MAT1202 SET THEORY: Week 1 · Introduction History of Set Theory 1.1 History of Set Theory Set theory is a branch of mathematical logic that studies sets, which informally are collections

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MAT1202 SET THEORY: Week 1

Thanatyod Jampawai, Ph.D.

Thanatyod Jampawai, Ph.D. MAT1202 SET THEORY: Week 1 1 / 41

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Outline for Week 1

Chapter 1 Introduction1.1 History of Set Theory1.2 Sets and Standard Notations1.3 Russell’s ParadoxChapter 2 The Axiomatic Set Theory2.1 The Axiom of Equality

ConclusionAssignment 1


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Introduction

Chapter 1 Introduction


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Introduction History of Set Theory

1.1 History of Set Theory

Set theory

is a branch of mathematical logic that studies sets, which informally are

collections of objects.

Although any type of object can be collected into a set, set theory is applied most often to objectsthat are relevant to mathematics.

The language of set theory can be used in the definitions of nearly all mathematical objects.

From Wikipedia


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Set theory is a branch of mathematical logic that studies sets, which informally are




From Wikipedia


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From Wikipedia


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From Wikipedia


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Georg Ferdinand Ludwig Philipp Cantor Julius Wilhelm Richard Dedekind

Cantor (19 February 1845 to 6 January 1918) was a German mathematician.

Dedekind (6 October 1831 to 12 February 1916) was a German mathematician.

The importance of one-to-one correspondence between the members of two sets (similar).

Defined infinite and well-ordered sets.

Proved that the real numbers are more numerous than the natural numbers.

The existence of an infinity of infinities.

Defined the cardinal and ordinal numbers and their arithmetic.


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Introduction Sets and Standard Notations

1.2 Sets and Standard Notations

LOGICAL SYMBOLS

NOTATIONS MEANING NAMES

p ∧ q p and q conjunction

p ∨ q p or q disjuction

p → q if p then q conditional statement

p ↔ q p if and only if q biconditional statement

¬p not p negation

p ⊢ q argument of p and q argument

∀x for all x universal quantifier

∃x for some x existential quantifier

∃!x there is one and only one x uniqueness quantifier


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STANDARD SETS

NOTATIONS SETS

N {1, 2, 3, ...} the set of all natural numbersNk {1, 2, 3, ..., k} the set of all natural numbers ≤ k

Z {0,±1,±2,±3, ...} the set of all integersZ− {−1,−2,−3, ...} the set of all negative integersZ+ {1, 2, 3, ...} the set of all positive integers

Q {ab: a, b ∈ Z, b ̸= 0} the set of all rational numbers

Q+ {x ∈ Q : x > 0} the set of all positive rational numbersQ− {x ∈ Q : x < 0} the set of all negative rational numbers

Qc {x : x /∈ Q} the set of all irrational numbers

R {all decimal numbers} the set of all real numbersR+ {x ∈ R : x > 0} the set of all positive real numbersR− {x ∈ R : x < 0} the set of all negative real numbers

C {a+ bi : a, b ∈ R, i =√−1} the set of all complex numbers


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NOTATIONS OF RELATION ON SETS

NOTATIONS MEANING

a ∈ A a is an element in A or a belong to A

a /∈ A a is not an element in A or a does not belong to A

A ⊆ B A is a subset of B

A ⊇ B A is a superset of B

A * B A is not a subset of B

A ⊂ B A is a proper subset of B

A = B A is equal to B

A ̸= B A is not equal to B


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NOTATIONS OF OPERATION ON SETS

NOTATIONS SETS

A ∪B {x ∈ U : x ∈ A ∨ x ∈ B} union of A and B

A ∩B {x ∈ U : x ∈ A ∧ x ∈ B} intersection of A and B

A−B {x ∈ U : x ∈ A ∧ x /∈ B} diffrent of A and B

Ac {x ∈ U : x /∈ A} complement of A


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A set is a collection of things.The things in the collection are called elements of the set.

A special notation called set-builder notation isused to describe sets that are too big orcomplex to list between braces. A set X written with set-builder notation has the syntax

X = { expression : rule }.

For examples,the set of all even numbers is

{n : n = 2k, k ∈ Z} or {2n : n ∈ Z}

the set of all odd numbers is

{n : n = 2k + 1, k ∈ Z} or {2n+ 1 : n ∈ Z}


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{n : n = 2k, k ∈ Z} or {2n : n ∈ Z}


{n : n = 2k + 1, k ∈ Z} or {2n+ 1 : n ∈ Z}


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{n : n = 2k, k ∈ Z} or {2n : n ∈ Z}


{n : n = 2k + 1, k ∈ Z} or {2n+ 1 : n ∈ Z}


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{n : n = 2k, k ∈ Z} or {2n : n ∈ Z}


{n : n = 2k + 1, k ∈ Z} or {2n+ 1 : n ∈ Z}


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Example (1.2.1)

Write out sets in builder form that equal the following sets.

1. {0}2. {1, 2, 3}3. {1, 4, 7, 10, ...}

4. {red, green, blue}5. {a, e, i, o, u}6. {1, 4, 9, 16, 25, ...}

Solution

1. {x ∈ Z : x = 0} or {x ∈ R : x2 = 0}2. {x ∈ N : x ≤ 3} or {x ∈ Z : 1 ≤ x ≤ 3}3. {3n− 2 : n ∈ N} or {n : n = 3k − 2, k ∈ N}4. {color : basic primary colors}5. {English alphabet : letters are vowels.}6. {n2 : n ∈ N} or {n : n = k2, k ∈ N}


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Example (1.2.1)


1. {0}2. {1, 2, 3}3. {1, 4, 7, 10, ...}


Solution

1. {x ∈ Z : x = 0} or {x ∈ R : x2 = 0}

2. {x ∈ N : x ≤ 3} or {x ∈ Z : 1 ≤ x ≤ 3}3. {3n− 2 : n ∈ N} or {n : n = 3k − 2, k ∈ N}4. {color : basic primary colors}5. {English alphabet : letters are vowels.}6. {n2 : n ∈ N} or {n : n = k2, k ∈ N}


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Example (1.2.1)


1. {0}2. {1, 2, 3}3. {1, 4, 7, 10, ...}


Solution

1. {x ∈ Z : x = 0} or {x ∈ R : x2 = 0}2. {x ∈ N : x ≤ 3} or {x ∈ Z : 1 ≤ x ≤ 3}

3. {3n− 2 : n ∈ N} or {n : n = 3k − 2, k ∈ N}4. {color : basic primary colors}5. {English alphabet : letters are vowels.}6. {n2 : n ∈ N} or {n : n = k2, k ∈ N}


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Example (1.2.1)


1. {0}2. {1, 2, 3}3. {1, 4, 7, 10, ...}


Solution

1. {x ∈ Z : x = 0} or {x ∈ R : x2 = 0}2. {x ∈ N : x ≤ 3} or {x ∈ Z : 1 ≤ x ≤ 3}3. {3n− 2 : n ∈ N} or {n : n = 3k − 2, k ∈ N}

4. {color : basic primary colors}5. {English alphabet : letters are vowels.}6. {n2 : n ∈ N} or {n : n = k2, k ∈ N}


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Example (1.2.1)


1. {0}2. {1, 2, 3}3. {1, 4, 7, 10, ...}


Solution

1. {x ∈ Z : x = 0} or {x ∈ R : x2 = 0}2. {x ∈ N : x ≤ 3} or {x ∈ Z : 1 ≤ x ≤ 3}3. {3n− 2 : n ∈ N} or {n : n = 3k − 2, k ∈ N}4. {color : basic primary colors}

5. {English alphabet : letters are vowels.}6. {n2 : n ∈ N} or {n : n = k2, k ∈ N}


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Example (1.2.1)


1. {0}2. {1, 2, 3}3. {1, 4, 7, 10, ...}


Solution

1. {x ∈ Z : x = 0} or {x ∈ R : x2 = 0}2. {x ∈ N : x ≤ 3} or {x ∈ Z : 1 ≤ x ≤ 3}3. {3n− 2 : n ∈ N} or {n : n = 3k − 2, k ∈ N}4. {color : basic primary colors}5. {English alphabet : letters are vowels.}

6. {n2 : n ∈ N} or {n : n = k2, k ∈ N}


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Example (1.2.1)


1. {0}2. {1, 2, 3}3. {1, 4, 7, 10, ...}


Solution

1. {x ∈ Z : x = 0} or {x ∈ R : x2 = 0}2. {x ∈ N : x ≤ 3} or {x ∈ Z : 1 ≤ x ≤ 3}3. {3n− 2 : n ∈ N} or {n : n = 3k − 2, k ∈ N}4. {color : basic primary colors}5. {English alphabet : letters are vowels.}6. {n2 : n ∈ N} or {n : n = k2, k ∈ N}


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Introduction Russell’s Paradox

1.3 Russell’s Paradox

Bertrand Arthur William Russell

18 May 1872 - 2 February 1970

He discovered the paradox in May or June 1901.

He was a British philosopher, logician, mathematician and historian.


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What is Russell’s paradox ?

A paradox is a statement that, despite apparently sound reasoning from true premises, leads toa self-contradictory or a logically unacceptable conclusion.

All the men in a village eithershave themselves or are shaved by a barber (himself a man fromthe village). The baber claims to shave only the male villagerwho do not shave themselves. So who shaves the baber ?

If he shave himself , then he do not shave himself .

If he do not shave himself , then he shave himself .

A = {X : X /∈ X}


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All the men in a village eithershave themselves or are shaved by a barber (himself a man fromthe village). The baber claims to shave only the male villagerwho do not shave themselves.

So who shaves the baber ?



A = {X : X /∈ X}


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A = {X : X /∈ X}


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A = {X : X /∈ X}


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A = {X : X /∈ X}


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A = {X : X /∈ X}


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Zermelo’s solution

BERTRAND RUSSELL confounded mathematicians when he published his famous paradox in1903. Bertrand Russell’s discovery of this paradox in 1901 dealt a blow to one of his fellowmathematicians.

Next, Zermelo’s solution to Russell’s paradox was to replace the axiom

for every formula p(x) there is a set y = {x : p(x)}

by the axiom (See axiom of Specification in section 2.1)

for every formula p(x) and every set A there is a set

B = {x : x is in A and p(x)}.


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Zermelo’s solution

BERTRAND RUSSELL confounded mathematicians when he published his famous paradox in1903. Bertrand Russell’s discovery of this paradox in 1901 dealt a blow to one of his fellowmathematicians.Next, Zermelo’s solution to Russell’s paradox was to replace the axiom

for every formula p(x) there is a set y = {x : p(x)}

by the axiom (See axiom of Specification in section 2.1)

for every formula p(x) and every set A there is a set

B = {x : x is in A and p(x)}.


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The Axiomatic Set Theory

Chapter 2 The Axiomatic Set Theory


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Ernst Friedrich Ferdinand Zermelo

27 July 1871 to 21 May 1953

He was German logician and mathematician.

He finished his doctorate in 1894 at the University of Berlin.

He work has major implications for the foundations of mathematics.

He is known for his role in Zermelo-Fraenkel axiomatic set theory.


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Abraham Halevi (Adolf) Fraenkel

17 February 1891 to 15 October 1965He was a German-born Israeli mathematicianHe studied mathematics at the University of Munich, University of Berlin, University ofMarburgHe became the first Dean of the Faculty of Mathematics, Hebrew University of Jerusalem.He is known for his contributions to axiomatic set theory or Zermelo-Fraenkel axiomaticset theory.


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What is ZFC and why is it important ?

ZFC → Zermelo-Fraenkel set theory with axiom of Choice


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The Axiomatic Set Theory The Axiom of Equality

2.1 The Axiom of Equality

Axiom 2.1.1 (The Existential Axiom)

There is a set at least one.

Axiom 2.1.2 (The Axiom of Extensionality)

Two sets are equal (are the same set) if they have the same elements.

∀x ∀y [∀z (z ∈ x ↔ z ∈ y) → (x = y)]

A = B ↔ ∀x (x ∈ A ↔ x ∈ B)

↔ ∀x[(x ∈ A → x ∈ B) ∧ (x ∈ B → x ∈ A)]

A ̸= B ↔ ∃x [(x /∈ A ∧ x ∈ B) ∨ (x ∈ A ∧ x /∈ B)]


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∀x ∀y [∀z (z ∈ x ↔ z ∈ y) → (x = y)]

A = B ↔ ∀x (x ∈ A ↔ x ∈ B)

↔ ∀x[(x ∈ A → x ∈ B) ∧ (x ∈ B → x ∈ A)]

A ̸= B ↔ ∃x [(x /∈ A ∧ x ∈ B) ∨ (x ∈ A ∧ x /∈ B)]


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∀x ∀y [∀z (z ∈ x ↔ z ∈ y) → (x = y)]

A = B ↔ ∀x (x ∈ A ↔ x ∈ B)

↔ ∀x[(x ∈ A → x ∈ B) ∧ (x ∈ B → x ∈ A)]

A ̸= B ↔ ∃x [(x /∈ A ∧ x ∈ B) ∨ (x ∈ A ∧ x /∈ B)]


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∀x ∀y [∀z (z ∈ x ↔ z ∈ y) → (x = y)]

A = B ↔ ∀x (x ∈ A ↔ x ∈ B)

↔ ∀x[(x ∈ A → x ∈ B) ∧ (x ∈ B → x ∈ A)]

A ̸= B ↔ ∃x [(x /∈ A ∧ x ∈ B) ∨ (x ∈ A ∧ x /∈ B)]


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∀x ∀y [∀z (z ∈ x ↔ z ∈ y) → (x = y)]

A = B ↔ ∀x (x ∈ A ↔ x ∈ B)

↔ ∀x[(x ∈ A → x ∈ B) ∧ (x ∈ B → x ∈ A)]

A ̸= B ↔ ∃x [(x /∈ A ∧ x ∈ B) ∨ (x ∈ A ∧ x /∈ B)]


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Theorem (2.1.3)

Let A,B and C be sets. Then

1. A = A, (Reflexive)

2. if A = B, then B = A, (Symmetric)

3. if A = B and B = C, then A = C. (Transitive)


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Theorem (2.1.3)






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Theorem (2.1.3)






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Proof .

1. A = A

Since ∀x (x ∈ A ↔ x ∈ A) is holds, A = A.

2. A = B → B = AAssume that A = B. Then

∀x (x ∈ A ↔ x ∈ B).

We obtain∀x (x ∈ B ↔ x ∈ A).

So, B = A.

3. (A = B) ∧ (B = C) → (A = C)

Assume that A = B and B = C. Then

∀x (x ∈ A ↔ x ∈ B) and ∀x (x ∈ B ↔ x ∈ C).

We obtain∀x (x ∈ A ↔ x ∈ C).

So, A = C.


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Proof .

1. A = ASince ∀x (x ∈ A ↔ x ∈ A) is holds, A = A.


∀x (x ∈ A ↔ x ∈ B).


So, B = A.

3. (A = B) ∧ (B = C) → (A = C)


∀x (x ∈ A ↔ x ∈ B) and ∀x (x ∈ B ↔ x ∈ C).


So, A = C.


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Proof .


2. A = B → B = A

Assume that A = B. Then∀x (x ∈ A ↔ x ∈ B).


So, B = A.

3. (A = B) ∧ (B = C) → (A = C)


∀x (x ∈ A ↔ x ∈ B) and ∀x (x ∈ B ↔ x ∈ C).


So, A = C.


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Proof .


2. A = B → B = AAssume that A = B.

Then∀x (x ∈ A ↔ x ∈ B).


So, B = A.

3. (A = B) ∧ (B = C) → (A = C)


∀x (x ∈ A ↔ x ∈ B) and ∀x (x ∈ B ↔ x ∈ C).


So, A = C.


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Proof .



∀x (x ∈ A ↔ x ∈ B).


So, B = A.

3. (A = B) ∧ (B = C) → (A = C)


∀x (x ∈ A ↔ x ∈ B) and ∀x (x ∈ B ↔ x ∈ C).


So, A = C.


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Proof .



∀x (x ∈ A ↔ x ∈ B).


So, B = A.

3. (A = B) ∧ (B = C) → (A = C)


∀x (x ∈ A ↔ x ∈ B) and ∀x (x ∈ B ↔ x ∈ C).


So, A = C.


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Proof .



∀x (x ∈ A ↔ x ∈ B).


So, B = A.

3. (A = B) ∧ (B = C) → (A = C)


∀x (x ∈ A ↔ x ∈ B) and ∀x (x ∈ B ↔ x ∈ C).


So, A = C.


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Proof .



∀x (x ∈ A ↔ x ∈ B).


So, B = A.

3. (A = B) ∧ (B = C) → (A = C)

Assume that A = B and B = C.

Then

∀x (x ∈ A ↔ x ∈ B) and ∀x (x ∈ B ↔ x ∈ C).


So, A = C.


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Proof .



∀x (x ∈ A ↔ x ∈ B).


So, B = A.

3. (A = B) ∧ (B = C) → (A = C)


∀x (x ∈ A ↔ x ∈ B) and ∀x (x ∈ B ↔ x ∈ C).


So, A = C.


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Proof .



∀x (x ∈ A ↔ x ∈ B).


So, B = A.

3. (A = B) ∧ (B = C) → (A = C)


∀x (x ∈ A ↔ x ∈ B) and ∀x (x ∈ B ↔ x ∈ C).


So, A = C.


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Theorem (2.1.4)

Let A1, A2, ..., An be sets, where n ∈ N with n ≥ 2.

If (A1 = A2) ∧ (A2 = A3) ∧ ... ∧ (An−1 = An), then A1 = An


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Proof .

Let n ∈ N such that n ≥ 2 and let P (n) be the statement

if (A1 = A2) ∧ (A2 = A3) ∧ ... ∧ (An−1 = An), then A1 = An.

Basic Step: It is easy to see that (A1 = A2) → (A1 = A2). So, P (2) holds.Inductive Step: Assume that P (k) holds where k ∈ N with k ≥ 2, i.e.,

if (A1 = A2) ∧ (A2 = A3) ∧ ... ∧ (Ak−1 = Ak) , then A1 = Ak .

Next, we will prove that P (k + 1) holds, i.e.,if (A1 = A2) ∧ (A2 = A3) ∧ ...∧(Ak = Ak+1), then A1 = Ak+1.

Supose that (A1 = A2) ∧ (A2 = A3) ∧ ... ∧ (Ak−1 = Ak) ∧(Ak = Ak+1).By induction hypothesis, A1 = Ak . By transitive law, it implies that

(A1 = Ak) ∧ (Ak = Ak+1) → (A1 = Ak+1).

Thus, P (k + 1) holds.


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Proof .



Basic Step:

It is easy to see that (A1 = A2) → (A1 = A2). So, P (2) holds.Inductive Step: Assume that P (k) holds where k ∈ N with k ≥ 2, i.e.,




(A1 = Ak) ∧ (Ak = Ak+1) → (A1 = Ak+1).



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Proof .



Basic Step: It is easy to see that (A1 = A2) → (A1 = A2). So, P (2) holds.

Inductive Step: Assume that P (k) holds where k ∈ N with k ≥ 2, i.e.,




(A1 = Ak) ∧ (Ak = Ak+1) → (A1 = Ak+1).



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Proof .



Basic Step: It is easy to see that (A1 = A2) → (A1 = A2). So, P (2) holds.Inductive Step:

Assume that P (k) holds where k ∈ N with k ≥ 2, i.e.,




(A1 = Ak) ∧ (Ak = Ak+1) → (A1 = Ak+1).



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Proof .







(A1 = Ak) ∧ (Ak = Ak+1) → (A1 = Ak+1).



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Proof .







(A1 = Ak) ∧ (Ak = Ak+1) → (A1 = Ak+1).



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Proof .







(A1 = Ak) ∧ (Ak = Ak+1) → (A1 = Ak+1).



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Proof .






Supose that (A1 = A2) ∧ (A2 = A3) ∧ ... ∧ (Ak−1 = Ak) ∧(Ak = Ak+1).

By induction hypothesis, A1 = Ak . By transitive law, it implies that

(A1 = Ak) ∧ (Ak = Ak+1) → (A1 = Ak+1).



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Proof .







(A1 = Ak) ∧ (Ak = Ak+1) → (A1 = Ak+1).



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Proof .







(A1 = Ak) ∧ (Ak = Ak+1) → (A1 = Ak+1).



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Axiom 2.1.5 (Axiom of Specification)

To every set A and to every condition p(x) there corresponds a set B whoseelements are exactly those elements x of A for which p(x) holds.

x ∈ B ↔ (x ∈ A ∧ p(x) holds )

In other word, for every set A, there is a set B satisfying

B = {x ∈ A : p(x) holds }.


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Axiom 2.1.5 (Axiom of Specification)

To every set A and to every condition p(x) there corresponds a set B whoseelements are exactly those elements x of A for which p(x) holds.

x ∈ B ↔ (x ∈ A ∧ p(x) holds )

In other word, for every set A, there is a set B satisfying

B = {x ∈ A : p(x) holds }.


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Theorem (2.1.6)

There is no a set A satisfying

every set x, x ∈ A.

We conclude that there is no a set of all sets.


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Theorem (2.1.6)

There is no a set A satisfying

every set x, x ∈ A.

We conclude that there is no a set of all sets.


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Proof .

We will prove that there is a set B such that B /∈ A for every set A.

Let A be a set and let p(x) be statement x /∈ x. By axiom of specification, there is a set

B = {x ∈ A : x /∈ x}.

Since (B ∈ B ∨B /∈ B) is a tautology, we have two tautologies

[B ∈ A ∧ (B ∈ B ∨B /∈ B)] ↔ B ∈ A

[(B ∈ A ∧B ∈ B) ∨ (B ∈ A ∧B /∈ B)] ↔ B ∈ A

Consider B ∈ A ∧B ∈ B

B ∈ B → B ∈ A ∧B /∈ B It is impossible.

Consider B ∈ A ∧B /∈ B

B ∈ A ∧B /∈ B → B ∈ B It is impossible.

Hence, B /∈ A.


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Proof .

We will prove that there is a set B such that B /∈ A for every set A.Let A be a set and let p(x) be statement x /∈ x.

By axiom of specification, there is a set

B = {x ∈ A : x /∈ x}.


[B ∈ A ∧ (B ∈ B ∨B /∈ B)] ↔ B ∈ A

[(B ∈ A ∧B ∈ B) ∨ (B ∈ A ∧B /∈ B)] ↔ B ∈ A





Hence, B /∈ A.


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Proof .

We will prove that there is a set B such that B /∈ A for every set A.Let A be a set and let p(x) be statement x /∈ x. By axiom of specification, there is a set

B = {x ∈ A : x /∈ x}.


[B ∈ A ∧ (B ∈ B ∨B /∈ B)] ↔ B ∈ A

[(B ∈ A ∧B ∈ B) ∨ (B ∈ A ∧B /∈ B)] ↔ B ∈ A





Hence, B /∈ A.


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Proof .


B = {x ∈ A : x /∈ x}.

Since (B ∈ B ∨B /∈ B) is a tautology,

we have two tautologies

[B ∈ A ∧ (B ∈ B ∨B /∈ B)] ↔ B ∈ A

[(B ∈ A ∧B ∈ B) ∨ (B ∈ A ∧B /∈ B)] ↔ B ∈ A





Hence, B /∈ A.


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Proof .


B = {x ∈ A : x /∈ x}.


[B ∈ A ∧ (B ∈ B ∨B /∈ B)] ↔ B ∈ A

[(B ∈ A ∧B ∈ B) ∨ (B ∈ A ∧B /∈ B)] ↔ B ∈ A





Hence, B /∈ A.


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Proof .


B = {x ∈ A : x /∈ x}.


[B ∈ A ∧ (B ∈ B ∨B /∈ B)] ↔ B ∈ A

[(B ∈ A ∧B ∈ B) ∨ (B ∈ A ∧B /∈ B)] ↔ B ∈ A





Hence, B /∈ A.


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Proof .


B = {x ∈ A : x /∈ x}.


[B ∈ A ∧ (B ∈ B ∨B /∈ B)] ↔ B ∈ A

[(B ∈ A ∧B ∈ B) ∨ (B ∈ A ∧B /∈ B)] ↔ B ∈ A





Hence, B /∈ A.


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Proof .


B = {x ∈ A : x /∈ x}.


[B ∈ A ∧ (B ∈ B ∨B /∈ B)] ↔ B ∈ A

[(B ∈ A ∧B ∈ B) ∨ (B ∈ A ∧B /∈ B)] ↔ B ∈ A





Hence, B /∈ A.


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Proof .


B = {x ∈ A : x /∈ x}.


[B ∈ A ∧ (B ∈ B ∨B /∈ B)] ↔ B ∈ A

[(B ∈ A ∧B ∈ B) ∨ (B ∈ A ∧B /∈ B)] ↔ B ∈ A





Hence, B /∈ A.


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Proof .


B = {x ∈ A : x /∈ x}.


[B ∈ A ∧ (B ∈ B ∨B /∈ B)] ↔ B ∈ A

[(B ∈ A ∧B ∈ B) ∨ (B ∈ A ∧B /∈ B)] ↔ B ∈ A





Hence, B /∈ A.


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Theorem (2.1.7)

There exists a unique set which has no elements.

Proof .

By the existential axiom , there is a set A. Let p(x) be statement x ̸= x. Theaxiom of specification implies that there is a set

B = {x ∈ A : x ̸= x}.

Since p(x) is not true, B has no element.

Next, we will show that it has unique. Let B1 and B2 be sets which have no elements.Then

∀x [x ∈ B1 ↔ x ∈ B2] is true.

Thus, B1 = B2.

Definition (2.1.8)

The empty set or null set denoted by ∅ or {} is a set which has no elements.


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Theorem (2.1.7)


Proof .

By the existential axiom , there is a set A.

Let p(x) be statement x ̸= x. Theaxiom of specification implies that there is a set

B = {x ∈ A : x ̸= x}.



∀x [x ∈ B1 ↔ x ∈ B2] is true.

Thus, B1 = B2.

Definition (2.1.8)



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Theorem (2.1.7)


Proof .

By the existential axiom , there is a set A. Let p(x) be statement x ̸= x.

Theaxiom of specification implies that there is a set

B = {x ∈ A : x ̸= x}.



∀x [x ∈ B1 ↔ x ∈ B2] is true.

Thus, B1 = B2.

Definition (2.1.8)



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Theorem (2.1.7)


Proof .


B = {x ∈ A : x ̸= x}.



∀x [x ∈ B1 ↔ x ∈ B2] is true.

Thus, B1 = B2.

Definition (2.1.8)



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Theorem (2.1.7)


Proof .


B = {x ∈ A : x ̸= x}.



∀x [x ∈ B1 ↔ x ∈ B2] is true.

Thus, B1 = B2.

Definition (2.1.8)



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Theorem (2.1.7)


Proof .


B = {x ∈ A : x ̸= x}.


Next, we will show that it has unique.

Let B1 and B2 be sets which have no elements.Then

∀x [x ∈ B1 ↔ x ∈ B2] is true.

Thus, B1 = B2.

Definition (2.1.8)



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Theorem (2.1.7)


Proof .


B = {x ∈ A : x ̸= x}.


Next, we will show that it has unique. Let B1 and B2 be sets which have no elements.

Then∀x [x ∈ B1 ↔ x ∈ B2] is true.

Thus, B1 = B2.

Definition (2.1.8)



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Theorem (2.1.7)


Proof .


B = {x ∈ A : x ̸= x}.



∀x [x ∈ B1 ↔ x ∈ B2] is true.

Thus, B1 = B2.

Definition (2.1.8)



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Theorem (2.1.7)


Proof .


B = {x ∈ A : x ̸= x}.



∀x [x ∈ B1 ↔ x ∈ B2] is true.

Thus, B1 = B2.

Definition (2.1.8)



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Theorem (2.1.7)


Proof .


B = {x ∈ A : x ̸= x}.



∀x [x ∈ B1 ↔ x ∈ B2] is true.

Thus, B1 = B2.

Definition (2.1.8)



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Take a BreakFor 10 Minutes


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Subsets

Definition (2.1.9)

Let A and B be sets. Then,

A is a subset of B, denoted by A ⊆ B , if all elements in A belong to B.

In orther word,

A ⊆ B ↔ ∀x [x ∈ A → x ∈ B]

A * B ↔ ¬∀x [x ∈ A → x ∈ B]

↔ ∃x [x ∈ A ∧ x /∈ B]


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Subsets

Definition (2.1.9)

Let A and B be sets. Then,

A is a subset of B, denoted by A ⊆ B , if all elements in A belong to B.

In orther word,

A ⊆ B ↔ ∀x [x ∈ A → x ∈ B]

A * B ↔ ¬∀x [x ∈ A → x ∈ B]

↔ ∃x [x ∈ A ∧ x /∈ B]


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Example (2.1.10)

Let A = {1, 2, 3, 4, 5}. Find all subsets of A satisfying the followingcondition.

1. a single element

2. two elements

3. two elements and product of them less than 6

4. three elements and sum of them is 9


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Solution

A = {1, 2, 3, 4, 5}

1. a single element

{1}, {2}, {3}, {4} and {5}

2. two elements{1, 2} {1, 3} {1, 4} {1, 5}{2, 3} {2, 4} {2, 5}{3, 4} {3, 5}{4, 5}


{1, 2}, {1, 3}, {1, 4} and {1, 5}


{1, 3, 5}, and {2, 3, 4}


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Solution

A = {1, 2, 3, 4, 5}

1. a single element

{1}, {2}, {3}, {4} and {5}

2. two elements{1, 2} {1, 3} {1, 4} {1, 5}{2, 3} {2, 4} {2, 5}{3, 4} {3, 5}{4, 5}


{1, 2}, {1, 3}, {1, 4} and {1, 5}


{1, 3, 5}, and {2, 3, 4}


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Solution

A = {1, 2, 3, 4, 5}

1. a single element

{1}, {2}, {3}, {4} and {5}

2. two elements{1, 2} {1, 3} {1, 4} {1, 5}{2, 3} {2, 4} {2, 5}{3, 4} {3, 5}{4, 5}


{1, 2}, {1, 3}, {1, 4} and {1, 5}


{1, 3, 5}, and {2, 3, 4}


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Solution

A = {1, 2, 3, 4, 5}

1. a single element

{1}, {2}, {3}, {4} and {5}

2. two elements

{1, 2} {1, 3} {1, 4} {1, 5}{2, 3} {2, 4} {2, 5}{3, 4} {3, 5}{4, 5}


{1, 2}, {1, 3}, {1, 4} and {1, 5}


{1, 3, 5}, and {2, 3, 4}


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Solution

A = {1, 2, 3, 4, 5}

1. a single element

{1}, {2}, {3}, {4} and {5}

2. two elements{1, 2} {1, 3} {1, 4} {1, 5}{2, 3} {2, 4} {2, 5}{3, 4} {3, 5}{4, 5}


{1, 2}, {1, 3}, {1, 4} and {1, 5}


{1, 3, 5}, and {2, 3, 4}


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Solution

A = {1, 2, 3, 4, 5}

1. a single element

{1}, {2}, {3}, {4} and {5}

2. two elements{1, 2} {1, 3} {1, 4} {1, 5}{2, 3} {2, 4} {2, 5}{3, 4} {3, 5}{4, 5}


{1, 2}, {1, 3}, {1, 4} and {1, 5}


{1, 3, 5}, and {2, 3, 4}


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Solution

A = {1, 2, 3, 4, 5}

1. a single element

{1}, {2}, {3}, {4} and {5}

2. two elements{1, 2} {1, 3} {1, 4} {1, 5}{2, 3} {2, 4} {2, 5}{3, 4} {3, 5}{4, 5}


{1, 2}, {1, 3}, {1, 4} and {1, 5}


{1, 3, 5}, and {2, 3, 4}


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Solution

A = {1, 2, 3, 4, 5}

1. a single element

{1}, {2}, {3}, {4} and {5}

2. two elements{1, 2} {1, 3} {1, 4} {1, 5}{2, 3} {2, 4} {2, 5}{3, 4} {3, 5}{4, 5}


{1, 2}, {1, 3}, {1, 4} and {1, 5}


{1, 3, 5}, and {2, 3, 4}


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Solution

A = {1, 2, 3, 4, 5}

1. a single element

{1}, {2}, {3}, {4} and {5}

2. two elements{1, 2} {1, 3} {1, 4} {1, 5}{2, 3} {2, 4} {2, 5}{3, 4} {3, 5}{4, 5}


{1, 2}, {1, 3}, {1, 4} and {1, 5}


{1, 3, 5}, and {2, 3, 4}


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Theorem (2.1.11)

The empty set is a subset of all sets.

Proof .

Since ∀x [x ∈ ∅ → x ∈ A] holds for all A, ∅ ⊆ A.

Theorem (2.1.12)


1. A ⊆ A, (Reflexive)

2. if A ⊆ B and B ⊆ C, then A ⊆ C. (Transitive)


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Theorem (2.1.11)


Proof .


Theorem (2.1.12)





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Theorem (2.1.11)


Proof .


Theorem (2.1.12)





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Proof .

1. Reflexive A ⊆ A

Since ∀x (x ∈ A → x ∈ A) is holds, A ⊆ A.

2. Transitive (A ⊆ B) ∧ (B ⊆ C) → (A ⊆ C)

Assume that A ⊆ B and B ⊆ C. Then

∀x (x ∈ A → x ∈ B) and ∀x (x ∈ B → x ∈ C).

We obtain∀x (x ∈ A → x ∈ C).

So, A ⊆ C. �


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Proof .





∀x (x ∈ A → x ∈ B) and ∀x (x ∈ B → x ∈ C).


So, A ⊆ C. �


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Proof .





∀x (x ∈ A → x ∈ B) and ∀x (x ∈ B → x ∈ C).


So, A ⊆ C. �


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Proof .




Assume that A ⊆ B and B ⊆ C.

Then

∀x (x ∈ A → x ∈ B) and ∀x (x ∈ B → x ∈ C).


So, A ⊆ C. �


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Proof .





∀x (x ∈ A → x ∈ B) and ∀x (x ∈ B → x ∈ C).


So, A ⊆ C. �


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Proof .





∀x (x ∈ A → x ∈ B) and ∀x (x ∈ B → x ∈ C).


So, A ⊆ C. �


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Proof .





∀x (x ∈ A → x ∈ B) and ∀x (x ∈ B → x ∈ C).


So, A ⊆ C. �


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Theorem (2.1.13)


If (A1 ⊆ A2) ∧ (A2 ⊆ A3) ∧ ... ∧ (An−1 ⊆ An), then A1 ⊆ An.

Proof. Assignment 1


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Theorem (2.1.13)


If (A1 ⊆ A2) ∧ (A2 ⊆ A3) ∧ ... ∧ (An−1 ⊆ An), then A1 ⊆ An.

Proof. Assignment 1


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Theorem (2.1.14)

Let A and B be sets. Then

A = B if and only if A ⊆ B and B ⊆ A.

Proof .


A = B ↔ ∀x[x ∈ A ↔ x ∈ B]

↔ ∀x[(x ∈ A → x ∈ B) ∧ (x ∈ B → x ∈ A)]

↔ ∀x(x ∈ A → x ∈ B) ∧ ∀x(x ∈ B → x ∈ A)

↔ A ⊆ B ∧B ⊆ A

Theorem (2.1.15)

For any set A, A = ∅ if and only if A ⊆ ∅.

Proof. By theorem 2.1.11 and 2.1.14, they imply this theorem. �


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Theorem (2.1.14)



Proof .

Let A and B be sets.

Then

A = B ↔ ∀x[x ∈ A ↔ x ∈ B]

↔ ∀x[(x ∈ A → x ∈ B) ∧ (x ∈ B → x ∈ A)]

↔ ∀x(x ∈ A → x ∈ B) ∧ ∀x(x ∈ B → x ∈ A)

↔ A ⊆ B ∧B ⊆ A

Theorem (2.1.15)




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Theorem (2.1.14)



Proof .


A = B ↔ ∀x[x ∈ A ↔ x ∈ B]

↔ ∀x[(x ∈ A → x ∈ B) ∧ (x ∈ B → x ∈ A)]

↔ ∀x(x ∈ A → x ∈ B) ∧ ∀x(x ∈ B → x ∈ A)

↔ A ⊆ B ∧B ⊆ A

Theorem (2.1.15)




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Theorem (2.1.14)



Proof .


A = B ↔ ∀x[x ∈ A ↔ x ∈ B]

↔ ∀x[(x ∈ A → x ∈ B) ∧ (x ∈ B → x ∈ A)]

↔ ∀x(x ∈ A → x ∈ B) ∧ ∀x(x ∈ B → x ∈ A)

↔ A ⊆ B ∧B ⊆ A

Theorem (2.1.15)




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Theorem (2.1.14)



Proof .


A = B ↔ ∀x[x ∈ A ↔ x ∈ B]

↔ ∀x[(x ∈ A → x ∈ B) ∧ (x ∈ B → x ∈ A)]

↔ ∀x(x ∈ A → x ∈ B) ∧ ∀x(x ∈ B → x ∈ A)

↔ A ⊆ B ∧B ⊆ A

Theorem (2.1.15)




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Example (2.1.16)

Let A,B and C be sets. Prove that

if A ⊆ B and B ⊆ C and C ⊆ A, then A = B = C.

Proof .

Let A,B and C be sets. Assume that A ⊆ B and B ⊆ C and C ⊆ A .By transitive law,

A ⊆ B ∧B ⊆ C → A ⊆ C.

ThenA ⊆ C ∧ C ⊆ A → A = C.

We obtainC ⊆ B ∧B ⊆ C → B = C.

Thus, A = B = C . �

In other word,

A ⊆ B ⊆ C ⊆ A → A = B = C


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Example (2.1.16)



Proof .

Let A,B and C be sets.

Assume that A ⊆ B and B ⊆ C and C ⊆ A .By transitive law,

A ⊆ B ∧B ⊆ C → A ⊆ C.

ThenA ⊆ C ∧ C ⊆ A → A = C.


Thus, A = B = C . �

In other word,

A ⊆ B ⊆ C ⊆ A → A = B = C


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Example (2.1.16)



Proof .

Let A,B and C be sets. Assume that A ⊆ B and B ⊆ C and C ⊆ A .

By transitive law,A ⊆ B ∧B ⊆ C → A ⊆ C.

ThenA ⊆ C ∧ C ⊆ A → A = C.


Thus, A = B = C . �

In other word,

A ⊆ B ⊆ C ⊆ A → A = B = C


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Example (2.1.16)



Proof .


A ⊆ B ∧B ⊆ C → A ⊆ C.

ThenA ⊆ C ∧ C ⊆ A → A = C.


Thus, A = B = C . �

In other word,

A ⊆ B ⊆ C ⊆ A → A = B = C


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Example (2.1.16)



Proof .


A ⊆ B ∧B ⊆ C → A ⊆ C.

ThenA ⊆ C ∧ C ⊆ A → A = C.


Thus, A = B = C . �

In other word,

A ⊆ B ⊆ C ⊆ A → A = B = C


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Example (2.1.16)



Proof .


A ⊆ B ∧B ⊆ C → A ⊆ C.

ThenA ⊆ C ∧ C ⊆ A → A = C.


Thus, A = B = C . �

In other word,

A ⊆ B ⊆ C ⊆ A → A = B = C


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Example (2.1.16)



Proof .


A ⊆ B ∧B ⊆ C → A ⊆ C.

ThenA ⊆ C ∧ C ⊆ A → A = C.


Thus, A = B = C . �

In other word,

A ⊆ B ⊆ C ⊆ A → A = B = C


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Example (2.1.16)



Proof .


A ⊆ B ∧B ⊆ C → A ⊆ C.

ThenA ⊆ C ∧ C ⊆ A → A = C.


Thus, A = B = C . �

In other word,

A ⊆ B ⊆ C ⊆ A → A = B = C


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Example (2.1.17)

Give an example(s) such that A ⊆ B and A ⊆ C and C ⊆ B.

1. A = ∅ and C = ∅ and B = {1, 2}

2. A = ∅ and C = {1} and B = {1, 2}

We can choose A,B and C such that A ⊆ C ⊆ B.Proof. Assignment 1


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Example (2.1.17)


1. A = ∅ and C = ∅ and B = {1, 2}

2. A = ∅ and C = {1} and B = {1, 2}



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Example (2.1.17)


1. A = ∅ and C = ∅ and B = {1, 2}

2. A = ∅ and C = {1} and B = {1, 2}



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Example (2.1.17)


1. A = ∅ and C = ∅ and B = {1, 2}

2. A = ∅ and C = {1} and B = {1, 2}

We can choose A,B and C such that A ⊆ C ⊆ B.

Proof. Assignment 1


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Example (2.1.17)


1. A = ∅ and C = ∅ and B = {1, 2}

2. A = ∅ and C = {1} and B = {1, 2}



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Example (2.1.18)

Let A be a set of all even numbers and

B be a set of all integers can be writen by sum of two odd numbers .

Show that A = B.

Solution

Set up A = {n : n = 2k, k ∈ Z}B = {a+ b : a = 2p+ 1, b = 2d+ 1, p, d ∈ Z}

(A ⊆ B)

Let n ∈ A. There exists k ∈ Z such that n = 2k. Choose p = k and d = −1. Then

n = 2k = (2k + 1)− 1 = (2k + 1) + (2(−1) + 1)

= (2p+ 1) + (2d+ 1) = a+ b ∈ B.

(B ⊆ A)

Let n ∈ B. There exists p, d ∈ Z such that a = 2p+ 1, b = 2d+ 1 and n = a+ b. Then

n = a+ b = (2p+ 1) + (2d+ 1) = 2(p+ d+ 1) = 2k ∈ A where k = p+ d+ 1.


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Example (2.1.18)



Show that A = B.

Solution


(A ⊆ B)


n = 2k = (2k + 1)− 1 = (2k + 1) + (2(−1) + 1)

= (2p+ 1) + (2d+ 1) = a+ b ∈ B.

(B ⊆ A)




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Example (2.1.18)



Show that A = B.

Solution


(A ⊆ B)


n = 2k = (2k + 1)− 1 = (2k + 1) + (2(−1) + 1)

= (2p+ 1) + (2d+ 1) = a+ b ∈ B.

(B ⊆ A)




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Example (2.1.18)



Show that A = B.

Solution


(A ⊆ B)

Let n ∈ A.

There exists k ∈ Z such that n = 2k. Choose p = k and d = −1. Then

n = 2k = (2k + 1)− 1 = (2k + 1) + (2(−1) + 1)

= (2p+ 1) + (2d+ 1) = a+ b ∈ B.

(B ⊆ A)




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Example (2.1.18)



Show that A = B.

Solution


(A ⊆ B)

Let n ∈ A. There exists k ∈ Z such that n = 2k.

Choose p = k and d = −1. Then

n = 2k = (2k + 1)− 1 = (2k + 1) + (2(−1) + 1)

= (2p+ 1) + (2d+ 1) = a+ b ∈ B.

(B ⊆ A)




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Example (2.1.18)



Show that A = B.

Solution


(A ⊆ B)

Let n ∈ A. There exists k ∈ Z such that n = 2k. Choose p = k and d = −1.

Then

n = 2k = (2k + 1)− 1 = (2k + 1) + (2(−1) + 1)

= (2p+ 1) + (2d+ 1) = a+ b ∈ B.

(B ⊆ A)




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Example (2.1.18)



Show that A = B.

Solution


(A ⊆ B)


n = 2k = (2k + 1)− 1 = (2k + 1) + (2(−1) + 1)

= (2p+ 1) + (2d+ 1) = a+ b ∈ B.

(B ⊆ A)




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Example (2.1.18)



Show that A = B.

Solution


(A ⊆ B)


n = 2k = (2k + 1)− 1 = (2k + 1) + (2(−1) + 1)

= (2p+ 1) + (2d+ 1) = a+ b ∈ B.

(B ⊆ A)




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Example (2.1.18)



Show that A = B.

Solution


(A ⊆ B)


n = 2k = (2k + 1)− 1 = (2k + 1) + (2(−1) + 1)

= (2p+ 1) + (2d+ 1) = a+ b ∈ B.

(B ⊆ A)

Let n ∈ B.

There exists p, d ∈ Z such that a = 2p+ 1, b = 2d+ 1 and n = a+ b. Then



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Example (2.1.18)



Show that A = B.

Solution


(A ⊆ B)


n = 2k = (2k + 1)− 1 = (2k + 1) + (2(−1) + 1)

= (2p+ 1) + (2d+ 1) = a+ b ∈ B.

(B ⊆ A)

Let n ∈ B. There exists p, d ∈ Z such that a = 2p+ 1, b = 2d+ 1 and n = a+ b.

Then



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Example (2.1.18)



Show that A = B.

Solution


(A ⊆ B)


n = 2k = (2k + 1)− 1 = (2k + 1) + (2(−1) + 1)

= (2p+ 1) + (2d+ 1) = a+ b ∈ B.

(B ⊆ A)




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Axiom 2.1.19 (Axiom of Pairing for Sets)

If x and y are sets, then there exists a set which contains x and y as elements.

∀x ∀y ∃z (x ∈ z ∧ y ∈ z)

In orther word,given sets A and B → {A,B} ⊆ P for some P.

For examples,

given {1} and {2} → {{1}, {2}}

given ∅ and {1} → {∅, {1}}

given Z+ and Z− → {Z+,Z−}

given {1} and {1} → {{1}, {1}}

given {1} and {{1}} → {{1}, {{1}}}


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∀x ∀y ∃z (x ∈ z ∧ y ∈ z)


For examples,

given {1} and {2} → {{1}, {2}}

given ∅ and {1} → {∅, {1}}


given {1} and {1} → {{1}, {1}}

given {1} and {{1}} → {{1}, {{1}}}


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∀x ∀y ∃z (x ∈ z ∧ y ∈ z)


For examples,

given {1} and {2} → {{1}, {2}}

given ∅ and {1} → {∅, {1}}


given {1} and {1} → {{1}, {1}}

given {1} and {{1}} → {{1}, {{1}}}


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∀x ∀y ∃z (x ∈ z ∧ y ∈ z)


For examples,

given {1} and {2} → {{1}, {2}}

given ∅ and {1} → {∅, {1}}


given {1} and {1} → {{1}, {1}}

given {1} and {{1}} → {{1}, {{1}}}


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∀x ∀y ∃z (x ∈ z ∧ y ∈ z)


For examples,

given {1} and {2} → {{1}, {2}}

given ∅ and {1} → {∅, {1}}


given {1} and {1} → {{1}, {1}}

given {1} and {{1}} → {{1}, {{1}}}


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∀x ∀y ∃z (x ∈ z ∧ y ∈ z)


For examples,

given {1} and {2} → {{1}, {2}}

given ∅ and {1} → {∅, {1}}


given {1} and {1} → {{1}, {1}}

given {1} and {{1}} → {{1}, {{1}}}


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Conclusion

The Existential Axiom There is a set at least one.The Axiom of Extensionality


∀x ∀y [∀z (z ∈ x ↔ z ∈ y) → (x = y)]

The Axiom of SpecificationTo every set A and to every condition p(x) there corresponds a set B whose elements areexactly those elements x of A for which p(x) holds.

x ∈ B ↔ (x ∈ A ∧ p(x) holds )

The Axiom of Pairing for SetsIf x and y are sets, then there exists a set which contains x and y as elements.

∀x ∀y ∃z (x ∈ z ∧ y ∈ z)

Subset

A ⊂ B ↔ ∀x(x ∈ A → x ∈ B)

A = B ↔ ∀x(x ∈ A ↔ x ∈ B)

A = B ↔ (A ⊆ B) ∧ (B ⊆ A)


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Conclusion

The Existential Axiom There is a set at least one.

The Axiom of ExtensionalityTwo sets are equal (are the same set) if they have the same elements.

∀x ∀y [∀z (z ∈ x ↔ z ∈ y) → (x = y)]


x ∈ B ↔ (x ∈ A ∧ p(x) holds )


∀x ∀y ∃z (x ∈ z ∧ y ∈ z)

Subset

A ⊂ B ↔ ∀x(x ∈ A → x ∈ B)

A = B ↔ ∀x(x ∈ A ↔ x ∈ B)

A = B ↔ (A ⊆ B) ∧ (B ⊆ A)


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Conclusion



∀x ∀y [∀z (z ∈ x ↔ z ∈ y) → (x = y)]


x ∈ B ↔ (x ∈ A ∧ p(x) holds )


∀x ∀y ∃z (x ∈ z ∧ y ∈ z)

Subset

A ⊂ B ↔ ∀x(x ∈ A → x ∈ B)

A = B ↔ ∀x(x ∈ A ↔ x ∈ B)

A = B ↔ (A ⊆ B) ∧ (B ⊆ A)


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Conclusion



∀x ∀y [∀z (z ∈ x ↔ z ∈ y) → (x = y)]


x ∈ B ↔ (x ∈ A ∧ p(x) holds )


∀x ∀y ∃z (x ∈ z ∧ y ∈ z)

Subset

A ⊂ B ↔ ∀x(x ∈ A → x ∈ B)

A = B ↔ ∀x(x ∈ A ↔ x ∈ B)

A = B ↔ (A ⊆ B) ∧ (B ⊆ A)


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Conclusion



∀x ∀y [∀z (z ∈ x ↔ z ∈ y) → (x = y)]


x ∈ B ↔ (x ∈ A ∧ p(x) holds )


∀x ∀y ∃z (x ∈ z ∧ y ∈ z)

Subset

A ⊂ B ↔ ∀x(x ∈ A → x ∈ B)

A = B ↔ ∀x(x ∈ A ↔ x ∈ B)

A = B ↔ (A ⊆ B) ∧ (B ⊆ A)


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Conclusion



∀x ∀y [∀z (z ∈ x ↔ z ∈ y) → (x = y)]


x ∈ B ↔ (x ∈ A ∧ p(x) holds )


∀x ∀y ∃z (x ∈ z ∧ y ∈ z)

Subset

A ⊂ B ↔ ∀x(x ∈ A → x ∈ B)

A = B ↔ ∀x(x ∈ A ↔ x ∈ B)

A = B ↔ (A ⊆ B) ∧ (B ⊆ A)


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Assignment 1 (30 Minutes)

1. Write out sets in builder form (only notation) that equal the following sets.

(a) (EVEN) {1, 2, 6, 24, 120, 720, ...}(b) (ODD) {0.1, 0.11, 0.111, 0.1111, ...}

2. Prove that(a) (EVEN) for any sets A1, A2, ..., An, where n ∈ N with n ≥ 2.

(A1 ⊆ A2) ∧ (A2 ⊆ A3) ∧ ... ∧ (An−1 ⊆ An) → A1 ⊆ An.(b) (ODD) for any sets A,B and C,

A ⊆ B ⊆ C ⊆ A → A = B = C.

3. (EVEN and ODD) Define

A = {n : n = 2k + 1, k ∈ Z} set of all odd numbersB = {a+ b : a = 2k, b = 2p+ 1, k, p ∈ Z} set of sum of even and odd numbers

Show that A = B.


Documents

MAT1202 SET THEORY: Week 1 · Introduction History of Set Theory 1.1 History of Set Theory Set theory is a branch of mathematical logic that studies sets, which informally are collections