Chapter 1 Introduction Definition. Let A, B be sets,
(i) We say A ⊂ B, if every element of A is in B ⇒ x ∈ A ⇒ x ∈ B (ii) We say A = B, if A ⊂ B & B ⊂ A
(iii) Define
A ∪ B := {x : x ∈ A or x ∈ B}
A ∩ B := {x : x ∈ A and x ∈ B}
A \ B := {x : x ∈ A and x 6∈ B}
Definition. Let ξ := {A α : α ∈ I} be a collection of sets.
Defint
S
α∈I A α := {x : ∃ α ∈ I, 3 x ∈ A α }
T
α∈I A α := {x : ∀ α ∈ I, x ∈ A α }
Theorem. (De Morgan’s laws)
Let A be a set & let ξ := {β α : α ∈ I} be a collection of sets.
Then
A \ α∈I S β α = α∈I T (A\β α ) A \ α∈I T β α = α∈I S (A\β α )
Definition. Let A, B be sets Decartes Define the cartesian product of A
& B to be A × B := { (x, y) : x ∈ A & y ∈ B }
Definition. Let X & Y be sets. Define the cartestion product of X & Y to be X × Y := { (x, y) : x ∈ X & y ∈ Y }
Definition. Let X, Y be sets. A relation < from X into Y is a subset of
X × Y . If X = Y then < is called a relation on X. If (x, y) ∈ R, then we
write x<y & x is said to be <-related to y.
Definition. Let X, Y be nonempty sets. A relation f from X into Y is called f function from X into Y , if ∀ x ∈ X ∃! y ∈ Y 3 (x, y) ∈ f (or xf y). In this case y is called the image(or value) of x under f & de- noted by y = f (x). In this case, we write f : X −→ Y , x is called the domain of f.
Let f : X −→ Y & let A ⊂ X & B ⊂ Y . Define f (A) := {y = f (x) : x ∈ A} called the image of A under f & f −1 (B) := {x ∈ X : y = f (x) ∈ B} called the inverse image(pre-imabe of B under f ).
Theorem. Let f : X −→ Y & let {A α : α ∈ I} be a collection of subsets of X. Then
(i) f ( α∈I S A α ) = α∈I S f (A α ) (ii) f ( α∈I T A α ) ⊂ α∈I T f (A α )
Remark : If f is 1-1 (or injective), then f ( α∈I T A α ) = α∈I T f (A α )
Theorem. Let f : X −→ Y & let {B α : α ∈ I} be a collection of subsets.
Then
(i) f −1 ( α∈I S B α ) = α∈I S f −1 (B α ) (ii) f −1 ( α∈I T B α ) = α∈I T f −1 (B α )
N := {1, 2, 3, . . .}
Z := {0, ±1, ±2, ±3, . . .}
Q := {x = p q : p, q ∈ Z & q 6= 0}
R := the set of all real numbers C := {z = a + bi : a, b ∈ R}
N Z Q R( C)
2
Postulatel (Field Axion)
There exist two operations. +, • : R −→ R satisfying.
F1 x + y = y + x, ∀x, y ∈ R
F2 (x + y) + z = x + (y + z), ∀x, y, z ∈ R F3 ∃! 0 ∈ R 3 x + 0 = x, ∀x ∈ R
F4 ∀ x ∈ R ∃! y ∈ R 3 x + y = 0 ( In which case, y is denoted by −x ) F5 xy = yx, ∀x, y ∈ R
F6 (xy)z = x(yz), ∀x, y, z ∈ R
F7 ∃! 1 ∈ R, 1 6= 0 3 x1 = x ∀x ∈ R F8 ∀ x 6= 0 in R. ∃! y ∈ R 3 xy = 1 ( In which case, y is denoted by x −1 ) F9 x(y + z) = xy + xz, ∀x, y, z ∈ R
Definition. ∀x, y ∈ R defin x − y = x + (−y) & x y = x · y −1 , provided the y 6= 0
Theorem. In R, we have : (i) x + y = x − z =⇒ y = z (ii) x · 0(= 0 · x) = 0 ∀x ∈ R (iii) −(−x) = x ∀x ∈ R
(iv) (−x)y = x(−y) = −(xy) ∀x, y ∈ R
(v) (−x)(−y) = xy ∀x, y ∈ R
[pf]:
(i) x + y = x + z
=⇒ (−x) + (x + y) = (−x) + (x + z)
=⇒ [(−x) + x] + y = [(−x) + x] + z
=⇒ 0 + y = 0 + z
=⇒ y = z
(ii) x · 0 = x(0 + 0) = x · 0 + x · 0 Thus, x · 0 + 0 = x · 0 = x · 0 + x · 0 by (i), x · 0 = 0
(iii) Since x + (−x) = 0, we have −(−x) = x
(vi) Observe that xy + (−x)y = (x + (−x))y = 0 · y = 0 so (−x)(−y) = −(x(−y)) = −(−(xy)) = xy
Theorem. Let a, b ∈ R (i) −(a + b) = (−a) + (−b)
(ii) if a 6= 0, b 6= 0, in R, then a b · d b = ab cd , a c + d b = ad+bc cd
Postulate 2 ( Order Axions )
There is a relation < on R satisfying he following conditions:
1 ( Trichotomy law )
for each x, y ∈ R, one & only one of the following holds : x < y, x = y, x > y( equivalently, y < x )
2 ( Transitive law ) x < y &y < z =⇒ x < z 3 ( Additive law )
x < y =⇒ x + z < y + z , ∀z ∈ R 4 ( Multiplicative law )
x < y =⇒ xz < yz , ∀z > 0
4
Theorem. In R
(i) x < y ⇐⇒ y − x > 0 (ii) x > 0 ⇐⇒ −x < 0
(iii) a < b & x < y =⇒ a + x < b + y (iv) a < b & c > 0 =⇒ ac > bc
Proof of (iii)
Since a < b, by 3, a + x < b + x
Since x < y, by 3 again, b + x < b + y, finally by 2, a + x < b + y
Theorem. ∀x 6= in R, x 2 > 0. In particular, taking x = 1 we obtain 1 > 0
[pf]:
by 1, x > 0 or x < 0 (−a)(−b) = ab
If x > 0, then by 4, x · x > 0 · x =⇒ x 2 > 0.
If x < 0, then −x > 0, by the preceding result, −(x) 2 > 0 =⇒ x 2 > 0.
Definition. ∀ x ∈ R, define the absolute value of x to be |x| =
( x , if x ≥ 0
−x , if x < 0
Theorem. In R
(i) |x| ≤ M, ⇐⇒ −M ≤ x ≤ M (ii) ( Triangle inequivalite )
|x + y| ≤ |x| + |y|. ∀x, y ∈ R
(iii) | |x| − |y| | ≤ |x − y|, ∀x, y ∈ R
[pf]:
(ii)
Observe that −|x| ≤ x ≤ |x| & − |y| ≤ y ≤ |y| so, −|x| − |y| ≤ x + y ≤
|x| + |y| and so |x + y| ≤ |x| + |y|
(iii) note that
|x| = |(x − y) + y|
≤ |x − y| + |y| so
|x| − |y| ≤ |x − y| similiarly
|y| − |x ≤ |x − y|| and so
−|x − y| ≤ |x| − |y| ≤ |x − y|. Therefore
| |x| − |y| | ≤ |x − y|
Definition. Let E ⊂ R
1. We day u ∈ R is an upper bound of E if
x ≤ u, ∀x ∈ E. In this case E is called bounded above (by u)
2. We say u ∈ R is a supermum ( or least upper bound )of E, if u is an upper bound of E and u ≤ any upper above of F.
In this case, u is denoted by supE. 3. We say ` ∈ R is a lower bound of E, if ` ≤ x , ∀x ∈ E.
4. We say ` ∈ R is an infimum ( or greatest lower bound ) of E if ` is a lower bound & ` ≥ any lower bound of E.
In this case, ` is denoted by inf E.
6
Ex:
1. Let E = [0, 1). Then supE = 1 , inf E = 0 2. Let E = [0, 1). Then supE = 1 , inf E = 0
3. Let E = [0, ∞). Then E has no upper bound, so supE does not exis.
4. Let E = {x ∈ R ; x 2 < 2}. Is E has bounded above? And: Yes!
Dose supE exist?? if exists, what is supE?
Postulate 3 ( Completeness axiom )
F let E ≤ R, if E is bounded above, then supE exists in R.
Theorem. ∀ c > 0 , in R ∃! a > 0 in R, 3 a 2 = c. ( In this case, a is denoted by √
c )
Poof:
Consider E := {x ∈ R 2 : x 2 < c}
Prove that
1. E 6= ø ( ∵ 0 ∈ E ∴ E 6= ø )
2. E has an upper bound ( by c + 1 )
∴ by completeness axiom, a = supE exist in R
Prove that a 2 = c
Real Number system R
P ostulate1 (Field Axioms) P ostulate2 (Order Axioms)
P ostulate3 (Completeness Axioms)
Every nonempty subset of R, which is bdd above, has a least upper bound.
Principle of Mathmatical Induction Well-Ordering Principle
Every nonempty subset of N has a least element.
Theorem. ( Archimedean Property )
∀x ∈ R ∃n ∈ N, 3 n > x
[pf]:
Given any x ∈ R, suppose n ≤ x, ∀n ∈ N, then N is bounded above ( by x ), as by completeness axiom. u = supN exist in R note that u − 1 < u, so u − 1 is not an upper bound of N, and so ∃n 0 ∈ N, 3 n 0 > u − 1,
=⇒ n 0 + 1 > u, which contradicts that u is an upper bound. Therefore,
∃ n ∈ N 3 n ≥ x.
Theorem. ( Density of Rationals )
∀ x, y ∈ R with x < y, ∃ r ∈ Q 3 x < r < y
[pf]:
There are three probability : 1. 0 ≤ x < y
2. x < 0 < y 3. x < y ≤ 0
Now, assume that 0 ≤ x < y.
By Archimedeam property, ∃ n 0 ∈ N 3 n 0 > y−x 1 i.e, x + n 1
0
< y Condider the set S := { m ∈ N : n m
0> x} which, by Archimean property again, is not empty. Then by Well-ordering princinple, S has a least element m 0 . Put r := m n
00
. Clearly r > x. By minimality of m 0 , we have m n
0−1
0
≤ x
=⇒ r = m n0
0≤ x + n 1
0
