Lecture 3: Feb 12

Criteria for good proofs:

  • Correct
  • Clear & well organized
  • Convincing
  • Enlightening
  • Clever
  • Elegant


WOP => PMI

Let S Í N st

  1. 0 Î S
  2. "n n Î S => n+1 Î S

Must show: S = N

Let T = N – S

Ether T ¹ Æ or T Ï Æ

Case#1: if T = Æ then S = N

Case#2: if T ¹ Æ then by WOP T has a least element x Î T.

Since 0 Î S, x > 0. By (b), and the fact x Ï S, it is true that x - 1 Ï S.

Therefore, x - 1Î T, contradicting the fact that x is a least element of T. Therefore T = Æ.


x – 1 < ë x û £ x £ é x ù < x + 1

Definition: lim® ¥ Sn = L iff " e > 0 $ Ne st

" n ³ Ne ç Sn –L ç ³ e

Property: Let x ¹ 1 be any real number st êx ê <1

 ¥
S x j = 1 / (1 - x)
j = 0

Proof: Let S : N ® R defined by

  n S(n) = S x j   whenever n Î N j = 0

We must show: lim® ¥ S(n) = 1 / (1 - x)

Given any e > 0, we want a Ne st " n ³ Ne êS(n) - 1 / (1 - x) ê £ e

(*) We want [1 / (1 - x)] - e £ S(n) £ [1 / (1 - x)] + e , for all n ³ Ne .

S(n) = ( 1 - x n+1) / ( 1 - x)

We want ( 1 – x n+1) / ( 1 – x ) £ [1 / (1 - x)] + e

x n+1 ³ - e ( 1- x )

n +1 ³ logxe ( x – 1 )

n ³ [logxe ( x – 1 )] – 1

then take max of two expressions.

Note: (*) holds for Ne = é logxe ( x – 1 )ù

 



Functions

Æ , {Æ }, {Æ ,{Æ }};

ê N ê = c o smallest cardinal number of the infinite set.

injection - " x,y f ( x ) = f ( y ) Þ x = y

surjection - " y $ x f ( x ) = y

bijection - both injection and bijection

function - f = ( f , A, B)

where f - rule of asigment

A - Source

B - target

f Í A´ B no two pairs have the same coordinate

The Image of the function

Image ( f ) = { x : ' y ( x , y ) = f }

The Domain of the function

Domain ( f ) = { x : ' y ( x , y )Î f }

Note: Functions can be composed with other functions or with themselves.

( f ° f ° f )

The function is computable only if there exists a program that can compute the function.

Note: not all functions are computable.

 

Proof Techniques

What if we need to show ê Aú = K (set has the exact cardinality K).

  1. Show that there is a bijection f : Zk® A

    2. Zk = {0, 1, 2, … k+1}ô Zkô =K

    we can break it into two parts.

    ô Aô £ K and ô Aô ³ K

    to show ô Aô ³ ô Bô need to show injection of A into B

    to show ô Aô =ô Bô need to show bijection f : A® B

  2. Show that there is an injection

    3. Claim: Z is countable.

    Must show: $ injection f : Z® N

    Let f : Z® N be defined as follows
    f ( x ) = { -2x
    2x-1    		 if x £ 0
    if x> 0

    must show f is injective

    Let x, y Î Z st f ( x ) = f ( y ), then

    case#1: If f ( x ) is even,

    then f ( x ) = -2x = -2y.

    Hence x = y.

    case#2: If f ( x ) is odd,

    then f ( x ) = 2x - 1 = 2y - 1.

    Hence x = y.


Suppose there was some bijection f : N -> R
Let d Î R whose binary representation is
d i = { 1
0		 if xjj = 0
if xjj = 1
Claim: d is not in Image( f ) were there some k Î N st d = f ( k ), then dk ¹ xkk, contradiction.