@汥瑀瑯步渠 Set Theoryjhc.sjtu.edu.cn/~hongfeifu/lecture7.pdf · 2019-10-15 · Herbert B....

Set Theory

Hongfei Fu

John Hopcroft Center for Computer ScienceShanghai Jiao Tong University

Oct. 15th, 2019

Hongfei Fu (SJTU JHC) Set Theory Oct. 15th, 2019 1 / 48

Previous Lecture

Finishing the Logic Part

functional completeness

prenex normal form

inference rules in predicate logic


Today’s Topic

Set Theory （集合论）

naive set theory

axiomatic set theory


Textbooks

main textbook:

Kenneth H. Rosen, Discrete Mathematics and Its Applications, 7thedition [R]

auxiliary textbook:

Herbert B. Enderton, Elements of Set Theory [E]


Naive Set Theory(朴素集合论)

main textbook, Page 115 – Page 125


Naive Set Theory

What is a set?

A set is a collection of objects treated as a single entity.

Key Points

a collection of objects

a single entity (object)


Naive Set Theory

Membership

a: an element/object

A: a set

We write that

a∈A if a is an element/member of A;

a 6∈A if a is not an element of A;

A Basic Principle

It should hold that either a∈A or a 6∈A but not both.


Naive Set Theory

Equalityx , y : sets

Two sets x and y are equal (i.e., they are the same set), written x = y , ifthey have the same members.

Logical Description

a: a variable whose domain is all objects

x , y : two variables whose domains are both all sets

Then we have that

∀x∀y [(x = y)↔∀a ((a∈ x)↔ (a∈ y))] .


Naive Set Theory

Example

x = {a ∈ R | a2 − 3 · a + 2 = 0}y = {1, 2}

Then x = y by our equality principle.


Naive Set Theory

Subsetsx , y : sets

Then we say that x is a subset of y , written x ⊆ y , if every element of x isan element of y .

Logical Description



Then we have that

∀x∀y [(x ⊆ y)↔∀a ((a∈ x)→ (a∈ y))] .

Proper Subsets

x is a proper subset of y , written x ⊂ y , if x ⊆ y and x 6= y .


Naive Set Theory

Examples

{2, 3}⊆{1, 2, 3, 5}{0, 1}⊆{x ∈ R | x2 − 2 · x ≤ 0}N⊆Q

An Important Property


We have that

∀x∀y [(x = y)↔ ((x ⊆ y) ∧ (y ⊆ x))] .


Naive Set Theory

Set Union （有限情况下的并集）x , y : sets

Then the union of sets x and y , written as x ∪ y , is the set consisting ofthe members of x together with the members of y .

Logical Description


x , y : two sets

Then we have that

∀a [(a∈ x ∪ y)↔ ((a∈ x)∨ (a∈ y))]


Naive Set Theory

Set Intersection （有限情况下的交集）x , y : sets

Then the intersection of sets x and y , written as x ∩ y , is the setconsisting of those objects that are members of both x and y .

Logical Description


x , y : two sets

Then we have that

∀a [(a∈ x ∩ y)↔ ((a∈ x)∧ (a∈ y))] .


Naive Set Theory

Set-Theoretic Difference （差集）x , y : sets

Then the (set-theoretic) difference of the set x w.r.t y , written as x − y orx \ y , is the set consisting of those elements of x that are not in y .

Logical Description


x , y : two sets

Then we have that

∀a [(a∈ x \ y)↔ ((a∈ x)∧ (a 6∈ y))] .


Naive Set Theory

The Empty Set （空集）

The set ∅ is the set that contains no elements.

Logical Description

∀a (a 6∈ ∅) .


Naive Set Theory

Notations for Sets

a1, . . . , an: n objects

{a1, . . . , an}: the set consisting of exactly the elements a1, . . . , an

logical description: ∀a (a∈{a1, . . . , an} ↔∨n

i=1 a = ai )

Examples

{2, 3, 5}{1}{1, 2, 2}(= {1, 2})


Naive Set Theory

Notations for Sets

P(a): a statement with the predicate P and the variable a

{a | P(a)}: the set consisting of exactly the elements b such thatP(b) holds

logical description: ∀b (b∈{a | P(a)} ↔ P(b))

Examples

[0, 1] = {a | a is a real number and 0 ≤ a ≤ 1}x ∪ y = {a | a ∈ x or a ∈ y}x ∩ y = {a | a ∈ x and a ∈ y}x \ y = {a | a ∈ x and a 6∈ y}


Naive Set Theory

Notations for Sets




Examples

[0, 1] = {a | a is a real number and 0 ≤ a ≤ 1}

x ∪ y = {a | a ∈ x or a ∈ y}x ∩ y = {a | a ∈ x and a ∈ y}x \ y = {a | a ∈ x and a 6∈ y}


Naive Set Theory

Notations for Sets




Examples

[0, 1] = {a | a is a real number and 0 ≤ a ≤ 1}x ∪ y = {a | a ∈ x or a ∈ y}x ∩ y = {a | a ∈ x and a ∈ y}x \ y = {a | a ∈ x and a 6∈ y}


Naive Set Theory

Ordered Pairs （序对）

a, b: objects

(a, b): the ordered pair such that (a, b) = (c , d) iff a = c and b = d

Cartisian Product （笛卡尔积）x , y : sets

the Cartisian product: x × y := {(a, b) | a ∈ x ∧ b ∈ y}

Examples

R× RZ× Z{0, 1, 2, 3} × {100, 150, 200}


Naive Set Theory

Power Set （幂集）x : a set

the power set 2x (or P(x)): the set consisting of all subsets of x

2x := {y | y ⊆ x}

Examples

2∅ = {∅}2{a,b} = {∅, {a}, {b}, {a, b}}


Naive Set Theory

Problem

Is naive set theory enough?


Naive Set Theory

Russell’s Paradox （罗素悖论）

X := {x | x 6∈ x};

the paradox:

X ∈ X implies X 6∈ X .X 6∈ X implies X ∈ X .

(optional) explanation:

The entity X conceptually exists, but is not a set.


Naive Set Theory

Russell’s Paradox （罗素悖论）

X := {x | x 6∈ x};the paradox:

X ∈ X implies X 6∈ X .X 6∈ X implies X ∈ X .

(optional) explanation:

The entity X conceptually exists, but is not a set.


Axiomatic Set Theory(公理化集合论)

[E], Page 17 – Page 33


Axiomatic Set Theory

The Principles

A set is a collection of objects treated as a single entity.

Every object is a set, and every set is an object.

It should hold that either a∈A or a 6∈A but not both.

A formal language is required for constructing meaningful statements.(will not be covered in the lecture)

Axioms are required for reasoning about sets.



Axioms

Axioms are statements that are assumed to be true.

Why do we need axioms?

Axioms are basic rules for establishing correct statements.



The Axiom of Extensionality （外延公理）

x , y , a: variables whose domains are sets

Then the axiom of extensionality says that

∀x∀y [(x = y)↔∀a ((a∈ x)↔ (a∈ y))] .

Impact

The only basic rule for judging whether two sets are equal!



The Empty-Set Axiom （空集存在公理）

x , y : variables whose domains are sets

Then the empty-set axiom says that

∃x∀y(y 6∈ x)

where the set x is denoted by ∅.

Exercise

Prove through the axiom of extensionality that the empty set is unique.



The Axiom of Set Union （有限情形下的并集公理）

x , y , z , a: variables whose domains are sets

Then the axiom of set union says that

∀x∀y∃z∀a [(a∈ z)↔ ((a∈ x)∨ (a∈ y))]

where such set z is denoted by x ∪ y .

Exercise

Prove through the axiom of extensionality that given any sets x , y , the setx ∪ y is unique.



The Pairing Axiom （无序对集合存在公理）

x , y , z , a: variables whose domains are sets

Then the pairing axiom says that

∀x∀y∃z∀a [(a∈ z)↔ ((a= x)∨ (a= y))]

where such set z is denoted by {x , y}.



Sets with Finitely Many Elements

a1, . . . , an: n objects

{a1, . . . , an}: the set consisting of exactly the elements a1, . . . , an

By the axioms of set unions and pairing, we can assert the existence of theset {a1, . . . , an}.



The Power-Set Axiom （幂集公理）

x , y , z : variables whose domains are sets

Then the power-set axiom says that

∀x∃z∀y(y ∈ z ↔ y ⊆ x)

where the set z is denoted by 2x (or P(x)).



The Subset Axioms （子集公理）

x , c : variables

φ(x): a predicate whose free variables are at most x

Then it is an axiom that

∀c ∃B ∀x (x ∈ B ↔ (x ∈ c ∧ φ(x)))

where B is denoted by {x ∈ c | φ(x)}.

The Role of c

The variable c represents the prescribed set over which the variable xranges over.



Examples

{x ∈ R | x3 − 3 · x2 + 4 · x − 2 ≤ 0}x ∩ y := {a ∈ x ∪ y | a ∈ x ∧ a ∈ y}x \ y := {a ∈ x ∪ y | a ∈ x ∧ a 6∈ y}{2 · n | n ∈ Z} = {n ∈ Z | ∃k ∈ Z (n = 2 · k)}



Examples

{x ∈ R | x3 − 3 · x2 + 4 · x − 2 ≤ 0}

x ∩ y := {a ∈ x ∪ y | a ∈ x ∧ a ∈ y}x \ y := {a ∈ x ∪ y | a ∈ x ∧ a 6∈ y}{2 · n | n ∈ Z} = {n ∈ Z | ∃k ∈ Z (n = 2 · k)}



Examples

{x ∈ R | x3 − 3 · x2 + 4 · x − 2 ≤ 0}x ∩ y := {a ∈ x ∪ y | a ∈ x ∧ a ∈ y}x \ y := {a ∈ x ∪ y | a ∈ x ∧ a 6∈ y}

{2 · n | n ∈ Z} = {n ∈ Z | ∃k ∈ Z (n = 2 · k)}



Examples

{x ∈ R | x3 − 3 · x2 + 4 · x − 2 ≤ 0}x ∩ y := {a ∈ x ∪ y | a ∈ x ∧ a ∈ y}x \ y := {a ∈ x ∪ y | a ∈ x ∧ a 6∈ y}{2 · n | n ∈ Z}

= {n ∈ Z | ∃k ∈ Z (n = 2 · k)}



Examples

{x ∈ R | x3 − 3 · x2 + 4 · x − 2 ≤ 0}x ∩ y := {a ∈ x ∪ y | a ∈ x ∧ a ∈ y}x \ y := {a ∈ x ∪ y | a ∈ x ∧ a 6∈ y}{2 · n | n ∈ Z} = {n ∈ Z | ∃k ∈ Z (n = 2 · k)}



Exercise

A,B,C : sets

Prove the De Morgan’s Law: C \ (A ∪ B) = (C \ A) ∩ (C \ B)

Prove the distributive Law: C ∩ (A ∪ B) = (C ∩ A) ∪ (C ∩ B)



Axiom of Extended Union （广义并集公理）

The axiom of extended union says that

∀A∃B ∀a (a ∈ B ↔ (∃A (A ∈ A ∧ a ∈ A)))

where the set B is denoted by⋃A.

Example⋃{{n} | n ∈ Z} = Z⋃{[x − 1, x + 1] | x ∈ [0, 1]} = [−1, 2]





∀A∃B ∀a (a ∈ B ↔ (∃A (A ∈ A ∧ a ∈ A)))


Example⋃{{n} | n ∈ Z}

= Z⋃{[x − 1, x + 1] | x ∈ [0, 1]} = [−1, 2]





∀A∃B ∀a (a ∈ B ↔ (∃A (A ∈ A ∧ a ∈ A)))


Example⋃{{n} | n ∈ Z} = Z

⋃{[x − 1, x + 1] | x ∈ [0, 1]} = [−1, 2]





∀A∃B ∀a (a ∈ B ↔ (∃A (A ∈ A ∧ a ∈ A)))


Example⋃{{n} | n ∈ Z} = Z⋃{[x − 1, x + 1] | x ∈ [0, 1]}

= [−1, 2]





∀A∃B ∀a (a ∈ B ↔ (∃A (A ∈ A ∧ a ∈ A)))


Example⋃{{n} | n ∈ Z} = Z⋃{[x − 1, x + 1] | x ∈ [0, 1]} = [−1, 2]



Extended Intersection （广义交集）

From the subset axiom, we also have extended intersection:

∀A(A 6= ∅ → ∃B ∀a (a ∈ B ↔ (∀A (A ∈ A → a ∈ A))))

where the set B is denoted by⋂A.

Proof

B = {a ∈ c | ∀A (A ∈ A → a ∈ A)} where c ∈ A .

Example⋂{[x − 1, x + 1] | x ∈ [0, 1]} = [0, 1]





∀A(A 6= ∅ → ∃B ∀a (a ∈ B ↔ (∀A (A ∈ A → a ∈ A))))


Proof


Example⋂{[x − 1, x + 1] | x ∈ [0, 1]}

= [0, 1]





∀A(A 6= ∅ → ∃B ∀a (a ∈ B ↔ (∀A (A ∈ A → a ∈ A))))


Proof


Example⋂{[x − 1, x + 1] | x ∈ [0, 1]} = [0, 1]



Indexed Notation⋃n An =

⋃{A1, . . . ,An, . . . } =

⋃∞n=1 An⋂

n An =⋂{A1, . . . ,An, . . . } =

⋂∞n=1 An

⋃i∈I Ai =

⋃{Ai | i ∈ I} (I is an index set.)⋂

i∈I Ai =⋂{Ai | i ∈ I}



Indexed Notation⋃n An =

⋃{A1, . . . ,An, . . . } =

⋃∞n=1 An⋂

n An =⋂{A1, . . . ,An, . . . } =

⋂∞n=1 An⋃

i∈I Ai =⋃{Ai | i ∈ I} (I is an index set.)⋂

i∈I Ai =⋂{Ai | i ∈ I}



Discussion

What is⋂∅?

Key Points

“vacuously truth”

“the set of all sets”

Answer⋂∅ is left undefined.



Theorem

There is no set to which every set belongs.

Proof

Suppose that there is such a set A. Then from the subset axiom, we candefine

B := {x ∈ A | x 6∈ x} .

Then we have that both B ∈ B and B 6∈ B holds. Contradiction.


Ordered Pairs and Cartesian Product

main textbook, Page 122 – Page 124auxiliary textbook, Page 35 – Page 38


Ordered Pairs

Ordered Pairs

example: coordinates (1, 2), (4.4, 7.3), . . .

key property:

(a, b) 6= (b, a) iff a 6= b(a, b) = (c , d) iff a = c and b = d

Definition

a, b: objects (sets)

Then we define that (a, b) := {{a}, {a, b}} .

Homework

Prove that (a, b) = (c , d) iff a = c and b = d .

Pay special attention to the case a = b.


Ordered Pairs

Tuples

(a, b, c) := ((a, b), c)

(a1, . . . , an) := ((a1, . . . , an−1), an)


Cartesian Product

Definitionx , y : two sets

in naive set theory: x × y := {(a, b) | a ∈ x ∧ b ∈ y}

in axiomatic set theory: x × y := {(a, b) ∈ 22x ∪ y | a ∈ x ∧ b ∈ y}

The Reasoning

{a} ⊆ x ∪ y and {a, b} ⊆ x ∪ y(a, b) = {{a}, {a, b}} ⊆ 2x ∪ y

(a, b) ∈ 22x ∪ y


Cartesian Product


in naive set theory: x × y := {(a, b) | a ∈ x ∧ b ∈ y}in axiomatic set theory: x × y := {(a, b) ∈ 22

x ∪ y | a ∈ x ∧ b ∈ y}

The Reasoning


(a, b) ∈ 22x ∪ y


Cartesian Product



x ∪ y | a ∈ x ∧ b ∈ y}

The Reasoning

{a} ⊆ x ∪ y and {a, b} ⊆ x ∪ y

(a, b) = {{a}, {a, b}} ⊆ 2x ∪ y

(a, b) ∈ 22x ∪ y


Cartesian Product



x ∪ y | a ∈ x ∧ b ∈ y}

The Reasoning


(a, b) ∈ 22x ∪ y


Cartisian Product

Cartisian Product of Multiple Sets

A× B × C := (A× B)× C

A1 × · · · × An := (A1 × · · · × An−1)× An

A1 × · · · × An = {(a1, . . . , an) | ak ∈ Ak for 1 ≤ k ≤ n}


Summary

naive set theory

axiomatic set theory


Textbooks

main textbook:

Kenneth H. Rosen, Discrete Mathematics and Its Applications, 7thedition [R]

auxiliary textbook:

Herbert B. Enderton, Elements of Set Theory [E]


Reading

[R], Page 115 – Page 134

[E], Page 1 – Page 38

(optional) 石纯一等，数理逻辑与集合论（第二版），第九章


Homeworks

[R], Page 125, Exercise 1(c), 10(c)(d)

[R], Page 126, Exercise 17, 18, 45

[R], Page 136, Exercise 24

[R], Page 137, Exercise 38(b), 40

[E], Page 26, Exercise 7(a)

[E], Page 38, Exercise 3

Note: In the homework, you can rely on Venn diagrams for intuition, butyou should write your homework using only formal proofs.


Homeworks

Homework Submission

submission time: the start of the class on Oct. 22nd

teaching assistant:

Peixin Wang: [email protected] Wang: [email protected]

submission:

written version: submit on the desk (preferred)electronic version: word or pdf version, send email with title

“离散数学+姓名+学号+第六周周二”

to the teaching assistants (Students from the classes F1903001 –F1903004, please send to Peixin Wang. All other students please sendto Jinyi Wang.)


@汥瑀瑯步渠 Set Theoryjhc.sjtu.edu.cn/~hongfeifu/lecture7.pdf · 2019-10-15 · Herbert B....

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