fact that each f (bn ) ≥ 0 (in fact > 0).
Similarly,
f (c) = lim f (an ) ≤ 0.
n→∞
Thus f (c) must be equal to zero, and the proof is complete.
Activity 6.10 Show that Theorem 6.15 implies Theorem 6.14. [Hint: consider g
defined by g(x) = f (x) − K.]
Open and closed sets of real numbers
Open sets
Very roughly speaking, a set U of real numbers is said to be an open set if round
every point of U there is some room to move in both the direction of increasing real
numbers and decreasing real numbers without leaving the set U . The formal definition
is as follows:
Definition 6.17 (Open set of real numbers) A set U ⊆ R is an open set (or
is open) if for every y ∈ R there is some = (y) > 0 such that (y − , y + ) ⊆ U .
We write = (y) in this definition to emphasise that can, and will, generally,
depend on y: that is, y is given and we then find a suitable .
Example: The open interval U = (1, 2) is open. To see this, let y ∈ U . Then y is
between 1 and 2. Provided we take to be no more than the smaller of y − 1 and
2 − y, then (y − , y + ) ⊆ U . Convince yourself of this! A similar argument shows
that any open interval (a, b) is open (which is a good thing, since we refer to it as an
‘open’ interval!).
Example: The interval U = (1, 2] is not open, because if we take x = 2 then no
matter how small is, the interval (2 − , 2 + ) contains numbers greater than 2 and
hence does not lie in U = (1, 2]. (Note that, although there is an open interval in U
around every other point of U , the fact that this fails to hold for the single point 2 is
enough to show that U is not open: to be open, we would need the condition to hold
for every point of U .)
The following result is easy to prove:
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Theorem 6.16 The union of any collection of open sets is again an open set.
It should not be thought that all open sets are open intervals: although every open
interval is open, there are many other types of open set. For example, the set (1, 2) ∪
(3, 4) is open, but is not an open interval.
Closed sets
We also have the notion of closed sets. But before defining what this means, we
need to clear up one source of potential confusion. By analogy with the use of the
words ‘open’ and ‘closed’ in everyday language, we might think that a given set of
real numbers must be either open or closed, and that if it is not open, it is closed.
This, unfortunately, will not be the case: as we shall see, sets can be open but not
closed, closed but not open, both open and closed, or neither open nor closed!
Here is the formal definition of a closed set:
Definition 6.18 A set C ⊆ R is a closed set (or is closed) if whenever (xn ) is a
convergent sequence and xn ∈ C for all n, then the limit of the sequence, lim xn , is
in C.
So a set C is closed if for any convergent sequence of members of C, the limit of the
sequence is in C. This is a tricky definition to work with, but as we shall shortly see,
there is another way of describing closed sets.
Example: The interval C = [0, 1] is closed. To see this, suppose that (xn ) is any
sequence in C, converging to a limit L. Then for each n, xn ∈ C, so 0 ≤ xn ≤ 1.
Now, it follows from this that 0 ≤ L ≤ 1 (prove this!), so L ∈ C, and hence C is
closed.
Example: The interval C = (0, 1] is not closed. Consider the sequence (xn ) where
xn = 1/n. For all n, xn ∈ C. The sequence converges to 0, but 0 is not in C. So C
is not closed.
We mentioned that ‘closed’ is not the ‘opposite’ of ‘open’, but the following result
linking open sets and closed sets is very useful.
Theorem 6.17 A set C of real numbers is closed if and only if its complement
R \ C is open.
Proof Because this is an ‘if and only if’ result, there are two things to prove here:
first, that if C is closed then R \ C is open; secondly, that if R \ C is open then C is
closed.
Suppose, first, that C is closed and consider its complement U = R \ C. We want to
show U is open. Suppose it isn’t. Then there is some y ∈ U such that for no > 0
do we have (y − , y + ) ⊆ U . In other words, for all > 0, the interval (y − , y + )
doe not lie entirely within U = R \ C and hence must contain points of C. For any
positive integer n, let’s take = 1/n. Then there is some xn ∈ (y − 1/n, y + 1/n) such
that xn ∈ C. Because |xn − y| < 1/n, we have that xn → y as n → ∞. So here we
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have a sequence (xn ) in C such that lim xn = y 6∈ C. But this cannot happen since
C is closed. So what’s gone wrong? Well, we supposed that R \ C was not open, and
this supposition must therefore be wrong. So R \ C is open.
Next, suppose that R \ C is open. To prove that C is closed, we need to show that
the limit of any convergent sequence of points of C is in C. So suppose (xn ) is a
convergent sequence, with xn ∈ C, and L = lim xn . We need to show L ∈ C. Suppose this isn’t so. Then L is in the open set R \ C, so there is some > 0 such that
(L − , L + ) ⊆ R \ C. Now, because xn → L, there is some N such that for n > N ,
|xn − L| < , that is xn ∈ (L − , L + ). But then for n > N , xn ∈ R \ C. This is a
contradiction to the fact that xn ∈ C. So we have gone wrong in assuming that L is
not in C. Therefore it is in C, and C is closed.
This theorem is an extremely useful characterisation of closed sets. In fact, we could,
if we had wanted, have taken the definition of a closed set to be a set C whose
complement R \ C is open, and many texts do this.
Example: Consider again the set C = [0, 1]. We showed this was closed by using
Definition 6.18. But we can also see that it is closed by considering its complement.
For,
R \ C = R \ [0, 1] = (−∞, 0) ∪ (1, ∞),
and this is open because it is the union of two open sets. So, since the complement
of C is open, C is closed.
Example: The set R of all real numbers is both open and closed. The interval (0, 1]
is neither open nor closed.
Activity 6.11 Convince yourself of the statements in the example just given.
Analysis on Rm
We now turn our attention to functions defined on Rm . (We use m rather than the
more usual n because we have used n in representing a typical member of a sequence
as xn .) All the above may be generalised considerably using the notion of a ‘metric’
or ‘distance function’.
Distance in Rm
The Euclidean distance (or simply distance) between x = (x1 , x2 , . . . , xm ) and
y = (y1 , y2 , . . . , ym ) in Rm is defined to be
v
u n
uX
kx − yk = t (xi − yi )2 .
i=1
(The case in which m = 1 corresponds to the distance |x − y| between two real
numbers.)
There is a certain mathematical attraction in defining distances on rather more abstract or unusual spaces, and this leads to the notion of a metric space. But in this
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subject, we are primarily interested in Rm , and we shall always use the Euclidean
distance.
The Euclidean distance between two vectors x, y is simply the norm of x − y, where
the norm is the one arising from the usual inner product on Rm . Consequently, the
distance measure has some nice properties. For example, by the triangle inequality
for norms, we have that for any x, y, z ∈ Rm ,
ky − zk = k(y − x) + (x − z)k ≤ ky − xk + kx − zk.
Let’s recall the definition of continuity of a function f : R → R. We say that f is
continuous at a if and only if given any > 0, there is δ > 0 such that
|x − a| < δ =⇒ |f (x) − f (a)| < .
This appears to use the fact that one can form the difference between any two real
numbers, However, as above, we can interpret |x − a| as the distance between the real
numbers x and a, in which case the above condition can be restated as
distance(x, y) < δ =⇒ distance(f (x), f (a)) < .
It should be clear from this that the condition doesn’t really use any algebraic properties of R, only ‘distance’ properties. This definition and many of its consequences
will remain if we have as domain and codomain Rm and Rk for any m, k ≥ 1.
Bounded subsets of Rm
A bounded subset of Rm is one in which there is some fixed number bounding the
distance between any two points in the set. Formally:
Definition 6.19 (Bounded subset of Rm ) B ⊆ Rm is bounded if there is K > 0
such that for all x, y ∈ B, kx − yk ≤ K.
Note that K is fixed: it does not depend on x, y. (The definition would be meaningless
if that were the case.)
Open balls
For m > 1, the counterpart in Rm to the open interval in R is the open ball.
Definition 6.20 (Open Ball) For x ∈ Rm and > 0, the open ball of radius around x is
B (x) = {y : kx − yk < } .
This is the set of y whose distance from x is less than .
Example: When m = 1, B (x) is exactly the open interval (x − , x + ).
Example: In R2 the open ball B (x) is the region enclosed by a circle of radius centred at x — note that the points on this circle do not lie in B (x).
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Continuity in general
We mentioned earlier that with a general concept of distance, we ought to be able to
generalise the definition of continuity to functions having as domain and codomain
Rm and Rk .
Definition 6.21 (Continuity of f : Rm → Rk ) Suppose that f : Rm → Rk and
that a ∈ Rm . Then, we say that f is continuous at a if given any > 0 there exists
δ > 0 such that if kx − ak < δ then kf (x) − f (a)k < .
(Note that in this definition we use, as is customary, the notation k.k to denote both
the distance function on Rm and that on Rk .)
This can also be phrased in terms of open balls.
Definition 6.22 The function f : Rm → Rk is continuous at a ∈ Rm if given any
open ball B (f (a)), there exists δ such that
f (Bδ (a)) ⊆ B (f (a)) .
For a subset X of Rm , we say that f is continuous on X if for all a ∈ Rm , f is
continuous at a. If f is continuous on Rm then we simply say that f is continuous.
The following result is sometimes useful.
Theorem 6.18 Suppose that f : Rm → Rk and that f1 , f2 , . . . , fk : Rm → R are
such that for all x ∈ Rm ,
f (x) = (f1 (x), f2 (x), . . . , fk (x))T .
Then f is continuous if and only if f1 , f2 , . . . , fk are continuous.
Open sets in Rm
We have already investigated in some depth the notion of open sets of real numbers.
All the ideas and results extend to Rm , and the proofs are very similar. You should
try them, following the proofs given earlier in this chapter.
The definition of open set
Definition 6.23 A subset U of Rm is open if for any y ∈ U , there is = (y) > 0
such that B (y) ⊆ U.
Informally, a set is open if, from any point of the set, we can move some positive
distance in any ‘direction’ without going outside the set.
It can easily be shown that any open ball is an open set, but there are other types of
open set. Just as for open sets in R, the union of any collection of open subsets of
Rm is again open.
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Sequences in Rm
A sequence (xn ) in Rm is simply an ordered list x1 , x2 , . . . of elements of Rm . (This
is a straightforward extension of the definition of a sequence of real numbers.) It’s
fairly simple to extend to Rm ideas about convergence of sequences. The following
definition is a straightforward extension of the earlier definition for convergence for
sequences of real numbers. (We use x rather than L just for cosmetic reasons.)
Definition 6.24 (Convergence of sequences in Rm ) Suppose that (xn ) is a sequence of points of Rm . Then we say that the sequence has limit x ∈ Rm if for every
> 0 there is N such that if n > N then kxn − xk < . Such a sequence is said to be
convergent and to converge towards x.
Having defined convergence of sequences in Rm , it is possible to describe an alternative
approach to the definition of continuity of functions from Rm to Rk . We have already
seen for the case of real functions on R that if a function is continuous and xn → x
then f (xn ) → f (x) and we have noted that this can in fact be used as the basis of a
definition of continuity: a function f : R → R is continuous at a ∈ R if and only
if for every sequence xn converging to a, f (xn ) → f (a). This can be generalised
for functions mapping from Rm into Rk (and this is the approach taken in Simon
and Blume). We have the following theorem (which may be taken as a definition of
continuity).
Theorem 6.19 Suppose that f : Rm → Rk and that a ∈ Rm . Then f is continuous
at a if and only if for every sequence (xn ) converging to a we have f (xn ) → f (a).
Therefore f is continuous (on the whole of Rm ) if for every convergent sequence (xn )
in Rm , we have lim f (xn ) = f (lim xn ).
Closed sets in Rm
Just as for subsets of R, we can define the notion of a closed subset of Rm .
Definition 6.25 A set C ⊆ Rm is a closed set (or is closed) if whenever (xn ) is
a convergent sequence and xn ∈ C for all n, then the limit of the sequence, lim xn , is
in C.
So a set C is closed if for any convergent sequence of members of C, the limit of the
sequence is in C.
As for the case m = 1 investigated above, we have the following result, the proof of
which is similar to the one given earlier.
Theorem 6.20 A set C ⊆ Rm is closed if and only if its complement Rm \ C is
open.
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Compactness
The idea of a compact set is extremely important in analysis. There are a number
of ways of defining what we mean by a compact set. The simplest way is as follows.
Definition 6.26 (Compact subset of Rm ) A subset C of Rm is said to be compact if and only if it is both closed and bounded.
Thus, for example, any closed ball
{y : ky − xk ≤ }
is compact, but the open ball B (x) = {y : ky − xk ≤ } is not.
The following result is useful. Earlier we mentioned that a continuous real function
on a closed interval [a, b] of the real numbers is bounded on the interval and attains
its maximum and minimum. That result can be seen as a special case of this one
(since the closed interval [a, b] is compact).
Theorem 6.21 Suppose that C is a compact subset of Rm and that the function
f : Rm → R is continuous on C. Then f is bounded on C and it achieves its
maximum and minimum; that is, explicitly, the set {f (x) : x ∈ C} is bounded and has
a maximum and a minimum: i.e., there are x1 , x2 ∈ C such that f (x1 ) = max{f (x) :
x ∈ C} and f (x2 ) = min{f (x) : x ∈ C}.
Compactness and subsequences
Consider the real number sequence
1 1 1 1 1
1, , , , , , . . . .
2 3 4 5 6
If we were to delete every second term our sequence would become
1 1 1
, , ,...
2 4 6
The resulting sequence is called a subsequence of the original sequence. The first
term of the new sequence is the second term of the orginal sequence; the second term
of the new sequence is the fourth term of the original sequence, and so on. Thus the
elements of the new subsequence can be found by looking in the appropriate position
in the original sequence. More precisely, the nth term of the subsequence is simply
the 2nth term of the original sequence. We may similarly consider subsequences of
sequences in Rm .
The formal definition of a subsequence is as follows.
m
Definition 6.27 (Subsequence) Let (xn )∞
and let f be
n=1 be a sequence in R
a strictly increasing function from N to N. The sequence (xf (n) )∞
n=1 is called a
subsequence of the sequence (xn )∞
.
n=1
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(Note that a function f is strictly increasing if f (n + 1) > f (n) for all positive n.)
In the example given above, we see that we have in fact obtained a subsequence
corresponding to the increasing function defined by f (n) = 2n. Note that this function
tells us precisely which terms to keep from the original sequence.
Often we will use shorthand to denote a sequence. For example, for a sequence
x1 , x2 , x3 , x4 , . . .
we may say that we wish to choose a subsequence
xk1 , xk2 , xk3 , xk4 , . . . .
We have written the increasing function f explicitly in terms of its value, by saying
exactly which terms of the original sequence to take; that is, we take terms k1 , k2 , k3
and so on. Note that this will be just for notation’s sake and in these cases the
underlying function has not disappeared; in fact the increasing function here is given
by f (i) = ki for i = 1, 2, 3, . . .. One result that is easy to show is:
Theorem 6.22 Let (xn ) be a sequence which tends to a limit L. Then any subsequence also tends to the limit L.
Activity 6.12 Prove this.
We have the following result on sequences lying in compact subsets of Rm . It is
known as the Bolzanno-Weierstrass Theorem. (You need not know the proof. 2 )
Theorem 6.23 (Bolzanno-Weierstrass Theorem) Let C be a compact subset of
Rm and let (xn ) be any sequence of points of C. Then (xn ) has a convergent subsequence whose limit is in C.
Consider the case m = 1. Since any bounded subset of the real numbers lies in some
closed interval (because if M = sup S and m = inf S then S ⊆ [m, M ]), it follows
that any bounded sequence can be assumed to be a sequence of numbers from some
compact set C. The theorem tells us, therefore, that any bounded sequence of real
numbers has a convergent subsequence. This is often useful in proving results in real
analysis.
Compactness and open covers
Another interesting aspect of compact sets is to do with open covers.
Definition 6.28 A cover for a subset B of Rm (or a cover of B) is a collection C
of sets such that
[
B⊆
U.
U ∈C
(That is, the union of all the sets in C contains B.) A subcover of C is a subcollection
D ⊆ C such that D itself is a cover for B. A cover is finite if it consists of a finite
number of sets.
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2
The proof can be found,
for the case m = 1, in
Bartle and Sherbert,
Section 3.4, for example.
Definition 6.29 (Open cover) An open cover for B ⊆ Rm is a cover for B
consisting of open sets.
The following result holds, and indeed is taken in many texts to be the definition
of a compact set.3 (It might seem weird to take it as the definition of compactness,
since the definition we have given is so much more straightforward. But there is a
bigger picture here, involving more general notions of ‘metric spaces’ and ‘topological
spaces’, in which it is appropriate to use the characterisation given below.)
Theorem 6.24 A subset B of Rm is compact if and only if any open covering of
B has a finite subcover. More explicitly, B is compactSif and only if whenever we are
given a family {UI : i ∈ I} of open sets such that B ⊆ I Ui , then there is some finite
subfamily
Ui1 , Ui2 , . . . , Uin
such that
B ⊆ Ui1 ∪ Ui2 ∪ . . . ∪ Uin .
There are some important points to be made on this theorem. First, we only consider
coverings by open sets. Secondly — and this is crucial! — it is important to realise
what the theorem does not say. It does not say that the set is compact if there is
a finite open cover for B. Indeed, that is always true, since we may simply take the
cover consisting of the single open set {Rm }. What it does say is something vastly
different: given any open cover C of B, there is a finite number of the sets in C which
are enough to cover B.
Learning outcomes
At the end of this chapter and the relevant reading, you should be able to:
• use existential and universal quantifiers
• explain exactly what is meant by the supremum and infimum and be able to
determine these
• explain what is meant by a sequence
• work with sequences defined by a formula or defined recursively
• explain precisely what it means to say that a sequence converges or tends to
infinity or minus infinity
• determine limits and prove results using the formal definition of the limit of a
sequence
• explain what is meant by a sequence being bounded, bounded above or bounded
below
• comprehend the links between boundedness and convergence
• prove and use the fact that bounded monotonic sequences converge
• calculate limits of sequences using the algebra of limits
106
3
For a proof of the result,
see, for example, Simon
and Blume, Section 29.5
• prove and use the Sandwich Theorem
• explain what is meant formally by the limit of a function at a point, or as x
tends to ∞ or −∞, and by one-sided limits
• determine limits and prove results using the formal definitions of the limit of a
function
• calculate limits of functions using the algebra of limits
• explain what is meant by continuity
• prove functions are continuous or discontinuous using the formal definition and
heredity results
• explain and be able to use the interaction between continuous functions and
convergent sequences
• use the fact that a continuous function has a maximum and a minimum value
on a closed bounded interval
• state and use the Intermediate Value Theorem
• explain what is meant by open and closed sets and be able to prove that sets
are open or closed or neither
• use the definitions and results concerning convergence, continuity, open and
closed sets, and compact sets in Rm .
Sample examination questions
Question 6.1 Prove that for any two real numbers x, y, |x − y| ≥ ||x| − |y||.
Question 6.2 For each of the following sets, find the sup, inf, max and min whenever
these exist.
[−1, ∞),
{x ∈ R : x2 − 2x − 1 < 0},
{x2 − 2x − 1 : x ∈ R}.
Question 6.3 Suppose A is a bounded subset of R. Let B be the set defined by
B = {b : b = 2a + 3, , a ∈ A}.
Prove that
sup B = 2 sup A + 3.
Question 6.4 Suppose that (xn ) is a sequence of real numbers that converges to a
limit L > 0. Show that there is N ∈ N such that for all n > N , xn > 0.
Question 6.5 Let (an ) be the sequence defined by
an =
4n2 + 9
.
+ 7n + 11
3n2
107
By considering the difference
an −
4 ,
3
show explicitly that
an →
4
as n → ∞.
3
Question 6.6 Show that if a sequence is decreasing and bounded below then it
converges.
Question 6.7 Let (xn ) be a sequence of non-negative real numbers (i.e. xn ≥ 0 for
each n). Suppose that the sequence converges to x. Prove that x ≥ 0. [Note, however,
that a sequence of positive terms need not have a positive limit; for example, (1/n)
is a sequence of positive terms converging to 0.]
Question 6.8 Find the limits as n → ∞ of:
4n − 5 5(32n ) − 1 1 + 2 + · · · + n n2 − 1
,
,
, 3
.
22n − 7 4(9n ) + 7
n2
n +1
Question 6.9 A sequence (xn ) is defined as follows. Let x1 be any positive real
number and, for n ≥ 1, let
x2 + k
xn+1 = n
,
2xn
where k is a fixed positive number.
2
2
(i) By using
√ the inequality a + b ≥ 2ab for real numbers a, b (or otherwise), prove
that xn ≥ k for n ≥ 2.
(ii) Prove that xn+1 ≤ xn for all n ≥ 2.
(iii) Deduce that (xn ) converges and find its limit.
Question 6.10 Find limn→∞ xn in the following cases:
xn =
xn =
2n3 + 1
,
3n3 + n + 2
n
(n + 1)
2n
+
+ ... + 2.
n2
n2
n
Question 6.11 Find
lim
n→∞
√
1
n2 + 1
+√
1
n2 + 2
+ ... + √
1
n2 + n
.
Question 6.12 Prove, from the formal definition of limit, that f defined on R \ {2}
by
5
f (x) =
(x − 2)2
tends to infinity as x → 2.
108
Question 6.13 Evaluate the following limit:
x3 + 5x + 7
.
x→1 x4 + 6x2 + 8
lim
Question 6.14 Let bxc denote the largest integer n such that n ≤ x. Determine
lim (x − bxc) and
x→1+
lim (x − bxc).
x→1−
Question 6.15 Suppose the real function f is continuous at c and f (c) > 0. Prove,
directly from the definition of continuity, that there is δ > 0 such that f is positive
in the interval (c − δ, c + δ).
Question 6.16 Suppose the real function f is continuous, positive and unbounded
on R and that inf{f (x) : x ∈ R} = 0. Use the Intermediate Value Theorem to prove
that the range of f is (0, ∞), the set of all positive real numbers. [It might be obvious,
but give a watertight proof. Explicitly, prove that for any y > 0 there is some c ∈ R
such that y = f (c).]
Question 6.17 Suppose that the real function f is continuous on the closed interval
[a, b] and that f maps [a, b] into [a, b]. By considering the function h(x) = f (x) − x,
show that there is c ∈ [a, b] with f (c) = c.
Question 6.18 Suppose that f : Rm → R is a continuous function and that f (x∗ ) >
0. Show that there is an open ball B = Bδ (x∗ ) such that f (x) > 0 for all x ∈ B.
Question 6.19 Suppose f : Rm → R is continuous at the point a = (a1 , . . . , am ).
Consider the function g : R → R defined by
g(t) = f (t, a2 , . . . , am ).
Show that g is continuous at a1 .
Question 6.20 Suppose (xn ) is a sequence in Rm and let xn = (x1n , x2n , . . . , xmn ).
Prove that xn → x = (x1 , . . . , xm ) if and only if, for i between 1 and m, xin → xi as
n → ∞.
Question 6.21 Let z ∈ Rm and > 0. Show that the closed ball
{x ∈ Rm : kx − zk ≤ }
is a closed subset of Rm .
Question 6.22 Prove that the intersection of any family of compact sets is compact.
109
Sketch answers or comments on selected questions
Question 6.1 We need to prove
−|x − y| ≤ |x| − |y| ≤ |x − y|.
So there are two inequalities to prove. The leftmost follows from the triangle inequality, as follows:
|y| = |(y − x) + x| ≤ |y − x| + |x|,
so
−|x − y| = −|y − x| ≤ |x| − |y|.
For the second, we observe that
|x| = |(x − y) + y| ≤ |x − y| + |y|.
Question 6.2 For the first set, min=inf=−1, and there is√no sup √
and no max. For
the second, we note that the
set
is
the
open
interval
(1
−
2,
1
+
2). There is no
√
√
max or min, but sup = 1 + 2, inf = 1 − 2. The third set equals [−2, ∞). No max
or sup, but inf=min=−2.
Question 6.3 There are several ways of proving the result. It is given that A is
bounded, so σA = sup A exists and, a ≤ σA for all a ∈ A. So, ∀a, 2a + 3 ≤ 2σa + 3.
So B is bounded above by 2σA + 3. Since B is bounded above, σB = sup B exists
and, since 2σA + 3 is an upper bound, we must have
σB ≤ 2σA + 3.
We need to show =, rather than ≤. We could now proceed to prove it by contradiction,
supposing that σB < 2σA + 3 and showing that, as a consequence of this supposition,
0
σA
= (σB − 3)/2 is an upper bound on A, which is less than σA —a contradiction.
Almost equivalently, we could simply observe that
A = {(b − 3)/2 : b ∈ B}
so, arguing as at the beginning,
σA ≤ (σB − 3)/2;
that is,
2σA + 3 ≤ σB .
Either way, we get what we need to complete the proof.
Question 6.4 Taking = L/2 > 0 in the definition of a limit, there is N so that
for n > N , |xn − L| < L/2. In particular, for n > N , xn − L > −L/2 and so
xn > L/2 > 0.
Question 6.5
4n2 + 9
4 |an − 4/3| = 2
−
3n + 7n + 11 3 12n2 + 27 − 12n2 − 28n − 44 = 3(3n2 + 7n + 11)
110
28n + 17
= 2
9n + 21n + 33 ≤
=
45n
9n2
5
.
n
This is < for n > N = 5/. [There is no need to solve a horrible quadratic for n:
make the approximation suggested and life is simple.]
Question 6.6 Suppose the sequence is (xn ) and let τ = inf{xn : n ∈ N} (which
exists, by boundedness). Let > 0. Since τ + cannot be a lower bound on the set
of xn , there’s N so that xN < τ + . But the sequence is decreasing, so for n > N ,
τ ≤ xn ≤ xN < τ + and hence |xn − τ | < . This shows the sequence converges to
τ.
Question 6.7 We show x can’t be negative. Suppose it was. Then −x/2 > 0, so
taking = −x/2 in the definition of limit, there is N so that for n > N ,
|xn − x| < −x/2, i.e., 3x/2 < xn < x/2.
But xn < x/2 says xn < 0 for n > N —a contradiction.
Question 6.8 We use the heredity results (that is, the algebra of limits).
4n − 5
1 − 5/4n
1
4n − 5
= n
=
→ = 1.
2n
2 −7
4 −7
1 − 7/4n
1
5(32n − 1
5 − 1/9n
5
=
→ .
4(9n ) + 7
4 + 7/9n
4
For the next one, the Sandwich theorem is no use (even though you might think so:
try it!). Instead:
1 + 2 + ··· + n
n(n + 1)/2
n2 /2 + n/2
1 + 1/(2n)
1
=
=
=
→ = 1.
n2
n2
n2
1
1
Finally,
n2 − 1
1/n − 1/n3
0
=
→ = 0.
n3 + 1
1 + 1/n3
1
Question 6.9 We have, for n ≥ 2,
√
√
x2n−1 + k
x2n−1 + ( k)2
2xn−1 k √
xn =
=
≥
= k.
2xn−1
2xn−1
2xn−1
Also, for n ≥ 1,
x2n + k
k − x2n
− xn =
≤ 0,
2xn
2xn
√
where we have used√the fact that xn ≥ k. So xn+1 ≤ xn . So the sequence is
bounded below (by k) and it is decreasing. These two facts mean that it converges.
Suppose xn → L. Then xn+1 → L too. But
xn+1 − xn =
xn+1 =
x2n + k
L2 + k
→
,
2xn
2L
and so L = (2L2 + k)/2L,
or L2 = k. Since xn is positive for all n, L ≥ 0, so L =
√
and we have xn → k.
111
√
k,
Question 6.10 First,
2n3 + 1
2 + 1/n3
2
=
→ .
3n3 + n + 2
3 + 1/n2 + 2/n3
3
For the second, although it looks like it’s worth trying, the Sandwich Theorem doesn’t
help much. Instead, we use
n + (n + 1) + · · · + 2n = (n + 1)(n + 2n)/2 = 3n2 /2 + 3n/2,
so
3n2 /2 + 3n/2
3/2 + 3/(2n)
=
→ 3/2.
n2
1
(Straightforward Sandwich theorem application only tells us that xn lies between 1
and 2.)
xn =
Question 6.11 Let xn denote the quantity we need to find the limit of. Then
n× √
1
n2
+n
≤ xn ≤ n × √
1
n2
+1
.
Now,
√
and
√
n
1
1
=p
→ = 1,
1
+n
1 + 1/n
n2
1
n
1
→ = 1.
=p
2
1
n2 + 1
1 + 1/n
By the Sandwich theorem, xn → 1.
Question 6.12 Let K > 0. We want to find δ so that if 0 < |x − 2| < δ then
f (x) > K. If |x − 2| < δ then
5
5
> 2,
2
(x − 2)
δ
p
so it suffices to have 5/δ 2 ≥ K, so we may take δ = 5/K.
f (x) =
Question 6.13
x3 + 5x + 7
13 + 5(1) + 7
13
=
=
.
x→1 x4 + 6x2 + 8
14 + 6(1)2 + 8
15
lim
Question 6.14 If 1 < x < 2 then x − bxc = x − 1, so
lim (x − bxc) = lim (x − 1) = 0.
x→1+
x→1+
If 0 < x < 1 then x − bxc = x − 0 = x, so
lim (x − bxc) = lim (x) = 1.
x→1−
x→1−
Question 6.15 Taking = f (c)/2 > 0 in the formal definition of continuity at c,
there is some δ such that if x ∈ (c − δ, c + δ) then |f (x) − f (c)| < f (c)/2. Thus, for
x ∈ (c − δ, c + δ), f (x) > f (c) − f (c)/2 = f (c)/2 > 0, so f is positive on that interval.
112
Question 6.16 Let y ∈ (0, ∞). We show that there is some c ∈ R such that f (c) = y.
This shows that the range is the whole of (0, ∞). (The fact that it is no larger follows
from the given fact that f is positive.) Now, f (R) = inf{f (x) : x ∈ R} = 0, so, since
y > 0, there must be some y1 ∈ f (R) with y1 < y. This means there is some x1 ∈ R
such that y1 = f (x1 ) < y. Similarly, because f is unbounded, which means f (R)
is unbounded, there must be some y2 ∈ f (R) with y2 > y, and there will be some
x2 ∈ R such that y2 = f (x2 ) > y. Then y lies between f (x1 ) and f (x2 ) and, since
f is continuous, the Intermediate Value Theorem shows that there is some c between
x1 and x2 with f (c) = y.
Question 6.17 The function h satisfies h(a) = f (a) − a ≥ 0, since f (a) ∈ [a, b]
implies f (a) ≥ a. Similarly, h(b) ≤ 0. So a < b and h(a) ≥ 0 ≥ h(b). Since f
is continuous, as is the function x 7→ x, it follows that h is continuous. By the
Intermediate Value Theorem, ∃c ∈ [a, b] with h(c) = 0; that is, f (c) = c.
Question 6.18 Let be such that 0 < < f (x∗ ). Such an can be chosen since
f (x∗ ) > 0. Because f is continuous at x∗ , there is δ > 0 such that
f (Bδ (x∗ )) ⊆ B (f (x∗ )) = (f (x∗ ) − , f (x∗ ) + ).
So, for x ∈ B = Bδ (x∗ ), we have f (x) > f (x∗ ) − > 0.
Question 6.19 Let > 0. Then there is δ > 0 such that if kx − ak < δ then
|f (x) − f (a)| < . If |t − a1 | < δ then
k(t, a2 , . . . , am ) − ak = k(t, a2 , . . . , am ) − (a1 , a2 , . . . , am )k
p
(t − a1 )2 + 0 + · · · + 0
=
= |t − a1 | < δ,
so we have
|f (t, a2 , . . . , am ) − f (a1 , a2 , . . . , am )| < ,
which says precisely that |g(t) − g(a1 )| < . This proves that g is continuous at a1 .
Question 6.20 Suppose xn → x and let > 0 be given. Then there is N such that
for n > N , kxn − xk < . But
v
um
uX
kxn − xk = t (xin − xi )2 ≥ |xin − xi |,
i=1
for any i between 1 and m, so for n > N , |xin − xi | < and hence xin → xi . On the
other hand, if, for each i, |xin − xi | < α, then
v
um
uX
kxn − xk = t (xin − xi )2
√
i=1
mα2
<
√
=
mα,
√
so if we let α = / m, we have:
√
|xin − xi | < / m, (i = 1, 2, . . . , m) =⇒ kxn − xk < .
√
If xin → xi for each i, we may take / m in the definition of limit√(in place of )
to see that there is some Ni such that for n > Ni , |xin − x√
i | < / m. Let N be
the largest of N1 , . . . , Nm . Then for n > N , |xin − xi | < / m for all i and hence
kxn − xk < . This shows that xn → x.
113
Question 6.21 There are a number of ways of proving this. We show that the
complement of B is open. Suppose that y ∈ Rm \ B. This complement is the set of
y such that ky − zk > . To prove that Rm \ B is open, we want to show that some
open ball Bδ (y) around y lies entirely in the complement of B. Leave δ undetermined
for now: we’ll choose it later. Let x ∈ Bδ (y), so that kx − yk < δ. Then
ky − zk ≤ ky − xk + kx − zk < δ + kx − zk,
so
kx − zk > ky − zk − δ.
Now, we know that ky − zk > and we want to choose δ so that kx − zk > (to show
that x 6∈ B). Because ky − zk is strictly bigger than , it is possible to find δ > 0
such that ky − zk − δ > . (Clearly we just need 0 < δ < ky − zk − .) With such a
δ, we see that Bδ (y) ⊆ Rm \ B and we are done.
Question 6.22 This is straightforward if we use the fact (indeed, the definition) that
a set is compact if and only if it is closed and bounded. Clearly the intersection of
any collection of bounded sets is bounded (convince yourself!). To show that the
intersection of any set of closed sets is closed can either be shown by considering their
open complements or (perhaps more easily) by simply noting that any convergent
sequence whose terms are in the intersection must (by closure of the sets) have a limit
which lies in each of the sets and hence in the intersection. The details are left to
you.
114