1 Set Theory and Functions

1.1 Basic De�nitions and Notation

� A set A is a collection of objects of any kind.

� We write a 2 A to indicate that a is an element of A:We express this as �a is contained

in A�.

� We write A � B if every element of A also is an element in B . We express this as �A

is a subset of B�

� We write A = B if A and B consist of exactly the same elements. Otherwise we write

A 6= B:

� If A � B and A 6= B we say that A is a proper subset of B:

� We write ? for the empty set, the set that contains no elements at all.

1.2 Operations on Sets

1. The union: Let A and B be two sets. Then, the union, denoted A [ B is the set of

all elements belonging to at least one of the two sets. That is,

A [B = fxjx 2 A or x 2 Bg :

Given an arbitrary collection of sets Ai; where i is some parameter such that i 2 I

(allows for �nite or in�nite numbers) we write [iAi for the union, which is the set of

all elements that belongs to at least one of the sets in the collection,

[iAi = fxj x 2 Ai for some i 2 I g :

2. The Intersection: Given A and B we write A \ B for its intersection- the set of all

elements belonging to both A and B:

A \B = fxjx 2 A and x 2 Bg :

1

The generalization for a collection of sets Ai is obviously

\iAi = fxj x 2 Ai for every i 2 I g

3. Disjoint Sets: A and B are said to be disjoint if they have no elements in common.

That is, if

A \B = ?:

4. The Complement (sometimes called di¤erence): We (KF uses di¤erent notation)

write AnB for the set of all elements in A that don�t belong to B;

AnB = fxjx 2 A and x =2 Bg

Remark 1 It follows immediately from the de�nitions that [ and \ satisfy communicative

and associative laws,

A [B = B [ A

A \B = B \ A

(A [B) [ C = A [ (B [ C)

(A \B) \ C = A \ (B \ C)

Hence, we can simply write A [B [ C:

In addition:

Proposition 1 [ and \ satisfy the distributive law

(A [B) \ C = (A \ C) [ (B \ C)

(A \B) [ C = (A [ C) \ (B [ C)

Proof. Exercise on Problem Set 1

Sometimes we will also use the Cartesian product:

De�nition 1 Given non empty sets A and B; the Cartesian product, denoted A�B; is the

set of all ordered pairs (a; b) such that a 2 A and b 2 B

2

Example 1 If A = fa1; a2; a3; a4g and B = fb1; b2g then

A�B = f(a1; b1) ; (a1; b2) ; (a2; b1) ; (a2; b2) ; (a3; b1) ; (a3; b2) ; (a4; b1) ; (a4; b2)g

Example 2 If A = [0; 1] and B = [0; 1] we write A� B = [0; 1]� [0; 1] or A� B = [0; 1]2 ;

where geometrically the Cartesian product of the two (unit) intervals is the (unit) square.

1.3 Functions

The terminology I�ll use is as follows:

De�nition 2 A function (or mapping) from a set X to Y (denoted f : X ! Y ) is a rule

that assigns a unique element y = f (x) 2 Y to every x 2 X:

Usually, X is referred to as the domain of f:

Remark 2 Sometimes functions are de�ned in terms of the graph G, which is a collection

of ordered pairs hx; f (x)i 2 X � Y:We say that G is the graph of a function if and only if

for every x 2 X there is a unique pair hx; f (x)i 2 G.

We call the element f (x) 2 Y the image of x (under f). This is generalized to subsets

as follows:

De�nition 3 If A � X we say that

f (A) � fy 2 Y jy 2 f (a) for some a 2 Ag

is the (direct) image of A:

Also,

De�nition 4 If B � Y we say that

f�1 (B) = fx 2 Xjf (x) 2 Bg

is the inverse image (or preimage) of B:

3

The image and the inverse image can be used to de�ne some important types of functions:

De�nition 5 f : X ! Y is said to be surjective (onto) if f (X) = Y

De�nition 6 f : X ! Y is said to be injective (one to one) if f (x1) 6= f (x2) for every pair

(x1; x2) 2 X such that x1 6= x2:

De�nition 7 f : X ! Y is said to be bijective if it is surjective and injective.

It is more or less immediate that:

Proposition 2 If f is a bijection, then there is a (uniquely de�ned) function g : Y ! X

de�ned by letting g (y) 2 X be given by g (y) = f�1 (y) for every y 2 Y: This function is

called the inverse of f (since no confusion can arise we will use f�1 also to denote the inverse

function in the rest of the course).

In some sense the conclusion is trivial. However, we�ll write down a proof to make sure

we understand what the de�nitions mean

Proof. Since f is injective there are two possibilities:

1. f�1 (y) = ?

2. there exists a unique element x 2 X such that fxg = f�1 (y)

To see this, assume that for contradiction that there exists two distinct elements x1 and

x2 such that x1 2 f�1 (y) and x2 2 f�1 (y) : Then,

fx1; x2g � f�1 (y) = fx 2 Xjf (x) 2 fygg = fx 2 Xjf (x) = yg

which contradicts the assumption that f is injective. Moreover, since f is surjective

Y = f (X) � fy 2 Y jy 2 f (x) for some x 2 Xg ;

which can be rephrased as saying that there exists some x 2 X such that f (x) = y for every

y 2 Y: This rules out f�1 (y) = ?; so the result follows

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De�nition 8 If f : X ! Y and g : Y ! Z the composition f � g is given by the function

h : X ! Y such that h (x) = g (f (x)) for every x 2 X:

Proposition 3 Suppose that f : X ! Y and g : Y ! Z are bijective, then the composition

f � g is bijective.

Proof. Fix any z0 2 Z: Since g : Y ! Z is surjective there exists y0 2 Y such that g (y0) = z0:

Since f is surjective there exists x0 2 X such that f (x0) = y0: Hence,

g (f (x0)) = g (y0) = z0

Since z0 2 Z was arbitrary g (f (X)) = Z:

Suppose that f � g is not injective. Then there exists x1 6= x2 such that g (f (x1)) =

g (f (x2)) : Since f is injective f (x1) = y1 6= y2 = f (x2) : Hence, there exists y1 6= y2 such

that g (y1) = g (y2) contradicting that g is an injection.

1.4 Sequences I

Just like functions can be de�ned in terms of ordered pairs (de�ning the function from its

graph) we may de�ne an ordered pair as a function

f : f1; 2g ! X:

Similarly, a �nite sequence can be de�ned as a mapping

f : f1; ::::; ng ! X:

Now, lettingN be the set of natural numbers we can de�ne: an in�nite sequence as a mapping

f : N ! X:

It is customary to depart somewhat from standard functional notation when dealing with

sequences and,

1. Write hxiini=1 instead of (f (1) ; f (2) ; :::; f (n)) or hf (i)ini=1

2. Write hxii1i=1 instead of hf (i)i1i=1

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1.5 Finite, In�nite, Countable and Uncountable Sets

1.5.1 Finite vs In�nite Sets

A �nite set is a set which is either empty or a set for which there exists some natural number

n such that the set has exactly n elements. That is:

De�nition 9 The set A is �nite if it is empty or if there exists some n 2 N and a bijection

f : f1; :::; ng ! A (this de�nes a set with exactly n elements too): A set A is in�nite if it is

not �nite.

Remark 3 Since the bijection f : f1; :::; ng ! A de�nes an inverse f�1 : A ! f1; :::; ng

�niteness could be de�ned as a bijection from A onto f1; :::; ng : Both conventions are used

to de�ne �niteness (and similarly for countability).

The remark above also proves part of the following useful claim:

Proposition 4 A set A is �nite if and only if there exists a �nite set B and a bijection

g : A! B

Proof. (=)) If A is �nite there exists some �nite n and a bijection f : f1; :::; ng ! A;

implying that g = f�1 : A! f1; :::; ng is well-de�ned. Letting B = f1; :::; ng proves the �rst

part.

((=) Suppose that there exists �nite set B and a bijection g : A! B: Since B is �nite

there exists some n 2 N and (using the remark) a bijection h : B ! f1; :::; ng : Since g and h

are bijective g � h : A! f1; ::::; ng is bijective. Hence, there is a bijection f : f1; :::; ng ! A

where

f (i) = g � h(i) for i = 1; ::::; n

We now prove a result sometimes used in combinatorics, which is referred to as the

�Pigeonhole principle�. If there are more pigeons than pigeonholes at least two pigeons need

to share a hole.

6

Proposition 5 Suppose that m;n 2 N: Then:

1. If m � n there exists an injection (onto map) f : f1; :::;mg ! f1; ::; ng

2. If m > n there is no injection f : f1; :::;mg ! f1; ::; ng :

Proof. For part 1, let f : f1; :::;mg ! f1; ::; ng be given by

f (j) = j for every j 2 f1; :::;mg � f1; ::; ng :

Obviously, f is injective

For the part 2, we begin by noting that if n = 1; then the only map from f : f1; :::;mg !

f1; ::; ng is f (j) = 1 for every j 2 f1; :::;mg ; which is not an injection for any m > 1.

Next, assume that there is no injection f : f1; :::;mg ! f1; ::; kg for any m > k:We need

to demonstrate that it follows that there is no injection f : f1; :::;mg ! f1; ::; k + 1g for any

m > k + 1; which will prove the claim by induction.

For contradiction, suppose such an injection would exist. Then, there must be some

j 2 f1; :::;mg such that f (j) = k+1 since otherwise h : f1; :::;mg ! f1; ::; kg is an injection

despite m > k + 1 > k; which violates the induction hypothesis. Moreover, it must be the

case that exactly one element j� 2 f1; :::;mg is such that f (j) = k+1 since obviously there

is no injection more than one element is mapped into k + 1 : It follows that the function

eh : f1; :::;mg n fj�g ! f1; ::; kg

de�ned as eh (j) = h (j) for every j 2 f1; :::;mg n fj�gmust be an injection. At this point we may either just assert that f1; :::;mg n fj�g and

f1; :::;m� 1g are equivalent sets or relabel by constructing the map g : f1; :::;m� 1g !

f1; ::; kg

g (j) =

8<: eh (j) = h (j) if j < j�eh (j � 1) = h (j � 1) if j > j�:

7

It is obvious that g is injective if and only if eh is injective, but g is not injective accordingto the induction hypothesis. The result follows

We can extend this result to

Proposition 6 Suppose that m 2 N: Then:

1. There exists an injection (onto map) f : f1; :::;mg ! N

2. There is no injection f : N ! f1; ::;mg

Proof. For part 1, let f : f1; :::;mg ! N be given by

f (j) = j for every j 2 f1; :::;mg � N:

Obviously, f is injective

For part 2, if f : N ! f1; ::;mg is an injection it follows that f (j) 6= f (i) for every

pair i; j 2 N; which implies that f (j) 6= f (i) for every pair i; j 2 f1; :::;m+ 1g ; which

contradicts the previous proposition

I will take the following for granted (see Bartle and Sherbert pages 61-62)

1.5.2 Countable vs. Uncountable Sets

Countably in�nite sets can be de�ned in similar fashion to the de�nition of �nite sets in

De�nition 9. However, it is more convenient to use the following de�nition:

De�nition 10 A set A is said to be countable if there exists an injective function f : A!

N:

Bartle and Sherbert use a di¤erent de�nition and establish existence of an injection

f : A! N as a proposition.

Example 3 Let A = N: Consider the identity function f : N ! N given by

f (i) = i for every i 2 N:

Obviously, this is injective, so N is countable.

8

Example 4 Let Z = f:::� 1; 0; 1; :::g be the set of positive and negative integers. Let f :

Z ! N be given by

f (i) =

8<: �2i if i < 0

2i+ 1 if i � 0;

which clearly is injective (as negatives map into even and positives map into odd numbers).

Example 5 Let A be �nite. Then there is some n 2 N and a bijection f : A ! f1; :::; ng :

De�ning f1 : A! N as

f1 (a) = f (a)

for each a 2 A is obvious that f1 is injective (but not surjective as there exists no a 2 A such

that f1 (a) = j when j > n). Hence, any �nite set is countable

Combining with previous results we conclude that:

Proposition 7 Let A be a countable set. Then, A is (countably) in�nite if and only if there

exists some bijective f : A! N:

Proof. ()) For contradiction, suppose f is surjective and that A is �nite. If f is bijective

f�1 : N ! A exists and is injective (and surjective). If A is �nite there exists n 2 N and

bijection g : A ! f1; :::; ng : Hence f�1 � g : N ! f1; :::; ng is injective (and surjective),

which contradicts Proposition 6. Hence, A cannot be �nite, implying that it is in�nite

(()Suppose that A is �nite and that there exists a bijection f : A! N: Then

f�1 : N ! A

is bijective. Moreover, since A is �nite there exists a bijection g : A ! f1; :::mg ; where

m 2 N: Hence,

f�1 � g : N ! f1; :::; ng

is bijective, which is impossible due to Proposition 6.

Theorem 1 If A is countable and B � A; then B is countable (if A is �nite, then B is

�nite).

9

Proof. First, consider the case with A �nite. Without loss, suppose B is non-empty (empty

set if �nite and therefore countable by default). If B is in�nite there exists bijection

f : B ! N

Let b0 2 B � A and consider map g : A :! B

g (a) =

8<: a if a 2 B

b0 if a =2 B:

As this map restricted on B is the identity map, this is obviously surjective. Moreover, since

A is �nite there exists n and bijective map h : f1; :::; ng ! A. Hence

h � g � f : f1; :::; ng ! N

is surjective, which which is impossible due to Proposition 6.

Now, suppose A is countably in�nite so that there exists bijection f : A ! N: Let

g : B ! A be given by

g (b) = b

for every b 2 B: This is obviously injective, so g � f : B ! N is injective.

Theorem 2 The union of a countable number of countable sets is countable.

Proof. Proof is very similar to proof of countability of rationals.

Without loss, suppose that fAigi is a collection of disjoint sets (otherwise look at

A1; A1nA2; A3n (A1 [ A2) :::::).

. Write faijgj for the elements. Let f : [iAi ! N be given by

f (aij) = 2i3fi(aij):

Which is well de�ned and injective.

[If not there is some aij 6= akl with

2i3fi(aij) = 2k3fk(akl) ,

2i�k = 3fk(akl)�fi(aij)

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LHS values

::::1

16;1

8;1

4;1

2; 1; 2; 4; 8; 16::::

RHS values1

81;1

27;1

9;1

3; 1; 3; 9; 27; 81; :::

Theorem 3 The set of rationals is countable.

Proof is very similar.

Theorem 4 The set of real numbers in the closed interval [0; 1] is uncountable.

Proof. If [0; 1] is countable we could �nd an injection f : N ! [0; 1] : Let fxigi2N denote

the implied sequence of elements in [0; 1] and write them in decimal form as

x1 = f (1) = 0:x11x12::::::x1n::::::::

x2 = f (2) = 0:x21x22::::::x2n::::::::

::::

xn = f (n) = 0:xn1xn2::::::xnn::::::::

:::::

Consider

y = 0:y1y2::::yn::::::

where every yi =2 f0; xii; 9g [ruling out 0 and 9 is to avoid dealing with 0:5 versus 0:499999::::]

Hence, y does not belong to fxigi2N : Since the sequence was arbitrary, no sequence contains

all the reals, which completes the proof.

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2 Sequences and Limits in R

2.1 Suprema and In�ma

We will not be careful about the axioms of the real number system, but rather proceed based

on our intuitive understanding of ordering properties and arithmetic. However, we will be

explicit about ONE crucial property, which relates to a more general completeness property

for metric spaces.

De�nition 11 Given a nonempty set A � R we say that u is an upper bound for A if x � u

for every x 2 A:

De�nition 12 If an upper bound for A exists we say that A is bounded above.

Lower bounds/bounded below are de�ned the same way.

De�nition 13 Given a nonempty set A � R we say that u is a least upper bound for A if:

1. u is an upper bound for A

2. x � u for any upper bound x for A

Example 6 The set of reals, R; has no upper or lower bound.

Example 7 Any x � 0 is an upper bound for (�1; 0j : The unique least upper bound for

(�1; 0j is 0:

Example 8 Any x � 0 is an upper bound for (�1; 0) : The unique least upper bound for

(�1; 0j is 0:

Proposition 8 If a least upper bound to A exists, then it is unique.

Proof. Suppose that u and u0 are both least upper bounds. Then, u � u0 since otherwise u

wouldn�t be a least upper bound and u0 � u for the same reason. Hence u = u0

In what follows we will call the least upper bound to a subset of R the supremum of

the set. The most crucial assumption about the reals is:

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Axiom 1 (Supremum property of R) For any non empty subset A of R which is bounded

above the supremum exists.

Idea is that there are no gaps among the set of reals.

To appreciate that the axiom is a bit more subtle than one way �rst think consider the

example:

Example 9 Let Q be the set of rational numbers and A = fx 2 Qjx2 < 2g : Intuitively we

then notice that

1. if u is an upper bound for A; then u2 = 2: (proof to follow below)

2. there is no rational number such that u2 = 2 (proof to follow below)

A consequence of the supremum property of the reals is that the in�num exists for sets

that are bounded below.

Proposition 9 If a nonempty subset of the reals A is bounded below, then it has a greatest

lower bound (in�num).

Proof. Problem set. Idea is that l is a lower bound of A if and only if �l is an upper bound

for

fx 2 Rj � x 2 Ag

We can then prove �Archimedes axiom�which can be thought of as either of the following

two statements:

(1) Given any number, you can always pick an integer that is larger than the original

number.

(2) Given any positive number, you can always pick an integer whose reciprocal is less

than the original number.

Proposition 10 The set of natural numbers N is not bounded above.

13

Proof. If N is bounded above, then u = supN exists by the completeness axiom. Let

n 2 N be picked arbitrary. We have that n � u as N is bounded above. But, n + 1 is also

bounded by u; so

n+ 1 � u, n � u� 1:

Since the argument holds for an arbitrary n 2 N it follows that u�1 is also an upper bound

for N; which contradicts the leastness of u:

Corollary 1 For any pair (x; y) 2 R such that x 6= y there exists a rational number in

between.

Proof. Without loss, assume that x < y: Suppose that x � 0: By Proposition 10 there

exists n such that

n >1

y � x;

implying that1

n< y � x

Moreover, there exists v 2 N such that (here we use fact that x � 0)

v � 1 � nx < v

since:

1. Set fm 2 N jm � nxg is nonempty (otherwise nx upper bound for N)

2. Set fm 2 N jm � nxg is bounded below by nx)exists a least element v 2 fm 2 N jm � nxg

x <v

n� x+ 1

n< x+ y � x = y:

Now, let x < 0: Suppose that y > 0: Then, the previous argument establishes that

there exists n; v 2 N such that

x < 0 <v

n< y:

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Finally, let x < y < 0: Then, the previous argument establishes that there exists n; v

such that

�y < v

n< �x;

implying that

x < �vn< y:

We can now get back to the example:

Example 10 (continued) CLAIM: if u is a least upper bound for A = fx 2 Qjx2 < 2g

then u2 = 2:

Proof. Obviously an upper bound exists (20 works...and is in Q).

CASE 1: Suppose the least upper bound is such that u2 < 2: By the Archimedan property

just proved we may pick n such that

0 <1

n<2� u24u

) 4u

n< 2� u2

and1

n< 2u;

But then (observing that we can trivially rule out the possibility that u is negative)

u2 <

�u+

1

n

�2= u2 +

2u

n+1

n2�1

n< 2u

�< u2 +

2u

n+2u

n= u2 +

4u

n�4u

n< 2� u2

�< u2 + 2� u2 = 2;

implying that u is not a lower bound.

CASE 2: Suppose that u2 > 2 . By the Archimedan property we can pick n such that

0 <1

n<u2 � 22u

)�u� 1

n

�2= u2 � 2u

n+1

n2> u2 � 2u

n> u2 �

�u2 � 2

�= 2;

implying that u� 1nis a smaller upper bound of A; contradicting the leastness of u:

15

Remark 4 CHECK: SUM OF RATIONAL S IS RATIONAL. HENCE THE EXAMPLE

SHOWS (ACCEPTING THATp2 IS IRRATIONAL) THAT THE SUPREMUM PROP-

ERTY DOES NOT APPLY TO THE SET OF RATIONAL NUMBERS.

Remark 5 By the supremum property of R the example also shows thatp2 exists in the

real number system.

2.2 Real Sequences

While formally a real sequence is a map f : N ! R; but we will usually write hx1; x2; :::; xi; ::::i

or hxni1n=1 when this cannot cause any confusion.

A crucial concept is that of convergence:

De�nition 14 hxii1i=1 converges to the (real number) x� if given any (real number) " > 0

there exists some �nite K (possibly dependent on ") such that

x� � " < xn < x� + "

for every n � K:

Example 11 Let hxni1n=1 be given by xn = n�1nfor every n 2 N: By the Archimedean

property of the naturals there exists K such that

0 <1

K< "

for every " > 0: Hence, if n � K

jxn � 1j =����n� 1n � 1

���� = �����1n���� = 1

n� 1

K:

We conclude that xn = n�1nconverges to 1:

Example 12 Sometimes limit not so easy to guess, as with

xn =

�1� 1

n

�n16

This can be motivated from probability theory: The chance of of k successes in n Bernoulli

trials is

pkjn =n!

k! (n� k)!pk (1� p)n�k

So, if the probability of success is 1n; then the probability of no success out of n draws is

n!

1� n!!

�1

n

�0�1� 1

n

�n=

�1� 1

n

�n:

Can show that�1� 1

n

�n ! 1e:

Proposition 11 If hxni1n=1 converges there is a unique limit.

Proof. Suppose that x� and x�� are limits of hxni1n=1 and let x� < x�� without loss of

generality. Let " = x���x�3: Because x� is a limit of hxni1n=1 there exists K� such that

xn < x� + " = x� +

x�� � x�3

=2

3x� +

1

3x��

for all n � K�: Symmetrically, as x�� is a limit of hxni1n=1 there exists K�� such that

xn > x�� � " = x�� � x

�� � x�3

=2

3x�� +

1

3x�

Hence,

2

3x�� +

1

3x� < xn <

2

3x� +

1

3x�� )

2

3(x�� � x�) <

1

3(x�� � x�) :

) 2 < 1:

De�nition 15 A sequence of real numbers hxni1n=1 is:

1. monotonic increasing if xn+1 � xn for every n 2 N

2. monotonic decreasing if xn+1 � xn for every n 2 N

3. monotonic if either monotonic increasing or decreasing

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De�nition 16 A sequence of real numbers hxni1n=1 is bounded if there exists a; b 2 R such

that a � xn � b for every n 2 N:

Proposition 12 Every bounded monotonic sequence is convergent.

Proof. Let X = fx1; x2:::; xn; ::::g � R be the set of elements of the sequence hxni1n=1

Remark 6 Formally, a real valued sequence is de�ned as a function f : N ! R; so X =

f (N) ; the direct image of N

Consider the case where the sequence is increasing. Then, X is bounded above by a: By

the completeness/supremum property of the real number system there exists a supremum of

X: Write x = supX:

Consider an arbitrary " > 0 and note that x� " can not be an upper bound of X; since

that would contradict that x is the LEAST upper bound. Hence, there exists K <1 such

that xK > x� "; which implies that

x � xn � xK > x� "

for every n � K: Hence

jxn � xj � "

for every n � K: Since " was arbitrary, x is the limit of hxni1n=1 :

Decreasing sequences-same argument.

2.3 Subsequences

De�nition 17 Let hxni1n=1 be a sequence and fkig1i=1 a strictly monotonic sequence with

ki 2 N for every i: Then, the sequence

hxknikn = hxk1 ; xk2 ; :::::; xkni

(constructed by setting xki = xn for every ki) is called a subsequence of hxni1n=1

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Example 13 Let hxni1n=1 be given by

xn =

8<: a if n is odd

b if n is even;

then two obvious subsequences are (a; a; :::; a; :::) and (b; b; :::; b; )

Theorem 5 (Bolzano & Weierstrass) Every bounded sequence of real numbers has a

convergent subsequence.

Proof. Let a0 and b0 be bounds for hxni1n=1 , so that

xn 2 [a0; b0]

for every n 2 N: Now let

L1 =

�a0;a0 + b02

�and let NL

1 =

�n 2 N jxn 2

�a0;a0 + b02

��H1 =

�a0 + b02

; b0

�and let NH

1 =

�n 2 N jxn 2

�a0 + b02

; b0

��There are now two possibilities:

1. NL1 is in�nite. If this is the case we let a1 = a0 and b1 = a0+b0

2and let N1 =�

n 2 N jxn 2�a0;

a0+b02

�2. NL

1 is �nite. If this is the case it follows that NH1 is in�nite. If this is the case we let

a1 =a0+b02

and b1 = b0, and let N1 = NH1 :

We conclude that there exists an interval [a1; b1] � [a0; b0] and an in�nite set N1 =

fn 2 N jxn 2 [a1; b1]g, where

b1 � a1 =1

2[b0 � a0]

We can then construct [a2; b2] ; [a3; b3] :::: and N2; N3::: by continuing to divide the intervals

in two halfs. We then have::

CLAIM: For every k 2 N there exists an interval [ak; bk] with

bk � ak =1

2k[b0 � a0]

19

and an in�nite set

N1 = fn 2 N jxn 2 [ak; bk]g

:Suppose claim is true for k 2 N: Let

Lk+1 =

�ak;ak + bk2

�and let NL

k+1 =

�n 2 N jxn 2

�ak;ak + bk2

��Hk+1 =

�ak + bk2

; bk

�and let NH

k+1 =

�n 2 N jxn 2

�ak + bk2

; bk

��Since

NLk+1 [NH

k+1 =

�n 2 N jxn 2

�ak;ak + bk2

��[�n 2 N jxn 2

�ak + b02

; bk

��= fn 2 N jxn 2 [ak; bk]g

is in�nite by the induction hypothesis we conclude that at either NLk+1 or N

Hk+1 is in�nite.

As

ak + bk2

� ak = bk �ak + bk2

=1

2(bk � ak) =

1

2k[b0 � a0]

by the induction hypothesis this proves the claim. Now:

1. ak 2 [a0; b0] for every k and ak � ak�1 for every k )limit a� exists (monotone bounded

convergence result)

2. bk 2 [a0; b0] for every k and bk � bk�1 for every k )limit b� exists (monotone bounded

convergence result)

3. ck = bk�ak 2 [0; b0 � a0] for every k and ck+1 = bk+1�ak+1 = 12(bk � ak) � (bk � ak) =

ck )limit c�exists (again)

4. c� = 0:

5. c� = b� � a�: Hence both fakg and fbkg has the same limit a� = b�:

20

Now, pick " > 0, let (very ine¢ cient choice)

k 2 inf

�n 2 N jn � 2 (b0 � a0)

"

�)

1

k� "

2 (b0 � a0)

And let

K (") = inf fn 2 N jxn 2 [ak; bk]g

Then for every n � K (") it follows that

xn � a� � bk � a� � bk � ak =1

2k[b0 � a0]

<1

k[b0 � a0] �

"

2 (b0 � a0)[b0 � a0] =

"

2;

which proves the result.

2.4 Open and Closed Sets in R

De�nition 18 Given any " > 0 we say that an open ball (" neighborhood) of x 2 R is given

by the set

B (x; ") = fy 2 Rj jx� yj < "g = (x� "; x+ ") :

De�nition 19 A set A � R is open if for every x 2 A there exists an open ball B (x; ") � A:

De�nition 20 A set A � R is closed if the complement B = RnA is open.

Example 14 Let A = R: Fix x and let " = 1: Then any y 2 (x� 1; x+ 1) 2 R; so R is

open.

Example 15 Consider the open interval A = (a; b) :Pick any x 2 (a; b) and let

" = min fx� a; b� xg

21

then for any y 2 B (x; ")

y > x� " = x�min fx� a; b� xg

� x+max fa� x; x� bg � x+ a� x = a

y < x+ " = x+min fx� a; b� xg

� x+ b� x = b:

Hence, (a; b) is an open set.

Example 16 Let A = (1; a) and B = (b;1) : Open, by same argument.

Example 17 Consider [a; b] : Not open since for every " > 0

B (a; ") \ [a; b] = (a� "; a]

in nonempty for every " > 0:

Note that

Rn [a; b] = (1; a) [ (b;1) :

By the previous argument, if x 2 (1; a) we can �nd a ball B (x; ") � (1; a) and if x is

in (b;1) we can also �nd a ball B (x; ") � (b;1) : Hence, there an open ball B (x; ") �

(1; a) [ (b;1) for every x 2 (1; a) [ (b;1) ; so (1; a) [ (b;1) : we conclude that [a; b] is

a closed set.

Example 18 ? is open (vacuously since there is no point in the set). Hence, R is closed

and open. Since R is also open, it follows that ? is open and closed.

Theorem 6 The union of any collection of open sets is open.

Proof. Let I be some indexing set and assume that Ai is open for every i 2 I: If x 2 A =

[i2IAi: Then there exists j 2 I and " > 0 such that B (x; ") � Aj: Hence,

B (x; ") � Aj � [i2IAi:

22

Theorem 7 The intersection of a �nite collection of open sets is open.

Proof. Let x 2 \ki=1Ai: Let "i > 0 be such that B (x; "i) � Ai; which exists since x 2

\ki=1Ai ) x 2 Ai for every i and each Ai is open. Let

" = min f"1; "2; ::; "kg > 0;

and note that

B (x; ") � B (x; "i) � Ai for every i

implying that

B (x; ") � \ki=1Ai:

Example 19 Let�0; 1

k

�and consider

\ki=1��1k; 1 +

1

k

�=

��1k; 1 +

1

k

�which is open for every �nite k: However

\1i=1��1k; 1 +

1

k

�= [0; 1] ;

which is not open.

Theorem 8 The intersection of an arbitrary collection of closed sets is closed.

Proof. Let I be some indexing set and assume that Ai is closed for every i 2 I: By de

Morgans laws

Rn (\i2IAi) = [i2IRnAi:

Each set RnAi is open, so (by previous theorem) [i2IRnAi is open, proving that Rn (\i2IAi)

is open, which by de�nition means that \i2IAi is closed.

Theorem 9 The union of a �nite collection of closed sets is closed.

23

Proof. By de Morgans laws

Rn�[ki=1Ai

�= \ki=1RnAi:

Each set RnAi is open and k is �nite, so (by previous theorem) \ki=1RnAi is open, proving

that Rn�[ki=1Ai

�is open, which by de�nition means that [ki=1Ai is closed.

De�nition 21 (ALT 1) Given set A � R; x 2 R is called a cluster point of A if for every

" > 0 there exists y 2 B (x; ") \ An fxg.

Proposition 13 A set A is closed if and only if it contains all its cluster points.

Proof. ()) Suppose not. Then, there exists an cluster point x 2 RnA; which is an open

set. Hence, there exists " > 0 such that B (x; ") � RnA, implying that B (x; ") \ A = ?;

contradicting that x is a cluster point of A:

(() Suppose that A contains all cluster points. Pick y 2 RnA: Then, y is not a cluster

point, so there exists open ball B (y; ") � RnA:True for all y 2 RnA; so RnA is open. Hence,

A is closed.

Almost the same result stated in terms of sequences.

Proposition 14 A is closed if and only if the limit of every converging sequence hxni with

xn 2 A for all n belongs to A:

Proof. We will prove this from the previous result by showing that

CLAIM: A contains all its cluster points,the limit of every converging sequence hxni

with xn 2 A for all n belongs to A:

()) Suppose that hxni converges to x�:Then, for every " > 0 there exists K such that

xn 2 B (x�; ") for n � K; and, by hypothesis, xn 2 A: Hence, x� is a cluster point of A: By

assumption A contains its cluster points, so limn!1 xn = x� 2 A:

Suppose that x� =2 A is a cluster point. Let "1 > 0 and pick x1 2 B (x�; "1) n fx�g

arbitrarily. As x1 6= x� there exists "2 > 0 such that x1 =2 B (x�; "1) ; implying that there

exists x2 2 B (x�; "2) � B (x�; "1) because x� is a cluster point: Recursively, if xk 6= x�

24

and xk 2 B (x�; "k) there exists "k+1 > 0 such that xk =2 B (x�; "1) : Hence, there exists

xk+1 2 B (x�; "k+1) � B (x�; "1) : This de�nes an in�nite sequence which obviously converges

to x�; so x� 2 A if the limit of every converging sequence is in A:

An alternative de�nition is.

Remark 7 The proof shows that given any set A � R; x 2 R is a cluster point of A if for

every " > 0 there exists an in�nite set X � B (x; ") \ A. This is an alternative de�nition

often used for a cluster point.

2.5 Compact Sets

De�nition 22 An open cover of a set A in R is a collection C = fCig of sets such that

A � [i2ICi

De�nition 23 C 0 is a subcover of C if each set Cj in C 0 also is in C and C 0 is a cover of A:

De�nition 24 A in R is said to be compact if every open cover C has a �nite open subcov-

ering C 0:

Theorem 10 (Heine Borel) A � R is compact if and only if A is closed and bounded.

Remark 8 Every analysis book says that result is "closely related to Bolzano Weierstrass",

yet proofs tend to not use that result. Here is an argument that makes explicit use of Bolzano

Weierstrass.

Proof. ()) Suppose that A is compact. We note that if Ci = (�i; i) then

[1i=1Ci

is an open cover of R: Hence,

A � R � [11 Ci

so f(�i; i) ji 2 Ngis an open cover of A as well. Since it is compact, there exists some �nite

K such that

A � [Ki=1Ci = (�K;K) ;

25

so A is bounded. Let y =2 A and let

Ci = Rn�y � 1

i; y +

1

i

�Obviously, Gi is open (

�y � 1

i; y + 1

i

�is closed) and we have that [1i=1Ci = R which covers

R and therefore also A: By compactness there is some �nite K such that

A � [Ki=1Ci = Rn�y � 1

K; y +

1

K

�)

A \�y � 1

K; y +

1

K

�= ?)

A \�y � 1

K; y +

1

K

�= ?

Hence, for every y =2 A , y 2 RnA there exists " > 0 such that�y � 1

K; y + 1

K

�� RnA; so

RnA is open. Hence, A is closed.

(() For contradiction, suppose A is closed and bounded, but that that there is an open

cover C = fC�g for which there is no �nite subcover.

For any n, consider the collection of open sets

Cn = fC�;ng where C�;n =�x 2 C�j inf

y2RnC�jx� yj > 1

n

�CASE 1: Suppose that there exists K such that CK is a subcover of C. Since A is bounded

by there exists m such that A � [�m;m] so let

x1 = �m and B�x1;

1

K

�=

��m� 1

K;�m+ 1

K

�x2 = �m+ 1

2Kand B

�x2;

1

K

�=

��m� 1

2K;�m+ 3

2K

�::::

xn = �m+ (n� 1)2K

and B�xn;

1

K

�=

��m+ (n� 3)

2K;�m+ (n+ 1)

2K

�::::

x4mK+1 = �m+ (4mK + 1� 1)2K

= �m+ 2m = m

Hence,�B�xn;

1K

�4mK+1n=1

is a �nite cover of A:Moreover, each xn 2 C�n;K by the hypothesis

that CK is a subcover of C. Pick any y 2 B�xn;

1K

�: Then,

infx2Rnxn2C�n;K

jx� yj � infxn2C�n;K

jx� xnj � jy � xnj > 0

26

since jy � xnj < 1Kas y is in B

�xn;

1K

�and infxn2C�n;K jx� xnj �

1Kby construction of

C�;K :

We conclude that B�xn;

1K

�� C�n;K : As

�B�xn;

1K

�4mK+1n=1

is an open cover it follows

that fC�n;Kg4mK+1n=1 is an open cover of A: Obviously, fC�n;Kg

4mK+1n=1 is a �nite subcover of

Cn = fC�;ng ; and by assumption Cn = fC�;ng is a subcover of C: Hence, A is compact.

CASE 2: Suppose A is not covered by Cn = fC�;ng for any n: Then there is a

sequence fxng with xn 2 A and xn 2 Rn[�C�n for every n 2 N: Let fxkng be a convergent

subsequence with limit x�: Then we note that:

1. Since A is closed x� 2 A

2. Since C is a covering of A there exists �� such that x� 2 C��

3. C�� is open so there exists " > 0 such that B (x�; ") � C��

4. Since there is no n such that x� 2 [�C�;n there is no n such that

x� 2 C��n =

�x 2 C��j inf

y2RnC��jx� yj > 1

n

�, there is no n such that B

�x�;

1

n

�� C��

[if B�x�; 1

n

�� C�� then jx� � yj � 1

n> 1

n+1for every y =2 C�� implying that x� 2

C��n+1]

5. Statements there exists " > 0 such that B (x�; ") � C�� and there exists no n such

that B�x�; 1

n

�� C�� contradict each other.

We conclude that there cannot be any convergent subsequence to fxng ; which (since

fxng is bounded) contradicts the Bolzano Weierstrass theorem. It follows that A must be

covered by Cn = fC�;ng for some n; which by analysis in CASE 1 means that A is compact.

2.6 Cauchy Convergence

An alternative way of thinking about convergent sequences is as a sequence with terms that

get closer and closer together further out in the sequence.

27

De�nition 25 hxni1n=1 is said to be a Cauchy sequence if given any " > 0 there exists K

such that

jxn � xmj < "

for every m;n � K:

The set of real numbers is a simple example of a complete metric space. Such spaces

have many important properties, one being that the two convergence criteria are equivalent

(which in the end gives a nice characterization of compactness).

The easy direction is:

Lemma 1 Every convergent sequence of real numbers hxni1n=1 is a Cauchy sequence.

Proof. Fix " > 0 and let hxni1n=1 have limit x�: As hxni1n=1 converges to x

� there exists

K�"3

�2 N such that

jxn � x�j <"

2

jxm � x�j <"

2

for every n;m � K�"2

�: Thus (triangle inequality "direct path shorter than any other path")

jxn � xmj = j(xn � x�) + (xm � x�)j

� j(xn � x�)j+ j(xm � x�)j

<"

2+"

2= "

To establish the other direction is a bit more work, but we may rely on some of the work

we have already done. In addition, the following fact is important.

Lemma 2 If hxni1n=1 is a Cauchy sequence, then there exists a; b such that a < xn < b for

every n:

28

Proof. Pick some " > 0 and let K be such that jxn � xmj < " for every n;m � K: Then

jxn � xK j < "

for every n; implying that

xn � xK < " and xK � xn < "

xK � " < xn < xK + "

for every n � K: The set

fx1; x2; :::; xK � "; xK + "g

is obviously bounded above and below and any bounds a; b to this set also bounds xK + ":

Theorem 11 hxni1n=1 converges if and only if it is a Cauchy sequence.

Proof. Let hxni1n=1 be a Cauchy sequence. By Lemma above it is then bounded above and

below. Applying the Bolzano-Weierstrass Theorem we know that there exists a convergent

subsequence hxknikn . Let the limit of the convergent subsequence be denoted by x�: Fix

" > 0: Since hxkni converges to x� there exists K1 2 fkig1i=1 such that

jxn � x�j <"

2

for every n � K1 such that there exists an element in fkig1i=1 with ki = n: This implies that

jxK1 � x�j <"

2:

But, since hxni1n=1 is Cauchy there exists K2 such that

jxn � xK1j <"

2

holds for every n � K2: Therefore

jxn � x�j = j(xn � xK1) + (xK1 � x�)j

� jxn � xK1j+ jxK1 � x�j < "

holds for every n � K2: We conclude that x� is the limit of hxni1n=1 :

29