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 [i Ai for the union, which is the set of
all elements that belongs to at least one of the sets in the collection,
[i Ai = 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
\i Ai = 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
set of all ordered pairs (a; b) such that a 2 A and b 2 B
2
B; is the
Example 1 If A = fa1 ; a2 ; a3 ; a4 g and B = fb1 ; b2 g 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
fy 2 Y jy 2 f (a) for some a 2 Ag
f (A)
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
fx1 ; x2 g
f
1
1
(y) : Then,
(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
4
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 z 0 2 Z: Since g : Y ! Z is surjective there exists y 0 2 Y such that g (y 0 ) = z 0 :
Since f is surjective there exists x0 2 X such that f (x0 ) = y 0 : Hence,
g (f (x0 )) = g (y 0 ) = z 0
Since z 0 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, letting N 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 hxi ini=1 instead of (f (1) ; f (2) ; :::; f (n)) or hf (i)ini=1
1
2. Write hxi i1
i=1 instead of hf (i)ii=1
5
1.5
1.5.1
Finite, In…nite, Countable and Uncountable Sets
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
de…ned as
e
h : f1; :::; mg n fj g ! f1; ::; kg
e
h (j) = h (j) for every j 2 f1; :::; mg n fj g
must 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
f1; ::; kg
g (j) =
8
<
:e
h (j
e
h (j) = h (j)
1) = h (j
7
if j < j
1) if j > j
:
1g !
It is obvious that g is injective if and only if e
h is injective, but g is not injective according
to 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
8
< a if a 2 B
g (a) =
:
: b if a 2
=B
0
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 fAi gi is a collection of disjoint sets (otherwise look at
A1 ; A1 nA2 ; A3 n (A1 [ A2 ) :::::).
. Write faij gj for the elements. Let f : [i Ai ! N be given by
f (aij ) = 2i 3fi (aij ) :
Which is well de…ned and injective.
[If not there is some aij 6= akl with
2i 3fi (aij ) = 2k 3fk (akl ) ,
2i
k
= 3fk (akl )
10
fi (aij )
LHS values
::::
1 1 1 1
; ; ; ; 1; 2; 4; 8; 16::::
16 8 4 2
RHS values
1 1 1 1
; ; ; ; 1; 3; 9; 27; 81; :::
81 27 9 3
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 fxi gi2N denote
the implied sequence of elements in [0; 1] and write them in decimal form as
x1 = f (1) = 0:x11 x12 ::::::x1n ::::::::
x2 = f (2) = 0:x21 x22 ::::::x2n ::::::::
::::
xn = f (n) = 0:xn1 xn2 ::::::xnn ::::::::
:::::
Consider
y = 0:y1 y2 ::::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 fxi gi2N : Since the sequence was arbitrary, no sequence contains
all the reals, which completes the proof.
11
Sequences and Limits in R
2
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
wouldn’t be a least upper bound and u0
u0 since otherwise u
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:
12
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 = sup N 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
1
<y
n
x
implying that
Moreover, there exists v 2 N such that (here we use fact that x
v
1
0)
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
x<
v
n
x+
1
<x+y
n
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<
14
v
< y:
n
nxg
Finally, let x < y < 0: Then, the previous argument establishes that there exists n; v
such that
y<
v
<
n
x;
implying that
x<
v
< y:
n
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
2 u2
4u
1
<
)
<2
n
4u
n
0 <
and
u2
1
< 2u;
n
But then (observing that we can trivially rule out the possibility that u is negative)
2
u
2
1
< 2u
n
4u
<2
n
u2
1
2u
1
<
u+
= u2 +
+ 2
n
n
n
2u 2u
4u
< u2 +
+
= u2 +
n
n
n
< 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<
u
implying that u
1
n
1
n
2
= u2
1
u2 2
<
)
n
2u
2u
1
+ 2 > u2
n
n
2u
> u2
n
u2
2 = 2;
is a smaller upper bound of A; contradicting the leastness of u:
15
Remark 4 CHECK: SUM OF RATIONAL S IS RATIONAL. HENCE THE EXAMPLE
p
SHOWS (ACCEPTING THAT 2 IS IRRATIONAL) THAT THE SUPREMUM PROPERTY DOES NOT APPLY TO THE SET OF RATIONAL NUMBERS.
Remark 5 By the supremum property of R the example also shows that
p
2 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 hxn i1
n=1 when this cannot cause any confusion.
A crucial concept is that of convergence:
De…nition 14 hxi i1
i=1 converges to the (real number) x if given any (real number) " > 0
there exists some …nite K (possibly dependent on ") such that
x
for every n
" < xn < x + "
K:
Example 11 Let hxn i1
n=1 be given by xn =
n 1
n
for every n 2 N: By the Archimedean
property of the naturals there exists K such that
0<
for every " > 0: Hence, if n
jxn
We conclude that xn =
n 1
n
1
<"
K
K
1j =
n
1
n
1 =
1
1
=
n
n
converges to 1:
Example 12 Sometimes limit not so easy to guess, as with
xn =
1
16
1
n
n
1
:
K
This can be motivated from probability theory: The chance of of k successes in n Bernoulli
trials is
pkjn =
n!
pk (1
k! (n k)!
p)n
k
So, if the probability of success is n1 ; then the probability of no success out of n draws is
n!
1 n!!
Can show that 1
1 n
n
1
n
0
n
1
n
1
=
1
1
n
n
:
! 1e :
Proposition 11 If hxn i1
n=1 converges there is a unique limit.
are limits of hxn i1
n=1 and let x < x
Proof. Suppose that x and x
generality. Let " =
x
x
3
: Because x is a limit of hxn i1
n=1 there exists K such that
xn < x + " = x +
for all n
without loss of
K : Symmetrically, as x
xn > x
x
x
3
2
1
= x + x
3
3
is a limit of hxn i1
n=1 there exists K
"=x
x
x
3
2
1
= x + x
3
3
Hence,
2
1
2
1
x + x < xn < x + x
3
3
3
3
2
1
(x
x) <
(x
x ):
3
3
) 2 < 1:
De…nition 15 A sequence of real numbers hxn i1
n=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
17
)
such that
De…nition 16 A sequence of real numbers hxn i1
n=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.
R be the set of elements of the sequence hxn i1
n=1
Proof. Let X = fx1 ; x2 :::; xn ; ::::g
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 = sup X:
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
for every n
for every n
xn
xK > x
jxn
xj
"
K: Hence
"
K: Since " was arbitrary, x is the limit of hxn i1
n=1 :
Decreasing sequences-same argument.
2.3
Subsequences
1
De…nition 17 Let hxn i1
n=1 be a sequence and fki gi=1 a strictly monotonic sequence with
ki 2 N for every i: Then, the sequence
hxkn ikn = hxk1 ; xk2 ; :::::; xkn i
(constructed by setting xki = xn for every ki ) is called a subsequence of hxn i1
n=1
18
Example 13 Let hxn i1
n=1 be given by
8
< a if n is odd
xn =
;
: 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 hxn i1
n=1 , so that
xn 2 [a0 ; b0 ]
for every n 2 N: Now let
L1 =
H1 =
a0 + b 0
a0 + b 0
and let N1L = n 2 N jxn 2 a0 ;
2
2
a0 + b 0
a0 + b 0
; b0 and let N1H = n 2 N jxn 2
; b0
2
2
a0 ;
There are now two possibilities:
1. N1L is in…nite. If this is the case we let a1 = a0 and b1 =
a0 +b0
2
and let N1 =
0
n 2 N jxn 2 a0 ; a0 +b
2
2. N1L is …nite. If this is the case it follows that N1H is in…nite. If this is the case we let
a1 =
a0 +b0
2
and b1 = b0 , and let N1 = N1H :
We conclude that there exists an interval [a1 ; b1 ]
fn 2 N jxn 2 [a1 ; b1 ]g, where
b1
a1 =
1
[b0
2
[a0 ; b0 ] and an in…nite set N1 =
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
[b0
2k
19
a0 ]
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 + b k
2
ak + b k
; bk
2
L
and let Nk+1
=
ak ;
Hk+1 =
H
and let Nk+1
=
ak + bk
2
ak + b k
n 2 N jxn 2
; bk
2
n 2 N jxn 2 ak ;
Since
ak + bk
2
= fn 2 N jxn 2 [ak ; bk ]g
L
H
Nk+1
[ Nk+1
=
n 2 N jxn 2 ak ;
[ n 2 N jxn 2
ak + b 0
; bk
2
L
H
is in…nite by the induction hypothesis we conclude that at either Nk+1
or Nk+1
is in…nite.
As
ak + b k
2
ak + b k
2
ak = b k
=
1
(bk
2
ak ) =
1
[b0
2k
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
bk
1
for every k )limit b exists (monotone bounded
convergence result)
2. bk 2 [a0 ; b0 ] for every k and bk
convergence result)
3. ck = bk ak 2 [0; b0
a0 ] for every k and ck+1 = bk+1 ak+1 = 21 (bk
ck )limit c exists (again)
4. c = 0:
5. c = b
a : Hence both fak g and fbk g has the same limit a = b :
20
ak )
(bk
ak ) =
Now, pick " > 0, let (very ine¢ cient choice)
k 2 inf n 2 N jn
1
k
2 (b0
a0 )
)
"
"
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
<
1
[b0
k
a
bk
ak =
1
[b0
2k
"
a0 ]
2 (b0
a0 )
[b0
a0 ]
"
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; ")
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
21
a; b
xg
A:
then for any y 2 B (x; ")
y > x
"=x
min fx
x + max fa
x; x
y < x + " = x + min fx
x+b
a; b
bg
a; b
xg
x+a
x=a
xg
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; ")
in (b; 1) we can also …nd a ball B (x; ")
(1; a) and if x is
(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 =
[i2I Ai : Then there exists j 2 I and " > 0 such that B (x; ")
B (x; ")
Aj
22
[i2I Ai :
Aj : Hence,
Theorem 7 The intersection of a …nite collection of open sets is open.
Proof. Let x 2 \ki=1 Ai : Let "i > 0 be such that B (x; "i )
Ai ; which exists since x 2
\ki=1 Ai ) x 2 Ai for every i and each Ai is open. Let
" = min f"1 ; "2 ; ::; "k g > 0;
and note that
B (x; ")
Ai for every i
B (x; "i )
implying that
\ki=1 Ai :
B (x; ")
Example 19 Let 0; k1 and consider
1
1
;1 +
k
k
\ki=1
=
1
1
;1 +
k
k
which is open for every …nite k: However
\1
i=1
1
1
;1 +
k
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 (\i2I Ai ) = [i2I RnAi :
Each set RnAi is open, so (by previous theorem) [i2I RnAi is open, proving that Rn (\i2I Ai )
is open, which by de…nition means that \i2I Ai is closed.
Theorem 9 The union of a …nite collection of closed sets is closed.
23
Proof. By de Morgans laws
Rn [ki=1 Ai = \ki=1 RnAi :
Each set RnAi is open and k is …nite, so (by previous theorem) \ki=1 RnAi is open, proving
that Rn [ki=1 Ai is open, which by de…nition means that [ki=1 Ai 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 hxn i 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 hxn i
with xn 2 A for all n belongs to A:
()) Suppose that hxn i 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
every " > 0 there exists an in…nite set X
R; x 2 R is a cluster point of A if for
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 = fCi g of sets such that
A
[i2I Ci
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 subcovering C 0 :
R is compact if and only if A is closed and bounded.
Theorem 10 (Heine Borel) A
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
[1
i=1 Ci
is an open cover of R: Hence,
A
R
[1
1 Ci
so f( i; i) ji 2 N gis an open cover of A as well. Since it is compact, there exists some …nite
K such that
A
[K
i=1 Ci = ( K; K) ;
25
so A is bounded. Let y 2
= A and let
1
1
;y +
i
i
Ci = Rn y
1
;y
i
Obviously, Gi is open ( y
+
1
i
is closed) and we have that [1
i=1 Ci = R which covers
R and therefore also A: By compactness there is some …nite K such that
1
1
;y +
K
K
1
1
;y +
K
K
A\ y
A\ y
1
1
;y +
)
K
K
[K
i=1 Ci = Rn y
A
= ?)
= ?
Hence, for every y 2
= A , y 2 RnA there exists " > 0 such that y
1
;y
K
+
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
;n g
where C
x 2 C j inf jx
=
;n
y2RnC
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 ;
x2 =
m+
1
K
=
m
1
1
and B x2 ;
2K
K
=
1
1
; m+
K
K
1
3
m
; m+
2K
2K
::::
xn =
m+
(n 1)
1
and B xn ;
2K
K
=
m+
(n 3)
(n + 1)
; m+
2K
2K
::::
x4mK+1 =
Hence, B xn ; K1
m+
(4mK + 1
2K
4mK+1
n=1
1)
=
m + 2m = m
is a …nite cover of A: Moreover, each xn 2 C
n ;K
that CK is a subcover of C. Pick any y 2 B xn ; K1 : Then,
inf
x2Rnxn 2C
n ;K
jx
yj
inf
xn 2C
n ;K
26
jx
xn j
jy
xn j > 0
by the hypothesis
since jy
C
xn j <
1
K
as y is in B xn ; K1
and inf xn 2C
n ;K
jx
1
K
xn j
by construction of
;K :
We conclude that B xn ; K1
C
n ;K
4mK+1
n=1
: As B xn ; K1
g4mK+1
is an open cover of A: Obviously, fC
n=1
that fC
n ;K
Cn = fC
;n g ;
and by assumption Cn = fC
CASE 2:
;n g
n ;K
g4mK+1
is a …nite subcover of
n=1
is a subcover of C: Hence, A is compact.
Suppose A is not covered by Cn = fC
sequence fxn g with xn 2 A and xn 2 Rn [ C
is an open cover it follows
n
;n g
for any n: Then there is a
for every n 2 N: Let fxkn g 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
3. C
such that x 2 C
is open so there exists " > 0 such that B (x ; ")
4. Since there is no n such that x 2 [ C
x
2 C
,
[if B x ; n1
C
n
there is no n such that
x2C j
inf
y2RnC
jx
yj >
there is no n such that B x ;
then jx
C
=
;n
C
yj
1
n
>
1
n+1
1
n
1
n
for every y 2
= C
C
implying that x 2
n+1 ]
5. Statements there exists " > 0 such that B (x ; ")
that B x ; n1
C
C
and there exists no n such
contradict each other.
We conclude that there cannot be any convergent subsequence to fxn g ; which (since
fxn g is bounded) contradicts the Bolzano Weierstrass theorem. It follows that A must be
covered by Cn = fC
2.6
;n g
for some n; which by analysis in CASE 1 means that A is compact.
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 hxn i1
n=1 is said to be a Cauchy sequence if given any " > 0 there exists K
such that
jxn
for every m; n
xm j < "
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 hxn i1
n=1 is a Cauchy sequence.
1
Proof. Fix " > 0 and let hxn i1
n=1 have limit x : As hxn in=1 converges to x there exists
K
"
3
2 N such that
jxn
jxm
for every n; m
K
"
2
"
2
"
xj <
2
xj <
: Thus (triangle inequality "direct path shorter than any other path")
jxn
xm j = j(xn
j(xn
<
x ) + (xm
x )j + j(xm
x )j
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 hxn i1
n=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
jxn
xm j < " for every n; m
K: Then
xK j < "
for every n; implying that
xK < " and xK
xn
xK
for every n
xn < "
" < xn < xK + "
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 hxn i1
n=1 converges if and only if it is a Cauchy sequence.
Proof. Let hxn i1
n=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 hxkn ikn . Let the limit of the convergent subsequence be denoted by x : Fix
" > 0: Since hxkn i converges to x there exists K1 2 fki g1
i=1 such that
jxn
for every n
x j<
"
2
K1 such that there exists an element in fki g1
i=1 with ki = n: This implies that
jxK1
"
x j< :
2
But, since hxn i1
n=1 is Cauchy there exists K2 such that
jxn
holds for every n
"
2
K2 : Therefore
jxn
x j = j(xn
jxn
holds for every n
xK1 j <
xK1 ) + (xK1
xK1 j + jxK1
x )j
x j<"
K2 : We conclude that x is the limit of hxn i1
n=1 :
29