Download UProperty 1

U
Mathematical Analysis
What is Analysis?
Calculus works with
0
0
all the time!

once thought paradoxical

now realized we just have to be careful and know exactly what we are doing (thus
university mathematics emphasize on rigour)

can think of Analysis as the art of dividing by zero 
We shall cover the following topics in this camp

What are real numbers:?

What is a limit?

What is a continuous function and what are its properties?
What is?
Start with the familiar set of natural numbers:   0,1,2,3, 
U
Property 1
Every natural number has a unique successor, i.e.
n    n  1 
------- (P)

(P) means that the set  is unbounded, i.e. there is no biggest member in 

The study of abstract algebra is motivated by the desire to understand the structure
of 

Some of the more important properties of  include
 Prime Factorization Theorem: every composite number can be written as a
unique (ignoring the order of multiplication) product of primes
 Euclid’s Theorem: there are an infinite number of primes

Famous unresolved problem
 Goldbach Conjecture: every even number bigger than or equal to 4 can be
written as the sum of two primes
We can visualize  by drawing a number line with marks at regular intervals
0
1
2
3
4
5 …..

Clearly there are lots of gaps on the line!

What do the points in the gaps represent? This issue has perplexed the ancient
Greeks for a long time 

Let  denote the totality of all numbers represented in the number line
U
Rational numbers
n


n, m  , m  0
m


is dense, i.e. given any two rational numbers,  (there exists) another rational
number in between

Once thought filled up the number line: Pythagoreans thought “All is Number”

Theory of harmonics
C=1
D = 16 : 9
E=8:5
F=3:2
G=4:3
A= 6 : 5
B = 16 : 15
A Greek Tragedy:
2
is not rational!
What is the length of the diagonal of a unit square?
2
1
1
A Pythagorean proved that

2
is not rational, in fact
Between any two rationals,  an irrational
2

Between any two irrationals,  a rational

So the number line was not filled up by  at all

There are far more irrationals than there are rationals, infinitely more! Technically,
we say the set of rationals is countable, while the set of irrationals is uncountable)

So the beautiful Pythagorean concept that everything is related to a ratio of natural
numbers has been undermined by a fellow Pythagorean using the Pythagorus
Theorem!
U
U
U
U
The problem then is, what else is there on the number line? But what are these
irrationals (at the moment, we just know that they are not rational numbers). In other
words, what exactly is the set of real numbers ? There have been many attempts to
define this set rigorously. All the study of calculus ultimately rests on a proper and
useful characterization of !
Least Upper Board and Greatest Lower Bound
Given any set of numbers S, we say it is bounded above if
 B such that x  S  x  B (B is a upper bound)
The Least Upper Bound (or LUB for short) of S satisfies
i) LUB is an upper bound of S
ii) if B is any upper bound of S, then LUB  B
Similarly, we can define a Greatest Lower Bound (GLB) in the same way. Given any
set of numbers S, we say it is bounded below if
 L such that x  S  x  L (L is a lower bound)
The Greatest Lower Bound (or GLB for short) of S satisfies
i) GLB is a lower bound of S
ii) if L is any lower bound of S, then GLB  L

clearly LUB/GLB, if it exists, is unique
Examples

Archimede’s Principle:
1

Let S   n  , n  0 , then given any   0 ,  n sufficiently large such
n

1
  . So, Archimede’s principle says that 0 is the GLB of this set S.
n
that




Let S  x   x 2  2 . What is its LUB?
LUB may not exist. For instance, in the example above, the LUB is 2 ,
which is the root to the algebraic equation x 2  2  0 . But 2 is not rational and
hence does not exist if we are only working in the field of rational numbers !

Many irrational numbers can be seen to be roots of an algebraic equation.
Can we define  by incorporating all such roots to? The answer, unfortunately,
is “No”!  transcendental numbers e ,  ,….. which are not roots of any
algebraic equation.


However, all such irrationals can be characterized as a LUB of a set
This lead to the concept of Completeness of : any non-empty bounded
subset of  has a LUB/GLB. This characterization of  sought to define the set
by imbuing the completeness property on the set.

Exercise
Prove that
p is irrational for any prime number p.
Limits
A sequence is a function f:   S, or you can think of a sequence as a set S with
elements labeled by natural numbers
S  xn   n  

A sequence S  xn   n   is said to have a limit l if given any   0 ,
for n sufficiently large, the elements of S are within  from l . I.e. xn  l   .
xn
x
x
l
x
x

x
x
x

n
If a sequence has a limit, it must be unique.
Examples




1
Let S   xn  n  , n  0 . From Archimede’s principle, we know that 0 is
n


the GLB of this set S. In fact, 0 is the limit of this sequence.
Zeno’s Paradox: In order to cover the distance of 1 unit, a runner must first cover
the first half of the distance, then half of the remaining distance, and so on ad
infinitum. Since this involves an infinite number of distances, it can never be
achieved.
But in terms of sequence, the distance covered by the n’th step is
2
xn 
U
3
1 1 1
    
2 2 2
n
1
1
    1  n . This sequence has a limit = 1.
2
2
Theorem 1
A bounded monotonic sequence has a limit.
U
Proof
Without loss of generality, suppose the sequence is monotonically increasing. Since
the sequence is a non-empty bounded set of real numbers, by the completeness
property, it has a LUB, say, l.
We claim that l is the limit of the sequence. For given any   0 , then l    l ,
hence l   cannot be an upper bound (since l is the least upper bound). Therefore,
for n sufficiently large, we have l    xn  l  xn  l   .

Example
n


1

Let S   xn  1   n    . S is clearly monotonic. And
n



1
1 n(n  1) 1
1
1
2
r 1
1
(1  ) n  1  n  
 ...  (1  )(1  )...(1 
)  ....  n
n
n
2! n 2
r!
n
n
n
n
1 1
1
  ... 
2! 3!
n!
1 1
1
 2    ...  n 1
2! 4!
2
1
 3 n  3
2
 2
i.e. S is bounded.
n
 1
Hence by the completeness property, S has a limit e  lim 1   .
n 
 n