OOK
AGENCY
same
time,
we
covered
abstract
manipulations
of
intuition
of
mathematics
 and
of
the
real
world),
often
in
a
very
satisfying
way.
By
the
time
they
were
assigned
the
To
he
expert
mathematician,
the
pace
of
this
book
may
seem
somewhat
slow,
especially
in
early
chapters,
as
there
is
a
heavy
emphasis
on
rigour
 except
the
text
and
can
beomitted.
transform,
continuous? differentiable?
\317\207lim xm.
m+l\342\200\224\321\216\320\276 m+1\342\200\224>oo
m+1\342\200\224>oo
But
if
m
oo,
then
m
m\342\200\224>oo
n\342\200\224>oo
and
thus
rrL
But
\320\230\321\202\320\266_,\320\276\320\276
in(rr),
make
the
change
of
variable
X\342\200\224\320\256\320\236
1/+7\320\223\342\200\224\320\256\320\236
\320\243\342\200\224\320\256\320\236 \321\203\342\200\224\320\256\320\236
Since
\320\230\321\202\321\205_\321\216\320\276
in(a:)
in(y)
we
thus
have
lim
sin(rr)
x\342\200\224\321\216\320\276
and hence
If
we
then
make
the
change
of
variables
\342\200\224
On
the
other
hand,
we
i=l
=
i=l
j=l
j=l
i=l
=
e~x dx
=
2/2
1\342\200\224\320\256\320\236
71\342\200\224\320\256\320\236\321\205_>1-
where
the
notation
2
oo,
we
should
obtain
1
cte
r^
dx
/-00
1
\321\203^^
dx
interchange
wo
partial
derivatives,
thus
one
expects
\320\2602/(\302\260>0)=<\320\271-(\302\260>0)\302\267
\320\264\321\205\320\264\321\203320\264\321\203\320\264\321\205
But
from
the
quotient
rule
we
have
df,
=
On
the
other
hand,
from
the
quotient
rule
again
320\2662
\321\2032
g x)
gf x)
\320\266\320\276>
\317\207\342\200\224\342\226\272\316\277
317\207
1
4x~3cos a:~4).
x\342\200\224>0 x\342\200\224\302\2730
The
first
limit
converges
to
zero
by
the
squeeze
est
  since
the
function
0,
and
by
the
squeeze
test
again.
oo,
Furthermore,
system,
which
of
course
is
an
extremely
convenient
way
to
manipulatenumbers,
as
you
might
think.
Why
\316\231,\316\231\316\231,\316\231\316\240
r
0++,
 0++)++,
  0-H-)++)++
  see
below)
created
we
have
yet
defined
what
a
is,
rest
of
this
chapter
we
and their
variable
can
now
become
false,
and
vice
versa.
So,
it
should have
the
same
as
zero,
even
if
it
hen
n-H-
0.5.
system
1/3
or
y/2
or
 None
n
a
system
number,
then
a
b++
Thus
than
one
of
the
statements
a
to
multiply
=
 
number
theory,
in
We
first
clarify
one
point:
we
nd
given
any
nonempty
set
A,
we
elements,
so
far).
The
next
larger
sets
=>
342\202\254
instead
of
 A
U
B)
U
and
so
forth.
(why?)
and
0
to
work
out
what
an
English
word
means.
A.
A and Au
 e
Associativity
=
\320\220\320\237 \320\222\320\237\320\241 .
 / Distributivity
 \316\247\\\316\221
nd
vice
versa.
 It
also
interchanges
X
take each
bject
z,
s
true
{8
between
0
and
5,
is
also
of
the
between
nd
5,
is
also
of
the
320\241
316\222
nd
of
talk
about
the
longer
allowed
primitive objects.
set,
then
theory)
that
the
universal
specification
axiom,
Axiom
3.8,
is
equivalent
to
an
axiom
postulating
the
existence
of
a
\ universal
set\
320\243rom one set
s
true,
namely
2 ,
but
.e.,
both
+2
and
explicit
inks
the
input
s,
a n .
Remark 3.3.6.
to ask whether
to
apply
=
\316\226,
\342\226\272
onto
function
depends
not
just
on
what
the
function
does,
but
also
/
if
=
s
in
particular
302\260
*-x->y
  b)
Show
that
if
  d)
Show
that
if
/
\316\260\316\263^\317\207\317\205\316\263
\320\264.
rom
a
set
X
We
call
{1,2,3}
(why?).
Example
3.4.6.
(Informal)
If
of all functions
e
sets.
Then
there
exists
a
set,
denoted
Yx,
which
consists
of
all
the
and
7
1;
the
function
that
maps
4
and
7
i->
0;
and
1
and
7
. The
reason
we
use
^a
by
first
choosing
some
element
316\221\316\261
nd
that
/^OW
\316\223\316\271(\317\205)\\\316\2234\316\275).
Exercise
3.4.5.
Let
a
a)
A^
A\ \302\267
ael
ael
gan s
set. This
{ 1,3 ,  1,4 ,
with
domain
\316\235,
i
Again,
this
definition
simply
postulates
Y[i<i<nXi
320\2603,
320\2543
316\2472
hen
we
have
{((\316\261\317\212,
320\254\320\267),
\320\260\320\267),(ai,
320\2602),320\2543),
320\2543)}
\320\2602,
\320\2542,320\2603)),
320\2603)),(\320\254\321\214\320\2602,
320\2543)),\320\254\321\214\320\2542,320\2603)),(\320\254\321\214\320\2542,
320\2543))}.
X\\ \320\245\320\2452)
321\205^\320\267,
X3,
and
so
forth;
we
will
s
just
X\\
 why?).
Also,
the
empty
Cartesian
product
ives,
singleton
set
other
words,
if
each
Xi
is
non-empty,
then
the
set
\316\240\316\271<\316\271<\316\267\316\247\316\257
s
also
non-empty.
Using
this
definition,
verify
that
{\320\243\320\263)\321\205<\320\263<\320\277
f and
317\207
e
non-empty
320\241
\320\241
317\207
\317\200\317\207\317\207\316\263->\317\207
1\321\205\320\243->1
and \317\200\317\207\317\207\316\275-y
I x
\320\243
e
the
maps
=
a: and
=
\317\200\317\207\317\207\316\263^\317\207
h
y_>y
the
direct
*s
emPty
if
^d
only
if
at
least
one
Exercise
3.5.10.
If
are
equal
if
and
only
if
they
have
the
same
graph.
Conversely,
if
G
is
any
subset
be a
or
=
=
or
all
(d)
N-H-
number
n'-H-'
Indeed,
the
Peanoaxiom
approach
treats
natural
numbers
by
noting
that
the
size
from
{8,9}.
One
way
to
define
sense
regardless
sets
have
equal
cardinality.
doesnot
preclude
one
even
natural
numbers,
Before
we
prove
this
proposition,
1}.
In
particular
X
e
finite
ave
320\222
from \320\222
 
 (|Jie
=
=
to
define
integers,
s in
here
f and
e
will
rectify
these
problems
later.
1In
the
language
of
set
theory,
what
=
e\342\200\224-/.
+
+
accept
these
definitions
(c\342\200\224d),
obtain
that
(a-^-b)
x
oth
sides
evaluate
to
(ac
in
the
same
way
isomorphism
hus
we
may
identify
the
natural
numbers
with
integers
this
thus
3
hen
it
will
also
be
equal
to
any
other
integer
for
instance
3
is
equal
not
only
to
etc.
We
can
o
a
positive
natural
number,
=
\321\201,
which is
=
or
some
positive
n,m,
=
\317\207
\320\245 \320\243\316\266)
=
use
Lemma
4.1.5nd
split
into
a
lot
nd \316\266
=
=
+
+
=
=
+
+
+
f
two
integers
by
the
formula
hen
a
integers.
We
say
that
+
\342\200\224 a f\342\200\2246 )so
equal integers
or
Two
rational
=
\317\207
=
normal
positive
iff
we
or some
bers
is
reflexive,
symmetric,
and
transitive.
 Hint:
for
transitivity,
use
Corollary
2.3.7.
Exercise
4.2.2.
Prove
the
nd
x/y
negative,
then
\\x\\
\342\200\224\320\266.316\231\316\220\317\207s
zero,
then
\\x\\
and
ff
<
Proof.
We
\316\265\316\264
s5)-close.
D
Remark
=
numbers,
and
^5
negative.
 c
rationals
are
arranged
with
respect
to
the
integers.
proposition
4.4.1
 Interspersing
of
integers
by
rationals).
Let
x
be
a
rational
1
is
sometimes
referred
to
as
the
integer
part
Proof.
x2
k
create a
\302\267
or
 i.e.,
ao
positive
rational
values
instead
of
natural
numbers?
Explain.
Exercise
4.4.3.
Fill
symbols
N,
Q,
and
R
stand
for
\ natural\ ,
\ quotient\ ,
and
\ real\
respectively.
was
equal
argument
and
with the
enotes
the
sequence
sequence
is said
standard
s
initial
members
of
a
sequence.
Definition
5.1.6
(Eventual
sequence
\320\260 ,\320\260\320\273\320\275-\321\212\320\260\320\273\320\263+2?
\302\267
302\267
302\267
efined
by
an
sequence
\316\261\316\271\316\277>\316\261\316\271\316\271>\316\2
i-e.,
1/10,1/11,1/12,...)
is
0.1-steady.
The
sequence
10,0,0,0,0,...
is
not
what it means
s
eventually
316\261<\316\271,...
only
if
it
is
eventually
efined
by
an
to be
\316\265-steady,
\316\261\316\267
is
unbounded.
 Can
you
prove
this?
Use
Proposition
4.4.1.
The
sequence
1,
bounded.
By
n
the
other
hand,
hus
\316\261\317\212,320\260\320\263,...,316\261\316\267,
nH_j-
is
bounded.
Proof.
is
iff
an
is
iff
|On
and
are \316\265-close. In
316\2612,...
Remark
5.2.4.
Again,
notations
for
and
and
\302\267302\267302\267
are
eventually
nd
s a
s a
nd
re
eventually
s
bounded
if
and
only
if
We shall
is a
re
said
to
be
equal
iff
sequence
1.5,1.42,1.415,1.4143,1.41422,...
then
\316\231\316\231\316\234\316\267-,\316\277\316\277\316\237\316\267
s
a
real
number,
and
real
number
as
LIMn
\342\200\224\321\216\320\276\320\254\321\202\321\214
ecause
are
equivalent
Cauchy
sequences:
320\253\320\234\320\277\342\200\224\321\216\320\276\320\260\320\277LIMn
We
to
\317\205\316\234\316\267_>\316\277\316\277\316\261\316\267
s the
LIMn
\342\200\224\321\216\320\276\320\260\320\277;
320\243
\342\200\224
LIMn\342\200\224\321\216\320\276\320\254\320\277,
nd
n
be
real
numbers.
Then,
with
the
above
Also,
if
e
s
s
eventually
s
eventually
s
is
eventually
s
s
IS
\317\207
\320\253\320\234\320\277_\321\216\320\276\320\260\320\277>
320\243
and
need
to
show
that
the
sequences
an
and
are
nd
(a'n
re
eventually
s
the
sequence
0.5,0.5,0.5,0.5,0.5,...
then
we
set
multiplication,
since
for
any
rational
numbers
a,
b
we
have
  LIMn_>oob)
317\207
 LIMn_>oob)
as the real numbers
\320\253\320\234\320\277_\321\216\320\276\320\260,
IMn-\321\216\320\276\320\254.
Also,
this
identification
  \342\200\2241)
-
\320\253\320\234\320\277-^\320\254\320\277,
nd
=
LlMn-\321\216\320\276\320\260\320\277\320\254\320\277
nd
xz
LIMn
-\321\216\320\276\320\230\320\277\320\241\320\277)
320\260,\321\2101o
\320\260\320\267,...
10,100,1000,10000,...
Cauchy
sequence
of
rational
limit of a
\317\207
cannot
be
eventually
or
every
s
a
Cauchy
sequence,
so
it
is
is
bounded
away
from
zero.
Actually,
what
it
does
s
eventually
let
us
fix
\316\271
is
a
Cauchy
sequence,
as
desired.
D
We
are
now
ready
be
a
Cauchy
sequence
bounded
away
which
limit,
On
the
other
hand,
since
\317\205\316\234\316\267_>\316\277\316\277\316\261\316\267
LIMn
\342\200\224\321\216\320\276\320\236\320\277,
e
can
write
\316\241
 \320\253\320\234\320\277-\321\216\320\276\320\260\ 1)
317\207
 LIMn^oobn)
 cf.
Proposition
5.3.10).
Comparing
e see that
\320\253\320\234\320\277-\321\216\320\276\32
LIMn
if
and
only
if
\320\253\320\234\320\277_\321\216\320\276\320\260LIMn_>oob
 i.e.,
the
Cauchy
sequences
a,
a,a,
a,...
and
6,
b,
6,
b...
if
and
only
if
how
that
number
should
be
positive,negative,
or
zero.
Since
such that
or
342\200\2241,1,
342\200\2241,...
positive
iff
it
can
be
written as \317\207
\320\253\320\234\320\277_\321\216\320\276\320\260\320\277
or
some
Cauchy
sequence
which
is
negatively
bounded
away
from
zero.
Proposition
5.4.4
(Basic
properties
of
positive
reals).
For
every
positive,
\321\203
=
y,
and
similarly
define
realsis consistent
1
is
negative,
a
contradiction.
Thus,
by
Proposition
5.4.4,
the
only
possibility
non-negative
rational
numbers.
Then
\320\253\320\234\320\277-\321\216\320\276^\320\277
s
a
non-negative
real
number.
Eventually,
we
we have \317\207
\320\253\320\234\320\277_\321\216\320\276\320\254\320\277
or
some
sequence
bn
which
is
negatively
bounded
away
from
0 such
and
re
not
eventually
c/2-close.
Since
c/2
\317\205\316\234\316\267_>\316\277\316\277\
LlMn-oobn\302\267
Cauchy
sequence
\317\207.
s
eventually
either
positively
bounded
away
from
zero
or
negatively
bounded
away
from
zero.)
Exercise
5.4.2.
Prove
the
remaining
x.
Similarly,
or
every
element
other
hand,
it
possible
or
any
number
be
an
upper
bound
of
E.
This
.motivates
the
following
definition:
Definition
5.5.5
 Least
upper
bound).
Let
upper
upper
bound;
of
integers,
number
Sas
5:=\320\253\320\234\320\277^\320\276\320\276^.
\316\267
upper
bound.
Let
342\200\224oo.
etting
this
equal
to
0
causes
some
problems.
Now
we
give
an
example
of
\317\2072
\317\2072 4\316\265
since \317\207
integer,
and
let
L
s
an
upper
bound
for
E,
but
that
(m
302\267302\267302\267
n
the
text
 one
can
define
these
nth
roots
once
one
defines
the
complexumbers,
but
we
shall
=
 
\320\260\320\224/(\320\254\320\276 )\320\262
\320\222\321\203
exponentiation;
Further,
Proposition
consistent
a
Cauchy
sequence
in
the
same
manner
\316\265-steady
bers
starting
at
some
integer
index
m.
Then
is a
5.1.8.
Now
suppose
that
number,
then
there
existsa
rational
ef
is
eventually
number,
and
let
f real
is
148
6.
Limits
of
sequence^
Proposition
6.1.7
  Uniqueness
of
limits).
Let
  \316\274\316\267)\342\204\242=\317\204\316\267
e
a
real
sequence
starting
was
converging
to
both
L
and
Z/.
Let
particular,
if
we
set
unique,
to denote this fact.
onverged
to
0,
where
an
is
eventually
find
an
is
converges
to
0.
D
Proposition
6.1.12
(Convergent
sequences
are
Cauchy).
Suppose
that
is a
is
not
a
Cauchy
sequence
(because
Proof
320\237-\320\256\320\236
(6)
T/ie
sequence
->oo
(c)
For
any
real
number
c,
the
sequence
(d)
T/ie
sequence
converges
to
n->oo
71\342\200\224\320\256\320\236
(/)
\317\200-\321\216\320\276
bn
n
(5)
T/ie
sequence
\321\202\320\260\321\205(\320\266,\321\203);
n
1\342\200\224KX)
n
other
words,
lim
min(an,bn)
71\342\200\224\320\256\320\236
1\342\200\224\302\27300
 
an
We have to
Then use
Exercise 5.4.8.
\320\260\320\263\320\265
seQuences
of
reals,
show
that
=
is
e
sometimes
write
notion
slightly.
Definition
6.2.6
  Supremum
of
sets
of
extended
reals).
=
\317\207
=
e
a
subset
o/R*.
\316\212\317\204\316\271\316\257 \316\225
  b)
Suppose
that
s the
s
the
sequence
1,
2,
3,
4,
=
hen
Proof
See
Exercise
6.3.3.
D
One
can
a
sequence
 \316\261\316\267 \342\204\242=\317\204\316\267
s
bounded
below
and
decreasing
 i.e.,
\316\261\316\267+\316\271\316\261\316\267 ,
hen it is
as
a
lower
bound
of
converges
to
some
_
converges
s
just
the
sequence
hifted
by
one,
iverges
when
e a
and
limit
points
161
\320\264\321\211\320\277\320\254\320\265\320\263.
We
say
ff
for
every
for
every
same
as
\ to
stick
to\ ;
hence
or
all
N
to
be
point
of
this
sequence.
imilarly
-1
is
a
limit
point
of
this
sequence;
owever
0
 say
is
be
a
sequence
which
converges
to
a
real
numberc.
Then
a
limit
point
\320\276/ \320\260\320\277 \342\204\242=\321\202,
nd
in
fact
it
is
the
only
limit
point
Proof.
See
Exercise
6.4.1.
D
Now
 
Example
6.4.7.
Let
320\260\320\267,...
enote
the
sequence
1,-2,3,-4,5,-6,7,-8,...
Then
a*,
aj,...
isthe
sequence
+oo,
and
so
the
limit
enote
the
sequence
1,-1/2,1/3,-1/4,1/5,-1/6,...
Then
enote
the
sequence
1,2,3,4,5,6,...
Then
a*,
a^,...
an,
and
\320\250\320\237\321\203^^\320\260\321\202\302\273
nstead of
342\200\224
which
in
our
new
notation
is
af.
Now
let
us
remove
the
first
element
<
then
 /
converges
to
c,
then
we
must
have
L+
Thus
\316\261^.
*
lim
sup
bn
n\342\200\224\321\216\320\276 \320\277-\321\216\320\276
lim
inf
an
and
or which
\320\230\321\202\320\277_,\320\276\320\276
320\260\320\277
be a
Cauchy sequence.
is
bounded;
by
Lemma
6.4.13
 or
Proposition
316\261\316\267
s
<
0,
and
L+
and
L~
superior
and
f
real
numbers
has
+oo
as
a
limit
point
as
a
limit
point
n
other
words,
the
limit
superior
is
the
largest
limit
point
of
a
sequence.
imilarly,
show
that
the
limit
Let \321\201
In
other
words,
limit
points
of
limit
points
are
themselves
limit
points
of
the
original
sequence.
e
clearly
have
lim
c
for
any
constant
Raising
By Proposition
6.1.11 we
thus have
321\205\320\277
=
which is
defined
by
is
a
subsequence
of
e
sequences
of
real
numbers.
Then
Furthermore,
is
a
subsequence
subsequence
superior
of
the
sequence
ince
n,
it
follows
from
the
compact.
The
are not the
\/bn
exists
and
is
equal
to
zero.
(Hint:
for
each
is also a
converges. By
being
that
qn>
gm,
 
=
-
Thus
such that
\\\321\205\320\263/\320\2721\\<\316\265\317\207~\316\234.
Now
since
-
\317\207\316\234\316\265\317\207~\316\234
\316\265
is
=
\320\245\320\257\320\277-\320\257\320\277\320\245\320\257\320\277),
Write
rn
s
eventually
to
1.
Since
\320\230\321\210^\320\276\320\276\320\226\ 1^
s
also
equal
to
1
by
Lemma
6.5.3,
is
convergent
to
0,
it
is
eventually
1/AT-close
o
0,
so
that
<
<
\321\205\320\223\320\277\317\207\316\273 \316\272
In particular
Qn,
where
the
limit
\320\235\321\202\320\277_>\320\276\320\276\320\2269\320\277
xists
by
ey
all
give
the
same
limit
by
Lemma
to
obtain
he
corresponding
results
for
reals.
Let
q
n
afld
r
\320\270\321\202\320\277-\321\216\320\276
n
\316\267\342\200\224\321\216\320\276
\317\200\342\200\224\321\216\320\276 \317\200\342\200\224\321\216\320\276
But
by
Lemma
320\260\320\263
identities
=
i=77i
i=m
this,
we
sometimes
express
316\243\342\204\242=\317\204\316\267\316\261>\3
ess
formally
as
l
=
\321\212=7\320\277
j=,m
ai
320\263=\321\202
\316\243\316\2610+
we
have
 /
Proof
\316\243\342\204\242=\317\204\316\267 \316
bA
simply
as
>i
R
be
a
function
from
 x)
first
select
\316\243/ *)==316\243/\320\276\320\274)\302\267
xex
be
the
 x)i
we
select
X,
e.g.,
\320\263=1
 \320\266),
=
\320\263=1
\320\263=1
R,
and
any
two
bijections
5,
fo 0)
 MO)\ \302\267
be
a
function,
and
let
g
 \316\243/ \316\2340 +*+[\316\255/ \316\234<+\316\220
\\\320\263=1 /
 7-2
But
is
true}
as
\316\243\317\207\316\262\316\247:\316\241 \317\207)
s true
is
no
chance
of
confusion.
For
instance,
\316\243\316\2676\316\235\302\2672<\316\267<4^ :\316\225)
0\316\223
is
Short-hand
for
 \320\226)
The
following
properties
 b)
If
is
a
bisection,
then
xex
yeY
 d)
 Substitution,
part
II)
Let
i=n
zexuY
\\xex /
be
functions.
Then
\321\205\320\261\320\245
\320\266\320\261\320\245\321\205\320\265\321\205
 \320\264)
R
be
a
function,
and
let
\316\243
=
+
[
(x,y)ex xY
/\320\276*.*))\302\267
{\321\205,\321\203)\320\265\320\245 \321\205\320\243
x,y)e{xo}xY
be
a
function.
Then
\316\243 \ 316\243/ * \302\267\302\273))= \316\243
f^y)
y\342\202\254Y
=
=
\316\243 \316\243/ *\302\267\302\273))\342\226\240
Proof.
In
light
of
Lemma
7.1.13,
\316\243
/ *>*>)\302\267
  \321\205,\321\203)\342\202\254\320\245\321\205\320\243
320\234\320\261\320\243\321\205\320\245
But
this
follows
from
Proposition
Example
1.2.5;
nc^
n(x))^Lm
(Convergence
of
series).
Let
\316\243\342\204\242=\317\200\316\271\316\261\316\
e
a
formal
infinite
series.
For
any
integer
N
\316\243\342\204\242=\317\200\316\271\316\
nd
say
that
L
is
the
316\277>\316\267\302\267
f
the
partial
sums
Sn
diverge,
then
we
say
bn
of
a
convergent
series.
Examples
7.2.4.
Consider
\316\267
>n
converges
if
and
only
if
for
every
real
number
or
a^
PiQ
\316\243\316\267=\317\201\316\261\316\267
*\316\267
IS
a
divergent
series.
(Note
howeverthat
1,1,1,1,...
isa
convergent
sequence]
convergence
of
series
diverges,
and
in
particular
does
not
converge
to
zero;
hus
the
series
\316\243\342\204\242=\316\271(\342\200\2241)\316\267
s
also
divergent.
/n
is
divergent
despite
Let
\316\243\342\204\242=\317\204\316\267\316\261\316\
e
a
316\231\316\261\342\204\2421
s
convergent.
In
order
to
distinguish
from
absolute
convergence,
we
sometimes
71=971
71=971
*s
conditionally
convergent\ ,
n
ls
316\261\316\267
n
converges
conditionally,
but
n0t
absolutely\ .
proposition
7.2.12
(Alternating
series
test).
Let
is
convergent
if
and
only
if
the
sequence
an
converges
to
0
as
if
\316\243\342\204\242=\317\200\316\271(\342\200\2241)\31
s a
(why?).
Thus
Sn
converges,
and
so
the
series
316\243\342\204\242=\317\204\316\267(\342\200\2241)\316\267\316\261\316\267
s
convergent.
D
zero. Thus
\316\243\302\243\317\213\316\271(~\316\212)\316\267/\316\267
S
convergent
(but
it
is
not
absolutely
convergent,
because
n
diverges
see
Corollary
7.3.7).
Thus
absolute
divergence
does
not
imply
conditional
divergence,
even
though
absolute
convergence
implies
conditional
convergence.
Some
basic
identities
n
=
n
be
a
series
of
real
numbers,
and
series
\316\243\342\204\242=\317\200\316\271\316\261\316\267
nd
are
convergent,
then
oo
z2an
be
a
series
of
real
numbers
converging
to
x,
and
n-fc
olso
converges
is.
There
be
a
sequence
of
real
numbers
which
converge
to
0,
i.e.,
\320\235\321\202\320\277_\321\216\320\276\320\260\320\277
0. Then
the
series
~
Proof
See
Exercise
7.2.6.
D
Exercise
7.2.1.
Is
the
series
\342\200\224
hould
be,
and
prove
your
assertion
using
induction.
7.3
Sums
of
non-negative
numbers
Now
we
specialize
the
preceding
discussion
sums
\316\243\342\204\242=\317\204\316\267\316\277>\316\267
here
all
n
is
:=
\316\243\316\267=\317\204\316\267\316\261\316\267
re
increasing,
i.e.,
words,
we
have
just
shown
Proposition
7.3.1.
Let
\316\243\342\204\242=\317\200\316\271\316\261\316\267
e
a
formal
series
of
negative
is
convergent,
then
s
absolutely
\316\243bn-
1=771
71=771
n>
m,
and
\316\243\342\204\242=\317\204\316\267\316\261\316\267
s
not
absolutely
convergent,
then
On
for
all
316\261\316\267
\320\260\320\263
316\261\316\267
ni
\320\2602\320\272
sums of
\320\225\320\272=\320\2762\320\272\320\2602\320\272\302\267
In
light
of
is
s
bounded.
To
do
this
we
need
the
following
claim:
Lemma
7.3.6.
For
any
which
is
clearly
true,
since
2S2k+i.
Also,
we
have
 using
Lemma
S2k+i
Similarly
we
have
2_^
Combining
these
inequalities
with
the
induction
hypothesis
we
obtain
as
desired.
This
proves
the
claim.
s
bounded.
Conversely,
if
/nQ
is
non-negative
and
decreasing
 by
Lemma
n9
n  also
n2
*s
convergent.
Remark
7.3.8.
The
quantity
/nq,
when
it
converges,
is
called
C ?)>
the
Riemann-zeta
function
of
q.
^e an
=
same.
For
instance,
+
=
\316\243\342\204\242=0\316\261\316\267
e
a
convergent
series
of
non-
negative
real
numbers,
f(m)
316\261\316\267
^e know
are
increasing.
Write
L
m=0
of
sum
is
 e
an
\an\\>
which
by
hypothesis
is
a
convergent
seriesof
non-
negative
numbers.
If
\316\270\317\200|>
a/(m)l
also
converges
af(m)
also
converges
M.
Since
\316\243\317\204\316\257=\316\277\316\231\316\261\342\204\242\316\231
S
converSent,
\\316\261\316\267\\
n
converges
to
Z/,
the
partial
sums
316\261\316\267
S
finite,
hence
bounded,
so
\316\243
\316\261\317\200
\316\243\316\261\316\267+\316\243\316\261\316\267
m=0
(why?).
Thus
\\\316\243\316\261\316\267\\<\316\243\\\316\261\316\267\\<316\243
\320\253^\320\244
by
our
choice
which
as
partial
sum.)
are all
316\261\316\267.)
Let
\316\243\342\204\242=\317\204\316\267\316\261\316\267
e
a
series
of
real
numbers,
the series
\316\243\342\204\242=\317\200\316\271\316\277>\316\267
s
y
Proposition.4.12(a),
<
by
Proposition
7.2.14(c)).
Thus
by
the
comparison
test,
we
see
that
\316\243\342\204\242=\316\275\316\237>\317\200
s
absolutely
convergent,
and
thus
is
not
1-close
to
0
for
316\261\316\267
Cn
71-\320\256\320\236
n
Proof.
There
or all \316\267
by
<
But
this
is
  why?
prove
by
contradiction),
as
desired.
D
Prom
Theorem
7.5.1nd
Lemma
7.5.2
  and
Exercise
7.5.3)
we
have
Corollary
7.5.3
  Ratio
test).
Let
\316\243\342\204\242=\317\200\316\271\316\261\316\267
e
a
series
of
1;
then
the
series
\316\243\342\204\242=\317\200\316\271\316\261>\
s
the series
\316\243\342\204\242=\317\200\316\271\316\277>\31
s
by
Proposition
6.1.11
1.
71\342\200\224\320\256\320\236 71\342\200\224\320\256\320\236 71\342\ 200\224KX)
The
claim
then
an
of
positive
numbers
an
such
that
=
316\261\316\231/\316\267
1,
and
give
an
example
n
of
positive
numbers
bn
such
that
inconclusive
even
when
the
\320\243
sets.
For
instance,
which
is
increasing,
in
in
Exercise
8.1.3.
We
now
define
a
sequence316\261\316\277,\316\261\316\271,\316\2612>\302\267\302\267\302\267
f
by
\317\207 \316\267or
then
must
have ?)
increasing
bijection
from
N
to
X
other
than
316\245
is
countable.
Proof.
Define
the
=
by
f(n,m)
nd \316\267 \316\267.
First
suppose
=
is
320\222
=
Q
be
the
function
/ a,
b
{\316\2610,\316\261\316\271,\316\2612,\316\2613,...}
such
that
every
=
be
 x
IS
absolutely
convergent
iff
for
some
bijection
g
 x
by
the
forrmila
oo
317\200=0
00
00
=
\317\200=0m=0
is to
show that
the series
\316\243\302\243\317\213\316\277(\316\243\342\204\242=\316\277
(n m))
converges
toL.
One
can
easily
show
that
\316\243(\316\267,\317\200\316\271)\316\265\317\207
(n>m)
{p>,
(\316\267>m)
\316\235\316\234
\316\243\316\243\316\257(\316\267,\317\204\316\267)<
316\243
oo
we
have
(by
limit
laws,
(\316\267>m)
a
finite
setlCNxN
such
that
\316\243(\316\267,\317\200\316\271)\316\265\317\207
(n>
=
n=0
771=0
\316\235\316\234
\302\243\302\243/ \320\277,\321\202)>\302\243\302\243/ \320\277,\321\202)>\302\243\302\243/ \320\277,\321\202)>\302\243-\320\265
n=0
ra=0
result for
be
a
function.
 x)
is
absolutely
convergent
if
and
only
if
sup
  x)
IS
absolutely
convergent
iff
sup
x)
is
equal
  x)
is
absolutely
convergent.
  x)
for
any
be
functions
  x)
and
  \317\207)
\321\205\320\265\321\205 xex
 
f x)
or
some
disjoint
sets
X\\
and
X2,
then
\316\243\317\207\342\202\254\316\247\316\271
 x)
and
 x)
are
xGXi \321\205^\320\245\321\207
Conversely,
if
h
\316\243\316\266\316\276-\316\247\316\212
{x)
and
 \321\205)
then
\316\243\317\207\316\266\317\2071\317\214\317\
320\234\320\266)
xGXi
XGX2
 d)
If
* \320\244 \321\203))
316\261\316\267
316\261\316\267
302\260\ \317\200
n
such
that
f(m)
316\261\316\267
A+
and
-.
and
\316\243\342\204\242=\316\277\316\261\316\257-(\342\204\242)
m
L,
or
320\260\321\211
or all
\320\263
or
all
320\260\320\277
320\236\321\211
m=N
m=N
sequence
316\261\316\277,\316\261\317\212,\316\261<\316\271
(Of
course,
by
s
an
absolutely
convergent
series
(by
Lemma
7.3.3),
of
this
set,
say
n0
+
10_\316\267)
\317\200<\317\200\316\277:\317\200\342\202\254-\316\221
-(
10_\316\267)
=
5Z
n>no:neB
ilO~no
\321\216-710-1
\320\224\321\214\302\243*\321\212\320\220
302\267
are
all
disjoint
from
each
other
 i.e.,
Dn
unintuitive
consequences
 for
instance
the
Banach-Tarski
paradox,
a
simplified
versionof
which
we
will
encounter
in
Section
18.3),
and
of choice
<*
set
*s
a
subset
=
ls
\316\240\316\271<\316\257<\317\204\316\275^
efined
in
Definition
3.5.7
 why?).
Recall
316\247*
<*
choices
-
but
with
the
Since
for
all
n.
But
then
we
have
\320\230\321\210\320\277-\321\216\320\276320\260\320\277
sup E
proposition
8.4.7.
Let
e
a
property
pertaining
to
an
object
s
true.
Then
there
exists
a
function
f
arbitrary
choice
of
sets
B\\
A,
then
the
axiom
of
choice
context,
Examples
8.5.2.
The
natural
numbers
N
together
with
the
usual
less-than-or-equal-to
relation
We
say
that
lement,
and
{2,3}
is
^\316\262\316\267
not
empty,
sincethe
subset
{xo}
of
X
is
clearly
good
  why?).
We
make
the
following
important
observation:
if
either
\320\243\321\201\320\223\320\276\320\263\320\223\321\201
316\245.
nd
hence
belongs
s
well-ordered
with
xo
as
its
minimal
element,
it
has
a
strict
upper
bound
hus we
  JjQ
be
the
map
efines
a
partial
ordering
on
f
the axiom
\320\260\320\273\320\266\320\265\320\243
uch
can be viewed
various
things
able
setof
all
the
adherent
points
of
X.
Lemma
9.1.11
 Elementary
properties
of
closures).
et
X
and
or
point
to
 a,b).
This
contradiction
[a,b]
as
claimed.
Q
b
are
real
numbers,
then
[a,b],
[a,+oo),  \342\200\224\316\277\316\277,\316\261],
nd
re
not.
\302\260f
X
as
well,
hen
every
element
of
I
\316\235,316\226,Q,
play
a
key
role
in
subsequent
and
bounded,
\320\260
from
the
real
line
to
the
real
this
R,
when
really
we
(This
distinction
makes
more
is a
theirdifference
by
their maximum
max /,g)
by
\321\202\320\260\321\205 /,\320\267) \320\266)=\321\202\320\260\321\205 / \320\266),\320\267 \320\266)),
their
minimum
min /,
g)
: X
by
=
\321\202\321\202 / \320\266),\320\267 \320\266)),
their
 or
and
 provided
that
is the function
is
the
function
is
the
function
is
the
function
 f
is
4\320\2662,
\320\2662
o
converge
to
a
be a
a
real
number.
We
say
be a
be4he
function
0 such
\320\230\321\202\320\266_,\320\245\320\276
 x)
would call
1^6\316\261.\316\277;\317\2076\302\243\\{:\316\225()}
(\320\266),
gxainple
9.3.8.
Let
be the function
proposition,
we
\ 1\316\257\317\200\316\271\317\207_>30
(x)
instance
we
now
know
that
xo
in
E.
Since
xq
is
an
adherent
point
of
E,
consisting
of
elements
also
converges
to
U.
But
this
onsisting
of
was an
arbitrary sequence
x\342\200\224>xq x\342\200\224*xq
lim
\\
X\342\200\224>XQ X\342\200\224*XQ
J
lim
min(f,g)(x)
J
lim
(fg)(x)
X\342\200\224*XQ X\342\200\224*XQ
lim
(where
we
have
dropped
the
restriction
\320\230\321\202\321\205_,\320\226\320\276
320\264(\321\205)
for
any
real
numbers
xo
lim
ex
lim x2
d
\316\247\342\200\224*\316\247\316\270',\316\247\302\243\302\261\316\233,
etc.,
where
c,
d
are
arbitrary
X\342\200\224>Xq\\x\302\243.L\302\267
if
and
only
if
lim
f(x)
x\342\200\224>xo\\x
316\262\316\225\316\267(\317\207\316\277-\316\264,\317\2070+\316\264)
Proof
f(x)\302\267
x^XQ\\xeE
\317\207^>\317\207\316\277;\317\207\342\202\254\316\225\316\223\\(\317\207\316\277-\316\264,\3
Thus
the
limit
of
a
function
at
xo,
if
it
exists,
only
depends
and
g
{1}.
Then
Proposition
9.3.14
does
apply,
and
we
have
\320\230\321\202\321\205^2;\321\205\320\261\320\232_{1}
\302\261f
\320\230\321\202\320\277_\321\216\320\276
 an
s
a
sequence
On the other
is
another
sequence
of
0.
n-\302\273oo n\342\200\224\321\216\320\276
Since
1
in
other
words,
the
limit
of
x\342\200\224>xq;x\342\202\254\302\243\\,
\342\200\224*xo\\xG\302\243\\,
thus
xq\342\202\254x;x
lim
sgn(rr)
z-a;a:eR
\320\266\342\200\224>1;\321\217\321\201\342\202\254(0.9\320\224.1)
lim
1
=
that
oo,
then
/ \320\266\320\276)
4x2
\342\200\224
defined
by
defined
by
=
\321\202\320\260\321\205 \320\266,\342\200\224321\205)
nd
the
functions
integer,
and
left
and
right
limits,
which
can
be
thought
hen we define
/ \317\207),
x\342\200\224\342\226\272\320\266\320\276;\320\266\342\202\254\320\220\320\237 :\320\263\320\2
provided
of
course
hen we
lim
Hx),
x\342\200\224>xo]xGXn \342\200\224\316\277\316\277,\316\266\316\277)
again
provided
that
will
not
be
defined.)
Sometimes
we
use
the
x\342\200\224>xo+
\317\207\342\200\224\321\216:\320\276;\320\266\302\243\320\245\320\237 :\320\263\320\276,\3
lim
x\342\200\224>xo-
\321\205\342\200\224^\321\205\320\276;\321\205\320\265\320\245\320\223\\ -\320\276\320\276,\
when
the
domain
X
of
x\342\200\224\342\226\272a:o;a:\302\243Rn 0,oo)
x\342\200\224\342\226\272a:o;z\342\202\254Rn 0,oo)
and
=
=
If
\316\247316\223\316\252
(\342\200\224\320\276\320\276,\320\266\320\276).
e*
f
x\302\243
316\233\316\240(\316\271\316\277\316\277\316\277)
and
lim
f(x)
L.
(9.2)
\320\266\342\200\224*\321\205\320\276]\321\205\302\243\320\245\320\223\\{\342\200\224\320\276\320\276,\320\266\32
Let
or
which
\\x
\316\247316\240
(\342\200\224\320\276\320\276,\320\266\320\276)
or which
signum
jump
discontinuity
or
bounded.
The
study
of
discontinuities
 also
called
singularities)
is
in
Thus
the
sequence
be a
342\202\254
sup(i?)
342\202\254
by
choosing,
for
each
nd
e have as before that
lim
J
J
 
which
is
which is
continuous, bounded,
which is
which has
or
/ a)
Clearly
Thus
/ \321\201)
nd
N
 since
c+
A
converges
342\200\224>
342\202\254
[\342\200\2242,2]
we have
take nth roots
be
[0,1]
be
a
continuous
function.
Show
that
there
exists
a
real
number
be
a
function.
We
say
that
[a,
b]
nd
hence
there
be
by
setting
g(q(n))
~\316\267
(r)
by
on a
(0,2),
is indeed
iff
an
is
are
\320\264\320\277\320\276\320\263\320\265
succinctly
using
our
language
of
limits:
and
and
and
 1/271)^
are
and
 / 1/271))^
are
not
equivalent
 why?
Use
Lemma
9.9.7
gain).
So
by
Proposition
9.9.8,
defined
by
and
(n+
be
a
uniformly
continuous
function.
Let
defined
by
f(x)
is
a
Cauchy
sequence
in
(0,2),
is
be a
(x)
exists
(in
particular,
it
is
a
real
number).
Proof.
See
Exercise
9.9.4.
D
We
and
\316\267>2
By
Proposition
6.6.5,
we
thus
have
Since
xni,
converges
to
L
as
342\200\242/fc
But
this
contradicts
 9.3)
 why?).
Prom
this
contradiction
we
conclude
that
e subsets of
316\245
316\226
316\226
to
xo
as
long
as
xo
adherent
points.
Remark
9.10.3
 Limits
at
infinity .
Let
X
be
laws
\320\246\321\210\320\277-\321\216\320\276
\ \320\277
\342\204\242\342\200\224\302\260o
where
the
notions
nstead
X
Basic
definitions
289
limit
point
of
X,
lim
 
^o
x\342\200\224>xo;z\302\243R\342\200\224{xo}
x
^0\
We
can
factor
the
Since
egitimately
342\200\224
xo
\317\207\342\200\224>\320\266\320\276\302\273
which
by
hus
g xo)
be
Um \320\253.
x~\302\260
x
and
right
limits.
The
right
limit
is
lim
1,
\316\266->0;\316\261:\342\202\254 0,\316\277\316\277)
316\247
\316\266->0;\316\261:\302\243 0,\316\277\316\277)
316\247
\316\266->0;\316\261:\316\276 0,\316\277\316\277)
while
the
left
limit
is
,im
M
1.
\320\266-^0;\320\266\320\261[0,\320\276\320\276)-{0}
\342\200\224
0
\316\266->0;\316\261:\342\202\254 0,\316\277\316\277)
316\247
be
a
function.
Iff
is
differentiable
at
xq,
then
f
isalso
continuous
at
xq.
Proof.
See
Exercise
10.1.3.
Q
Definition
10.1.11
  Differentiability
  e)
f
is
differentiable
at
xq
and
to
show
that
for
all
x0
e
R-
R
js
differentiable
at
xo,
and
(gofy(xo)
=
visually
appealing
manner
is
also
be
or all \317\207
R,
the
restriction
\316\264
Remark
10.2.2.
If
be
by
which
is
continuous
which
is
differentiable,
which
which
is
strictly
monotone
increasing
and
differentiable,
but
whose
derivative
at
0
is
zero.
316\245
X,
what
can
we
say
about
the
differentiability
of
/_1?
This
will
be
useful
for
many
applications,
for
316\245
f f
[0, oo)
1
\317\207\316\267
\320\220\320\266\~\320\233\320\266\320\2
js
q
non-zero.
Also,
by
hypothesis
f(xo)
is
non-zero.
So
by
limit
laws
nd
e
lim
=
2/n
as desired.
be
the
function
f(x)
=
be
the
function
=
<*.
\320\263\321\213
\317\207^\317\207\316\277\\\317\207<\316\236(\317\207\316\267(\317\207\316\277-\316\264,\317\207\316\277+\316\264))-{\317\2070}
(x)
s
not
necessarily
defined
at
all
points
in
X
ani
\320\235\321\210\321\217._\320\260;\321\217.\320\265(\320\260|\321\206
^fj-
(a,
6],
and
\320\230\321\202\320\266_>\320\260;\321\217.\320\265(\320\260|5]
\320\246
eziste and
the
conditions
of
the
proposition
hold
(in
particular,
that
f(a)
show that
\320\230\321\202\321\205^\320\260.\321\205\320\265(\320\260\320\264
^j
g(xn)
for
taking
values
in
(a,
b]
which
converges
nd
equals
0
at
both
Since
2/n
or
all
n,
and
rrn
converges
oo,
we
see
from
the
squeeze
test
(Corollary
6.4.14)
hat
yn
also
converges
to
a
as
o.
Thus
^rn(
converges
to
L,
converges
to
L,
as
bounded interval.
be
of
f
with
respect
to
the
partition
is
rather
artificial,
but
be
=
c[^^)|[l53)| \320\230-\321\201\321\207:\32 \267>|{3}|
\320\232\320\276(3,4]1\320\241\320\227,4]|
=
let
F
Exercise 11.2.2.Prove Lemma
and
5
be
a
bounded
function
defined_on
defined
by
stant,
hence
piecewise
constant,
and
majorizes
consider
unbounded
be
and
g
: J
be
Riemann
integrable
functions
on
J.
 a)
The
function
s Riemann
be
the
function
and
and
g
R
and
and
are also
he
case
of
min /,
g
being
Basic
properties
of
the
Riemann
integral
323
Inparticular,
if
h :
<
\321\202\320\260\321\205(/(\320\266),\320\267(\320\266))
h(x).
Inserting
/
=
also
Riemann
integrable.
Proof.
This
split /
for
all
=
7+(x)(9+-g\302\261)(x)+9\302\261(x)(7+-f\302\261(x))
<Mi<3+-
g+)(x)
\316\2342 2\316\265 .
Again,
R
arbitrary.
By
uniform
continuity,
xeJk
and
N
/>\316\243(\317\211/(*\316\270)\316\220\316\233\316\220
so
in
particular
I
f-
I
f<
]\302\243\321\205\320\265\320\223\\[\320\260
\320\265,\320\254-\320\265]
Clearly
h
is
piecewise
constant
\320\270 \321\214
//* >+\342\200\224\321\214+1\302\273\342\200\224\302\267
l
+
=
 n >
 \316\267 >
be
the
discontinuous
function
\\
be
a
monotone
increasing
function,
and
let
be
the
function
One
be
a
function
be the
when \320\266
be
the
function
2)]
be
the
identity
function
in
the
proposition
with
p.c.
Jfp
f
be
the
signum
function
{1
be a
be
a
Riemann
integrable
function.
Let
F
6].
Now
Now
let
oo,
\316\264]\316\240
342\202\254
and
thus
which was
by
be
a
 
is
an
antiderivative
of
f,
then
for
every
partition
JeP:J?9xeJ
and
G
be
differentiable
functions
on
[a,
b]
such
be
a
monotone
increasing
be
a
monotone
increasing
function,
and
suppose
 
[0 a),
0 6)]
/.
For
each
interval
J,
let
[0 a),
0 b)]
is
Riemann-Stieltjes
integrable
with
respect
to
316\231
f
\316\221\317\206\316\234,\316\246\316\246)}
Since
\302\243\316\277\317\206\316\254\317\206<
\316\257
\316\257\316\277\317\206\316\254\317\206
J[\ M
J-[a,b]
\316\257\316\277\317\206\316\254\317\206>
316\231
J[a,b]
\317\212\316\277\317\206\316\254\317\206<
316\257
316\257
J-[a,b]
J
WM \316\221\316\246\316\234,\316\246\316\237\302\
Since \316\265
L[at>)
[0 a),
0 b)]
be a
be
a
Riemann
integrable
function
on
is
Riemann
integrable
on
[a,b],
and
J[a,b] J[4> a),<
Exercise 11.10.1. Prove
numbers,
how
that
g
is
also
Riemann
integrable,
nd
Jj_b
for
non-mathematicians
as
well.
So
saying
that
Many
of
you
have
probably
written
ill-formed
or
otherwise
inaccurate
statements
in
your
mathematical
work,
while
intending
to
meansome
other,
falsity
can
is ill-defined.
\320\243
s a
is
false\ ,
or
\ It
is
the
case
that
X\ ,
s
automatically
true
 think
re all
s
equal
to
2,
then
by
\316\267
as
X.)
Thus
one
could
ay
\ X
of
\ If
X
is
true,
=
and
fast
rule.
Logicians
often
use
320\243\
implies
D
implies
D,
we
D
Again,
from
a
logical
point
of
view
true
this
hat
D
is
true;
which
truth
value.
Sometimes
variable.
At
other
times,
we
set
a
For
instance,
the
statement
 x
2
does
not
imply
that
and
uniqueness
of
suchan
x.)
To
prove
such
=
1,
or
for
Proof.
e
only
does
what
you
\316\265ntil later in
0 such
=
Note that the
that
for
sm x)
sin rr)
=
=
hus
we
have
as
0++
and
  0++)++.
The
basicreason
for
this
is
\316\261\316\267\316\261\316\267_\316\271...
316\261\316\277
\316\261\316\267\316\261\316\267-\316\271...
316\261\316\277
Also,
a
single
i=0
hus
by
the
strong
induction
hypothesis,
the
number
an...
ao
has
only
one
decimal
representation,
which
means
that
nf
must
equal
given
by
ratio,
e.g.,
6/4
=
..
=
..
0
uch
that
317\207
his
identity
and
induction,
we
Thus
B.2.1. If
..
316\261\316\271
^\320\277\321\201\321\206\320\276\320\277^
58
functions,
329
of
uniformly
continuous
functions,
326
Riemann
integral,
318
upper
and
lower,
317