MAM2000W 2RA Notes (Real Analysis)

REAL ANALYSIS
PREP

, THE WELL-ORDERING PRINCIPLE
the well sanderling principle

AXIOM 2 [ THE WELL -
ORDERING PRINCIPLE ] :




Every non -



empty subset of 1N has a smallest element .




THEOREM 3 [ THE PRINCIPLE OF MATHEMATICAL INDUCTION ] :


For each K C- IN
,
let PCKI be a
logical proposition concerning
the number K ( Pllc) is true / false
depending
the value of K) If Pll ) is true and for each C- IN there of that
on . n
,
is some
way showing
Pln ) true that Plnti ) Therefore PCKI true for all
being implies is true .
is KEN .




THEOREM : 2 is not a rational number

Proof [ proof contradiction ]
: if 52 is rational by
↳ : .
52 =
Pig pig C- Z

↳ Assume 9 C- IN
q > 0 : .




↳ : 52 is rational → s= {q c- IN : Eq C- Z } ≠ ∅
↳ has smallest element [ Well
s a
Ordering ] go .




↳ : 1 ( 2 [ 4 i. I < T2 ( 2


↳ Oc S2 -
I Ll


↳ 9 , =
152-1190
i. 0C 91 (
go


↳ But
go C- S
,
so 5290 C- Z

i.
9 5290 C- Z
go
-

, =




↳ But 0cg , 91 be C- IN and
cqo i. must 91<90
↳ However : 5291 =
5215290 -90 ] 290-5290 = .
!
9, C- S



↳ Contradicts go is the smallest element of 5



NUMBER SYSTEMS

IN -


natural numbers { 1,2 , 3,4 ,
. . . }
Z -


the
integers { . . .

, -3 , -2 , -1
, 0,112,3 }
④
-


rational numbers { mln : MEZ in C- IN }
IR real numbers { }
-

-


x.ro




Algebraically ,
it is difficult to
distinguish R and 02 . Thus we need another axiom .




Natural numbers

↳ Every subset has a smallest element CW-0.PT
↳ Natural lm.nl
separated
"
numbers are
" :
Mtn =) ≥ I

↳
They are bounded below : n so for all n C- IN


A subset ,
A ≤ IR is bounded below it there exists an M C- IR sit .
m≤ x for all ✗ C- A

i. We call m a lower bound for A.

lower bound
↳
Any
"

than
"

number smaller M is also a .

,THE COMPLETENESS AXIOM OF IR
THE COMPLETENESS AXIOM Ii


Any non -



empty subset
of IR that is bounded below has a
greatest lower bound .




To be the lower bound for A must
greatest a set
, a number m
satisfy two conditions :
① m ≤ x for all sc C- A

② For each number n with men ,
there is an x EA with x Cn .




THE COMPLETENESS AXIOM 2 :


of that above has least
Any non -




empty subset IR is bounded a
upper bound .




A c- IR is bounded above it there exists a number m such that x ≤ m for all ✗ C- A .




We call M for
an
upper bound A.

It A is bounded above → -
A = § -

x KEA } is bounded below
, .




Greatest lower bound = infimum
least upper bound supremum
=




MAXIMUM AND SUPREMUM


Maximum [ definition] We b the maximum of the set A it
say
: :
is

① for C- A a ≤ b
every a
,




② b C- A

→ Max IA ) =
maximum of A



Supremum [ definition] We that b is the least bound of A it :

say
:
•
upper
① for C- A a ≤ b
every a ,

② whenever a ≤ c for every a C- A
,
then b ≤ C.

→ sup (A) supremum of A
=
.




* A set need not have a maximum . [ even it it is bounded above]
* A set which is bounded above
always has a
supremum .




* It Max (A) exists , Max / A) C- A , sap (A) does not need to be an element of A .




* It above : 1A )
Max / A) exists
,
A is bounded SUPIA ) = Max




MINIMUM t INFIMUM


Minimum [ definition] We b the of the set A it
say
: :
is minimum
① for C- A a ≥ b
every a
,




② b C- A

→ mail.AT =
maximum of A


b the of it
Infimum [ definition] we that is
greatest lower bound A :

say
:



① for C- A a ≥ b
every a ,

② whenever a
≥ c for every a C- A
,
then b ≥ c.

infimum (A)
'


→ = int


* A set need not have a minimum [ even it it is bounded below ]
A has
* set which is bounded below always a infimum
* It Min / A) exists , min (A) C- A int CA ) does not need to be an element of A
,
.




* It min / A) exists
,
A is bounded above : int (A) = min .LA )

, BOUNDED SUBSETS

A subset A of IR is
bounded it it is bounded both above and below .




A is bounded iff .
there is a number B such that :
1×1<13 for all x C- A

Example : 1- =
{ sin ≥ : KEIR } where B=I




CONSEQUENCES OF THE
COMPLETENESS AXIOM
THE EXISTENCE OF TE


let A = { x c- IR : x2 C2 }
↳ A- =/ ☒ since 0 C- A




__ 2
y
- - - - - -




y
-
- -




1

I

:
• •



A
42




Claim : The number 2 is bound for A
an
upper

Proof : ① Consider number 0C with >2
any
a


i. x2 > 2x 74 Thus x & A
,




② The Completeness Axiom tells us that A has a least
upper bound ,
S .




that 0 's ≤ 2 and
such
sup (A) 2
=

<




Claim ; s =
52



Proof first
:
① Suppose that SZCZ

② For h 20 , Csth / 2 =
S2 tzsh th
?

any
③ It 0 < hcl ,
then hzch ,
so Csthl ? L S2 1- 4h th [ 5=2 ]

i. Csth )2 ( S2 1- 5h



④ By choosing h
sufficiently small
, we can make s2 1- 5h 22

But then 5th C- A :S isn't an
upper bound for A