MAT1202 SET THEORY :Week2 · Thanatyod Jampawai, Ph.D. MAT1202 SET THEORY :Week2 4 / 1..... The...

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MAT1202 SET THEORY :Week2

Thanatyod Jampawai, Ph.D.

Thanatyod Jampawai, Ph.D. MAT1202 SET THEORY :Week2 1 / 1

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The Aixiomatic Set Theory

Review Week 1

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)]

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 )

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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The Aixiomatic Set Theory

Outline for Week 2

Chapter 2 The Axiomatic Set Theory2.2 The Axiom of Operations2.3 The Power set Axiom

ConclusionAssignment 2


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The Aixiomatic Set Theory The Axiom of Operations

2.2 The Axiom of Operations

Theorem (2.2.1)

Let A and B be sets. Then there is a unique set C satisfying

x ∈ C ↔ (x ∈ A ∧ x ∈ B).

Proof .

Let A and B be sets. Let p(x) be statement x ∈ B. By the axiom of specification, there is a set Csuch that

x ∈ C ↔ (x ∈ A ∧ x ∈ B).

Let C2 be a set satisfyingx ∈ C2 ↔ (x ∈ A ∧ x ∈ B).

Thenx ∈ C ↔ x ∈ C2 for all x.

Thus, C = C2.


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Theorem (2.2.1)


x ∈ C ↔ (x ∈ A ∧ x ∈ B).

Proof .

Let A and B be sets.

Let p(x) be statement x ∈ B. By the axiom of specification, there is a set Csuch that

x ∈ C ↔ (x ∈ A ∧ x ∈ B).



Thus, C = C2.


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Theorem (2.2.1)


x ∈ C ↔ (x ∈ A ∧ x ∈ B).

Proof .

Let A and B be sets. Let p(x) be statement x ∈ B.

By the axiom of specification, there is a set Csuch that

x ∈ C ↔ (x ∈ A ∧ x ∈ B).



Thus, C = C2.


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Theorem (2.2.1)


x ∈ C ↔ (x ∈ A ∧ x ∈ B).

Proof .

Let A and B be sets. Let p(x) be statement x ∈ B. By the axiom of specification, there is a set Csuch that

x ∈ C ↔ (x ∈ A ∧ x ∈ B).



Thus, C = C2.


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Definition (2.2.1)

The set C in theorem 2.1.1 is intersection of A and B, denoted A ∩B ,i.e.,

A ∩B = {x : x ∈ A ∧ x ∈ B}

orx ∈ A ∩B ↔ (x ∈ A ∧ x ∈ B).


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Theorem (2.2.2)

Let A,B and C be sets. Then

1. A ∩∅ = ∅

2. A ∩A = A (idempotent law)

3. A ∩B = B ∩A (commutative law)

4. A ∩ (B ∩ C) = (A ∩B) ∩ C (associative law)


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Proof.

Let A,B and C be sets.

1. A ∩ ∅ = ∅For any x

x ∈ ∅ ↔ x ∈ ∅↔ x ∈ A ∧ x ∈ ∅ (∵ x ∈ ∅ is false)↔ x ∈ A ∩ ∅

Thus, A ∩ ∅ = ∅.2. Idempotent law A ∩A = A

For any x

x ∈ A ↔ x ∈ A

↔ x ∈ A ∧ x ∈ A

↔ x ∈ A ∩A

Thus, A ∩A = A.


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Proof.


1. A ∩ ∅ = ∅

For any x



For any x

x ∈ A ↔ x ∈ A

↔ x ∈ A ∧ x ∈ A

↔ x ∈ A ∩A

Thus, A ∩A = A.


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Proof.


1. A ∩ ∅ = ∅For any x



For any x

x ∈ A ↔ x ∈ A

↔ x ∈ A ∧ x ∈ A

↔ x ∈ A ∩A

Thus, A ∩A = A.


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Proof.


1. A ∩ ∅ = ∅For any x


Thus, A ∩ ∅ = ∅.

2. Idempotent law A ∩A = AFor any x

x ∈ A ↔ x ∈ A

↔ x ∈ A ∧ x ∈ A

↔ x ∈ A ∩A

Thus, A ∩A = A.


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Proof.


1. A ∩ ∅ = ∅For any x



For any x

x ∈ A ↔ x ∈ A

↔ x ∈ A ∧ x ∈ A

↔ x ∈ A ∩A

Thus, A ∩A = A.


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3. Commutative law A ∩B = B ∩A

For any x

x ∈ A ∩B ↔ x ∈ A ∧ x ∈ B

↔ x ∈ B ∧ x ∈ A

↔ x ∈ B ∩A

Hence, A ∩B = B ∩A.

4. Associative law A ∩ (B ∩ C) = (A ∩B) ∩ C

For any x

x ∈ A ∩ (B ∩ C) ↔ x ∈ A ∧ x ∈ B ∩ C

↔ x ∈ A ∧ (x ∈ B ∧ x ∈ C)

↔ (x ∈ A ∧ x ∈ B) ∧ x ∈ C

↔ x ∈ (A ∩B) ∩ C

So, A ∩ (B ∩ C) = (A ∩B) ∩ C.


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3. Commutative law A ∩B = B ∩AFor any x

x ∈ A ∩B ↔ x ∈ A ∧ x ∈ B

↔ x ∈ B ∧ x ∈ A

↔ x ∈ B ∩A

Hence, A ∩B = B ∩A.

4. Associative law A ∩ (B ∩ C) = (A ∩B) ∩ C

For any x

x ∈ A ∩ (B ∩ C) ↔ x ∈ A ∧ x ∈ B ∩ C

↔ x ∈ A ∧ (x ∈ B ∧ x ∈ C)

↔ (x ∈ A ∧ x ∈ B) ∧ x ∈ C

↔ x ∈ (A ∩B) ∩ C

So, A ∩ (B ∩ C) = (A ∩B) ∩ C.


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Theorem (2.2.3)

Let A and B be sets. Then

1. A ∩B ⊆ A and A ∩B ⊆ B

2. A ∩B = A if and only if A ⊆ B

Proof .


1. A ∩B ⊆ A x ∈ A ∩B ↔ (x ∈ A ∧ x ∈ B) → x ∈ A

A ∩B ⊆ B x ∈ A ∩B ↔ (x ∈ A ∧ x ∈ B) → x ∈ B

2. A ∩B = A ↔ A ⊆ B

(→) Assume that A ∩B = A. By 1, A = A ∩B ⊆ B. So A ⊆ B.(←) Assume that A ⊆ B.(A ∩B ⊆ A) By 1, A ∩B ⊆ A.

(A ⊆ A ∩B) Let x ∈ A. Then x ∈ B. So x ∈ A ∩B or A ⊆ A ∩B.


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Theorem (2.2.3)


1. A ∩B ⊆ A and A ∩B ⊆ B


Proof .


1. A ∩B ⊆ A

x ∈ A ∩B ↔ (x ∈ A ∧ x ∈ B) → x ∈ A

A ∩B ⊆ B x ∈ A ∩B ↔ (x ∈ A ∧ x ∈ B) → x ∈ B

2. A ∩B = A ↔ A ⊆ B




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Theorem (2.2.3)


1. A ∩B ⊆ A and A ∩B ⊆ B


Proof .


1. A ∩B ⊆ A x ∈ A ∩B ↔ (x ∈ A ∧ x ∈ B) → x ∈ A

A ∩B ⊆ B x ∈ A ∩B ↔ (x ∈ A ∧ x ∈ B) → x ∈ B

2. A ∩B = A ↔ A ⊆ B




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Theorem (2.2.3)


1. A ∩B ⊆ A and A ∩B ⊆ B


Proof .


1. A ∩B ⊆ A x ∈ A ∩B ↔ (x ∈ A ∧ x ∈ B) → x ∈ A

A ∩B ⊆ B

x ∈ A ∩B ↔ (x ∈ A ∧ x ∈ B) → x ∈ B

2. A ∩B = A ↔ A ⊆ B




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Theorem (2.2.3)


1. A ∩B ⊆ A and A ∩B ⊆ B


Proof .


1. A ∩B ⊆ A x ∈ A ∩B ↔ (x ∈ A ∧ x ∈ B) → x ∈ A

A ∩B ⊆ B x ∈ A ∩B ↔ (x ∈ A ∧ x ∈ B) → x ∈ B

2. A ∩B = A ↔ A ⊆ B




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Theorem (2.2.3)


1. A ∩B ⊆ A and A ∩B ⊆ B


Proof .


1. A ∩B ⊆ A x ∈ A ∩B ↔ (x ∈ A ∧ x ∈ B) → x ∈ A

A ∩B ⊆ B x ∈ A ∩B ↔ (x ∈ A ∧ x ∈ B) → x ∈ B

2. A ∩B = A ↔ A ⊆ B

(→)

Assume that A ∩B = A. By 1, A = A ∩B ⊆ B. So A ⊆ B.(←) Assume that A ⊆ B.(A ∩B ⊆ A) By 1, A ∩B ⊆ A.



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Theorem (2.2.3)


1. A ∩B ⊆ A and A ∩B ⊆ B


Proof .


1. A ∩B ⊆ A x ∈ A ∩B ↔ (x ∈ A ∧ x ∈ B) → x ∈ A

A ∩B ⊆ B x ∈ A ∩B ↔ (x ∈ A ∧ x ∈ B) → x ∈ B

2. A ∩B = A ↔ A ⊆ B

(→) Assume that A ∩B = A.

By 1, A = A ∩B ⊆ B. So A ⊆ B.(←) Assume that A ⊆ B.(A ∩B ⊆ A) By 1, A ∩B ⊆ A.



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Theorem (2.2.3)


1. A ∩B ⊆ A and A ∩B ⊆ B


Proof .


1. A ∩B ⊆ A x ∈ A ∩B ↔ (x ∈ A ∧ x ∈ B) → x ∈ A

A ∩B ⊆ B x ∈ A ∩B ↔ (x ∈ A ∧ x ∈ B) → x ∈ B

2. A ∩B = A ↔ A ⊆ B

(→) Assume that A ∩B = A. By 1, A = A ∩B ⊆ B.

So A ⊆ B.(←) Assume that A ⊆ B.(A ∩B ⊆ A) By 1, A ∩B ⊆ A.



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Theorem (2.2.3)


1. A ∩B ⊆ A and A ∩B ⊆ B


Proof .


1. A ∩B ⊆ A x ∈ A ∩B ↔ (x ∈ A ∧ x ∈ B) → x ∈ A

A ∩B ⊆ B x ∈ A ∩B ↔ (x ∈ A ∧ x ∈ B) → x ∈ B

2. A ∩B = A ↔ A ⊆ B

(→) Assume that A ∩B = A. By 1, A = A ∩B ⊆ B. So A ⊆ B.

(←) Assume that A ⊆ B.(A ∩B ⊆ A) By 1, A ∩B ⊆ A.



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Theorem (2.2.3)


1. A ∩B ⊆ A and A ∩B ⊆ B


Proof .


1. A ∩B ⊆ A x ∈ A ∩B ↔ (x ∈ A ∧ x ∈ B) → x ∈ A

A ∩B ⊆ B x ∈ A ∩B ↔ (x ∈ A ∧ x ∈ B) → x ∈ B

2. A ∩B = A ↔ A ⊆ B

(→) Assume that A ∩B = A. By 1, A = A ∩B ⊆ B. So A ⊆ B.(←) Assume that A ⊆ B.

(A ∩B ⊆ A) By 1, A ∩B ⊆ A.



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Theorem (2.2.3)


1. A ∩B ⊆ A and A ∩B ⊆ B


Proof .


1. A ∩B ⊆ A x ∈ A ∩B ↔ (x ∈ A ∧ x ∈ B) → x ∈ A

A ∩B ⊆ B x ∈ A ∩B ↔ (x ∈ A ∧ x ∈ B) → x ∈ B

2. A ∩B = A ↔ A ⊆ B

(→) Assume that A ∩B = A. By 1, A = A ∩B ⊆ B. So A ⊆ B.(←) Assume that A ⊆ B.(A ∩B ⊆ A)

By 1, A ∩B ⊆ A.



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Theorem (2.2.3)


1. A ∩B ⊆ A and A ∩B ⊆ B


Proof .


1. A ∩B ⊆ A x ∈ A ∩B ↔ (x ∈ A ∧ x ∈ B) → x ∈ A

A ∩B ⊆ B x ∈ A ∩B ↔ (x ∈ A ∧ x ∈ B) → x ∈ B

2. A ∩B = A ↔ A ⊆ B




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Theorem (2.2.3)


1. A ∩B ⊆ A and A ∩B ⊆ B


Proof .


1. A ∩B ⊆ A x ∈ A ∩B ↔ (x ∈ A ∧ x ∈ B) → x ∈ A

A ∩B ⊆ B x ∈ A ∩B ↔ (x ∈ A ∧ x ∈ B) → x ∈ B

2. A ∩B = A ↔ A ⊆ B


(A ⊆ A ∩B)

Let x ∈ A. Then x ∈ B. So x ∈ A ∩B or A ⊆ A ∩B.


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Theorem (2.2.3)


1. A ∩B ⊆ A and A ∩B ⊆ B


Proof .


1. A ∩B ⊆ A x ∈ A ∩B ↔ (x ∈ A ∧ x ∈ B) → x ∈ A

A ∩B ⊆ B x ∈ A ∩B ↔ (x ∈ A ∧ x ∈ B) → x ∈ B

2. A ∩B = A ↔ A ⊆ B


(A ⊆ A ∩B) Let x ∈ A.

Then x ∈ B. So x ∈ A ∩B or A ⊆ A ∩B.


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Theorem (2.2.3)


1. A ∩B ⊆ A and A ∩B ⊆ B


Proof .


1. A ∩B ⊆ A x ∈ A ∩B ↔ (x ∈ A ∧ x ∈ B) → x ∈ A

A ∩B ⊆ B x ∈ A ∩B ↔ (x ∈ A ∧ x ∈ B) → x ∈ B

2. A ∩B = A ↔ A ⊆ B


(A ⊆ A ∩B) Let x ∈ A. Then x ∈ B.

So x ∈ A ∩B or A ⊆ A ∩B.


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Theorem (2.2.3)


1. A ∩B ⊆ A and A ∩B ⊆ B


Proof .


1. A ∩B ⊆ A x ∈ A ∩B ↔ (x ∈ A ∧ x ∈ B) → x ∈ A

A ∩B ⊆ B x ∈ A ∩B ↔ (x ∈ A ∧ x ∈ B) → x ∈ B

2. A ∩B = A ↔ A ⊆ B




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Theorem (2.2.4)


1. A ⊆ B ∩ C if and only if A ⊆ B and A ⊆ C

2. if A ⊆ B, then (A ∩ C) ⊆ (B ∩ C)


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Proof .


1. A ⊆ B ∩ C ↔ (A ⊆ B) ∧ (A ⊆ C)

A ⊆ B ∩ C ↔ ∀x [x ∈ A→ x ∈ B ∩ C]

↔ ∀x [x ∈ A→ (x ∈ B ∧ x ∈ C)]

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

↔ A ⊆ B ∧A ⊆ C

2. A ⊆ B → (A ∩ C) ⊆ (B ∩ C) Assume that A ⊆ B. For any x

x ∈ A ∩ C → x ∈ A ∧ x ∈ C

→ x ∈ B ∧ x ∈ C (∵ A ⊆ B)

→ x ∈ B ∩ C

Therefore, A ∩ C ⊆ B ∩ C.


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Proof .


1. A ⊆ B ∩ C ↔ (A ⊆ B) ∧ (A ⊆ C)

A ⊆ B ∩ C ↔ ∀x [x ∈ A→ x ∈ B ∩ C]

↔ ∀x [x ∈ A→ (x ∈ B ∧ x ∈ C)]

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

↔ A ⊆ B ∧A ⊆ C

2. A ⊆ B → (A ∩ C) ⊆ (B ∩ C)

Assume that A ⊆ B. For any x

x ∈ A ∩ C → x ∈ A ∧ x ∈ C

→ x ∈ B ∧ x ∈ C (∵ A ⊆ B)

→ x ∈ B ∩ C



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Proof .


1. A ⊆ B ∩ C ↔ (A ⊆ B) ∧ (A ⊆ C)

A ⊆ B ∩ C ↔ ∀x [x ∈ A→ x ∈ B ∩ C]

↔ ∀x [x ∈ A→ (x ∈ B ∧ x ∈ C)]

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

↔ A ⊆ B ∧A ⊆ C

2. A ⊆ B → (A ∩ C) ⊆ (B ∩ C) Assume that A ⊆ B.

For any x

x ∈ A ∩ C → x ∈ A ∧ x ∈ C

→ x ∈ B ∧ x ∈ C (∵ A ⊆ B)

→ x ∈ B ∩ C



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Proof .


1. A ⊆ B ∩ C ↔ (A ⊆ B) ∧ (A ⊆ C)

A ⊆ B ∩ C ↔ ∀x [x ∈ A→ x ∈ B ∩ C]

↔ ∀x [x ∈ A→ (x ∈ B ∧ x ∈ C)]

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

↔ A ⊆ B ∧A ⊆ C

2. A ⊆ B → (A ∩ C) ⊆ (B ∩ C) Assume that A ⊆ B. For any x

x ∈ A ∩ C → x ∈ A ∧ x ∈ C

→ x ∈ B ∧ x ∈ C (∵ A ⊆ B)

→ x ∈ B ∩ C



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Theorem (2.2.5)

Let A,B,C and D be sets. Then

1. if A ⊆ B and C ⊆ D, then A ∩ C ⊆ B ∩D

2. if A = B and C = D, then A ∩ C = B ∩D


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Proof .

Let A,B,C and D be sets.

1. A ⊆ B ∧ C ⊆ D → A ∩ C ⊆ B ∩D

Assume that A ⊆ B and C ⊆ D. Then

x ∈ A ∩ C → x ∈ A ∧ x ∈ C

→ x ∈ B ∧ x ∈ D (∵ A ⊆ B and C ⊆ D)

→ x ∈ B ∩D

Hence, A ∩ C ⊆ B ∩D.

2. A = B ∧ C = D → A ∩ C = B ∩D

A = B ∧ C = D → A ⊆ B ∧ C ⊆ D

→ A ∩ C ⊆ B ∩D ( By 1 )

A = B ∧ C = D → B ⊆ A ∧D ⊆ C

→ B ∩D ⊆ A ∩ C ( By 1 )

Therefore, A ∩ C = B ∩D.


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Proof .


1. A ⊆ B ∧ C ⊆ D → A ∩ C ⊆ B ∩D

Assume that A ⊆ B and C ⊆ D.

Then

x ∈ A ∩ C → x ∈ A ∧ x ∈ C

→ x ∈ B ∧ x ∈ D (∵ A ⊆ B and C ⊆ D)

→ x ∈ B ∩D


2. A = B ∧ C = D → A ∩ C = B ∩D

A = B ∧ C = D → A ⊆ B ∧ C ⊆ D

→ A ∩ C ⊆ B ∩D ( By 1 )

A = B ∧ C = D → B ⊆ A ∧D ⊆ C

→ B ∩D ⊆ A ∩ C ( By 1 )



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Proof .


1. A ⊆ B ∧ C ⊆ D → A ∩ C ⊆ B ∩D


x ∈ A ∩ C → x ∈ A ∧ x ∈ C

→ x ∈ B ∧ x ∈ D (∵ A ⊆ B and C ⊆ D)

→ x ∈ B ∩D


2. A = B ∧ C = D → A ∩ C = B ∩D

A = B ∧ C = D → A ⊆ B ∧ C ⊆ D

→ A ∩ C ⊆ B ∩D ( By 1 )

A = B ∧ C = D → B ⊆ A ∧D ⊆ C

→ B ∩D ⊆ A ∩ C ( By 1 )



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Example (2.2.1)

Give counter examples for convering of the statements in theorem 2.2.5

Solution

1. if A ⊆ B and C ⊆ D, then A ∩ C ⊆ B ∩D

Converse: A ∩ C ⊆ B ∩D → A ⊆ B ∧ C ⊆ D

A = {1}, C = {2}, B = {3}, D = {4} → A ∩ C = ∅ and B ∩D = ∅

2. if A = B and C = D, then A ∩ C = B ∩D

Converse: A ∩ C = B ∩D → A = B ∧ C = D

A = {1}, C = {2}, B = {3}, D = {4} → A ∩ C = ∅ and B ∩D = ∅


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Theorem (2.2.6)

Let A1, A2, ..., An, B1, B2, ..., Bn be sets. Then, for any n ∈ N,

1. if A1 ⊆ B1 ∧A2 ⊆ B2 ∧ ... ∧An ⊆ Bn, then

(A1 ∩A2 ∩ ... ∩An) ⊆ (B1 ∩B2 ∩ ... ∩Bn),

2. if A1 = B1 ∧A2 = B2 ∧ ... ∧An = Bn, then

(A1 ∩A2 ∩ ... ∩An) = (B1 ∩B2 ∩ ... ∩Bn).


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Proof .

Let n ∈ N and let A1, A2, ..., An, B1, B2, ..., Bn be sets.

1. A1 ⊆ B1 ∧A2 ⊆ B2 ∧ ... ∧An ⊆ Bn → (A1 ∩A2 ∩ ... ∩An) ⊆ (B1 ∩B2 ∩ ... ∩Bn)

Let P (n) be this statement where n ∈ N.

Basic step : Since A1 ⊆ B1 → A1 ⊆ B1 is true, P (1) holds.Inductive step : Assume that P (k) holds where k ∈ N. Then

A1 ⊆ B1 ∧A2 ⊆ B2 ∧ ...∧Ak ⊆ Bk → (A1 ∩A2 ∩ ...∩Ak) ⊆ (B1 ∩B2 ∩ ...∩Bk)

Suppose that A1 ⊆ B1 ∧A2 ⊆ B2 ∧ ... ∧Ak ⊆ Bk∧Ak+1 ⊆ Bk+1. By inductionhypothesis,(A1 ∩A2 ∩ ...∩Ak) ⊆ (B1 ∩B2 ∩ ...∩Bk) and ∧Ak+1 ⊆ Bk+1. By theorem 2.2.5,

(A1 ∩A2 ∩ ... ∩Ak)∧Ak+1 ⊆ (B1 ∩B2 ∩ ... ∩Bk)∧Bk+1

Hence, P (k + 1) holds. �

2. A1 = B1 ∧A2 = B2 ∧ ... ∧An = Bn → (A1 ∩A2 ∩ ... ∩An) = (B1 ∩B2 ∩ ... ∩Bn)

Assignment 2


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Proof .




Basic step : Since A1 ⊆ B1 → A1 ⊆ B1 is true, P (1) holds.

Inductive step : Assume that P (k) holds where k ∈ N. Then



(A1 ∩A2 ∩ ... ∩Ak)∧Ak+1 ⊆ (B1 ∩B2 ∩ ... ∩Bk)∧Bk+1



Assignment 2


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Proof .




Basic step : Since A1 ⊆ B1 → A1 ⊆ B1 is true, P (1) holds.Inductive step : Assume that P (k) holds where k ∈ N.

Then



(A1 ∩A2 ∩ ... ∩Ak)∧Ak+1 ⊆ (B1 ∩B2 ∩ ... ∩Bk)∧Bk+1



Assignment 2


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Proof .







(A1 ∩A2 ∩ ... ∩Ak)∧Ak+1 ⊆ (B1 ∩B2 ∩ ... ∩Bk)∧Bk+1



Assignment 2


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Proof .






Suppose that A1 ⊆ B1 ∧A2 ⊆ B2 ∧ ... ∧Ak ⊆ Bk∧Ak+1 ⊆ Bk+1.

By inductionhypothesis,(A1 ∩A2 ∩ ...∩Ak) ⊆ (B1 ∩B2 ∩ ...∩Bk) and ∧Ak+1 ⊆ Bk+1. By theorem 2.2.5,

(A1 ∩A2 ∩ ... ∩Ak)∧Ak+1 ⊆ (B1 ∩B2 ∩ ... ∩Bk)∧Bk+1



Assignment 2


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Proof .






Suppose that A1 ⊆ B1 ∧A2 ⊆ B2 ∧ ... ∧Ak ⊆ Bk∧Ak+1 ⊆ Bk+1. By inductionhypothesis,(A1 ∩A2 ∩ ...∩Ak) ⊆ (B1 ∩B2 ∩ ...∩Bk) and ∧Ak+1 ⊆ Bk+1.

By theorem 2.2.5,

(A1 ∩A2 ∩ ... ∩Ak)∧Ak+1 ⊆ (B1 ∩B2 ∩ ... ∩Bk)∧Bk+1



Assignment 2


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Proof .







(A1 ∩A2 ∩ ... ∩Ak)∧Ak+1 ⊆ (B1 ∩B2 ∩ ... ∩Bk)∧Bk+1



Assignment 2


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Axiom 2.2.1 (Axiom of Union)

Let A and B be sets. There is a set C satisfying

x ∈ C ↔ (x ∈ A ∨ x ∈ B).

Theorem (2.2.7)

Let A and B be sets. There is a unique set C satisfying

x ∈ C ↔ (x ∈ A ∨ x ∈ B).

Proof .

Let C1 and C2 be sets satisfying condition of theorem 2.2.7. Then

x ∈ C1 ↔ (x ∈ A ∨ x ∈ B)

x ∈ C2 ↔ (x ∈ A ∨ x ∈ B)

∴ x ∈ C1 ↔ x ∈ C2

Therefore, C1 = C2.


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x ∈ C ↔ (x ∈ A ∨ x ∈ B).

Theorem (2.2.7)


x ∈ C ↔ (x ∈ A ∨ x ∈ B).

Proof .

Let C1 and C2 be sets satisfying condition of theorem 2.2.7.

Then

x ∈ C1 ↔ (x ∈ A ∨ x ∈ B)

x ∈ C2 ↔ (x ∈ A ∨ x ∈ B)

∴ x ∈ C1 ↔ x ∈ C2

Therefore, C1 = C2.


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x ∈ C ↔ (x ∈ A ∨ x ∈ B).

Theorem (2.2.7)


x ∈ C ↔ (x ∈ A ∨ x ∈ B).

Proof .

Let C1 and C2 be sets satisfying condition of theorem 2.2.7. Then

x ∈ C1 ↔ (x ∈ A ∨ x ∈ B)

x ∈ C2 ↔ (x ∈ A ∨ x ∈ B)

∴ x ∈ C1 ↔ x ∈ C2

Therefore, C1 = C2.


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Definition (2.2.2)

The set C in theorem 2.2.7 is union of A and B, denoted A ∪B , i.e.,

A ∪B = {x : x ∈ A ∨ x ∈ B} or x ∈ A ∪B ↔ (x ∈ A ∨ x ∈ B).

Theorem (2.2.8)


1. A ∪∅ = A

2. A ∪A = A (idempotent law)

3. A ∪B = B ∪A (commutative law)

4. A ∪ (B ∪ C) = (A ∪B) ∪ C (associative law)


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Proof.


1. A ∪ ∅ = A

For any x

x ∈ A ↔ x ∈ A

↔ x ∈ A ∨ x ∈ ∅ (∵ x ∈ ∅ is false)↔ x ∈ A ∪ ∅

So, A ∪ ∅ = A.2. Idempotent law A ∪A = A

For any x

x ∈ A ↔ x ∈ A

↔ x ∈ A ∨ x ∈ A

↔ x ∈ A ∪A

Thus, A ∪A = A.


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Proof.


1. A ∪ ∅ = AFor any x

x ∈ A ↔ x ∈ A

↔ x ∈ A ∨ x ∈ ∅ (∵ x ∈ ∅ is false)↔ x ∈ A ∪ ∅


For any x

x ∈ A ↔ x ∈ A

↔ x ∈ A ∨ x ∈ A

↔ x ∈ A ∪A

Thus, A ∪A = A.


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Proof.



x ∈ A ↔ x ∈ A

↔ x ∈ A ∨ x ∈ ∅ (∵ x ∈ ∅ is false)↔ x ∈ A ∪ ∅

So, A ∪ ∅ = A.

2. Idempotent law A ∪A = AFor any x

x ∈ A ↔ x ∈ A

↔ x ∈ A ∨ x ∈ A

↔ x ∈ A ∪A

Thus, A ∪A = A.


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Proof.



x ∈ A ↔ x ∈ A

↔ x ∈ A ∨ x ∈ ∅ (∵ x ∈ ∅ is false)↔ x ∈ A ∪ ∅


For any x

x ∈ A ↔ x ∈ A

↔ x ∈ A ∨ x ∈ A

↔ x ∈ A ∪A

Thus, A ∪A = A.


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3. Commutative law A ∪B = B ∪A

For any x

x ∈ A ∪B ↔ x ∈ A ∨ x ∈ B

↔ x ∈ B ∨ x ∈ A

↔ x ∈ B ∪A

Hence, A ∪B = B ∪A.

4. Associative law A ∪ (B ∪ C) = (A ∪B) ∪ C

For any x

x ∈ A ∪ (B ∪ C) ↔ x ∈ A ∧ x ∈ B ∪ C

↔ x ∈ A ∨ (x ∈ B ∨ x ∈ C)

↔ (x ∈ A ∨ x ∈ B) ∨ x ∈ C

↔ x ∈ (A ∪B) ∪ C

Therefore, A ∪ (B ∪ C) = (A ∪B) ∪ C.


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3. Commutative law A ∪B = B ∪AFor any x

x ∈ A ∪B ↔ x ∈ A ∨ x ∈ B

↔ x ∈ B ∨ x ∈ A

↔ x ∈ B ∪A

Hence, A ∪B = B ∪A.

4. Associative law A ∪ (B ∪ C) = (A ∪B) ∪ C

For any x

x ∈ A ∪ (B ∪ C) ↔ x ∈ A ∧ x ∈ B ∪ C

↔ x ∈ A ∨ (x ∈ B ∨ x ∈ C)

↔ (x ∈ A ∨ x ∈ B) ∨ x ∈ C

↔ x ∈ (A ∪B) ∪ C

Therefore, A ∪ (B ∪ C) = (A ∪B) ∪ C.


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Theorem (2.2.9)


1. A ⊆ A ∪B and B ⊆ A ∪B

2. A ∪B = B if and only if A ⊆ B

Proof .


1. A ⊆ A ∪B x ∈ A → (x ∈ A ∨ x ∈ B) → x ∈ A ∪B

B ⊆ A ∪B x ∈ B ↔ (x ∈ A ∨ x ∈ B) → x ∈ A ∪B

2. A ∪B = B ↔ A ⊆ B

(→) Assume that A ∪B = B. By 1, A ⊆ A ∪B = B. So A ⊆ B.(←) Assume that A ⊆ B.(A ∪B ⊆ B) Let x ∈ A ∪B. Then x ∈ A or x ∈ B.

If x ∈ A, then x ∈ B because A ⊆ B. Thus, A ∪B ⊆ B.If x ∈ B, it is done. Thus, A ∪B ⊆ B.(B ⊆ A ∪B) By 1, B ⊆ A ∪B.


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Theorem (2.2.9)


1. A ⊆ A ∪B and B ⊆ A ∪B


Proof .


1. A ⊆ A ∪B

x ∈ A → (x ∈ A ∨ x ∈ B) → x ∈ A ∪B

B ⊆ A ∪B x ∈ B ↔ (x ∈ A ∨ x ∈ B) → x ∈ A ∪B

2. A ∪B = B ↔ A ⊆ B




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Theorem (2.2.9)


1. A ⊆ A ∪B and B ⊆ A ∪B


Proof .


1. A ⊆ A ∪B x ∈ A → (x ∈ A ∨ x ∈ B) → x ∈ A ∪B

B ⊆ A ∪B x ∈ B ↔ (x ∈ A ∨ x ∈ B) → x ∈ A ∪B

2. A ∪B = B ↔ A ⊆ B




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Theorem (2.2.9)


1. A ⊆ A ∪B and B ⊆ A ∪B


Proof .


1. A ⊆ A ∪B x ∈ A → (x ∈ A ∨ x ∈ B) → x ∈ A ∪B

B ⊆ A ∪B

x ∈ B ↔ (x ∈ A ∨ x ∈ B) → x ∈ A ∪B

2. A ∪B = B ↔ A ⊆ B




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Theorem (2.2.9)


1. A ⊆ A ∪B and B ⊆ A ∪B


Proof .


1. A ⊆ A ∪B x ∈ A → (x ∈ A ∨ x ∈ B) → x ∈ A ∪B

B ⊆ A ∪B x ∈ B ↔ (x ∈ A ∨ x ∈ B) → x ∈ A ∪B

2. A ∪B = B ↔ A ⊆ B




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Theorem (2.2.9)


1. A ⊆ A ∪B and B ⊆ A ∪B


Proof .


1. A ⊆ A ∪B x ∈ A → (x ∈ A ∨ x ∈ B) → x ∈ A ∪B

B ⊆ A ∪B x ∈ B ↔ (x ∈ A ∨ x ∈ B) → x ∈ A ∪B

2. A ∪B = B ↔ A ⊆ B

(→) Assume that A ∪B = B.

By 1, A ⊆ A ∪B = B. So A ⊆ B.(←) Assume that A ⊆ B.(A ∪B ⊆ B) Let x ∈ A ∪B. Then x ∈ A or x ∈ B.



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Theorem (2.2.9)


1. A ⊆ A ∪B and B ⊆ A ∪B


Proof .


1. A ⊆ A ∪B x ∈ A → (x ∈ A ∨ x ∈ B) → x ∈ A ∪B

B ⊆ A ∪B x ∈ B ↔ (x ∈ A ∨ x ∈ B) → x ∈ A ∪B

2. A ∪B = B ↔ A ⊆ B

(→) Assume that A ∪B = B. By 1, A ⊆ A ∪B = B.

So A ⊆ B.(←) Assume that A ⊆ B.(A ∪B ⊆ B) Let x ∈ A ∪B. Then x ∈ A or x ∈ B.



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Theorem (2.2.9)


1. A ⊆ A ∪B and B ⊆ A ∪B


Proof .


1. A ⊆ A ∪B x ∈ A → (x ∈ A ∨ x ∈ B) → x ∈ A ∪B

B ⊆ A ∪B x ∈ B ↔ (x ∈ A ∨ x ∈ B) → x ∈ A ∪B

2. A ∪B = B ↔ A ⊆ B

(→) Assume that A ∪B = B. By 1, A ⊆ A ∪B = B. So A ⊆ B.

(←) Assume that A ⊆ B.(A ∪B ⊆ B) Let x ∈ A ∪B. Then x ∈ A or x ∈ B.



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Theorem (2.2.9)


1. A ⊆ A ∪B and B ⊆ A ∪B


Proof .


1. A ⊆ A ∪B x ∈ A → (x ∈ A ∨ x ∈ B) → x ∈ A ∪B

B ⊆ A ∪B x ∈ B ↔ (x ∈ A ∨ x ∈ B) → x ∈ A ∪B

2. A ∪B = B ↔ A ⊆ B

(→) Assume that A ∪B = B. By 1, A ⊆ A ∪B = B. So A ⊆ B.(←) Assume that A ⊆ B.

(A ∪B ⊆ B) Let x ∈ A ∪B. Then x ∈ A or x ∈ B.If x ∈ A, then x ∈ B because A ⊆ B. Thus, A ∪B ⊆ B.If x ∈ B, it is done. Thus, A ∪B ⊆ B.(B ⊆ A ∪B) By 1, B ⊆ A ∪B.


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Theorem (2.2.9)


1. A ⊆ A ∪B and B ⊆ A ∪B


Proof .


1. A ⊆ A ∪B x ∈ A → (x ∈ A ∨ x ∈ B) → x ∈ A ∪B

B ⊆ A ∪B x ∈ B ↔ (x ∈ A ∨ x ∈ B) → x ∈ A ∪B

2. A ∪B = B ↔ A ⊆ B

(→) Assume that A ∪B = B. By 1, A ⊆ A ∪B = B. So A ⊆ B.(←) Assume that A ⊆ B.(A ∪B ⊆ B)

Let x ∈ A ∪B. Then x ∈ A or x ∈ B.If x ∈ A, then x ∈ B because A ⊆ B. Thus, A ∪B ⊆ B.If x ∈ B, it is done. Thus, A ∪B ⊆ B.(B ⊆ A ∪B) By 1, B ⊆ A ∪B.


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Theorem (2.2.9)


1. A ⊆ A ∪B and B ⊆ A ∪B


Proof .


1. A ⊆ A ∪B x ∈ A → (x ∈ A ∨ x ∈ B) → x ∈ A ∪B

B ⊆ A ∪B x ∈ B ↔ (x ∈ A ∨ x ∈ B) → x ∈ A ∪B

2. A ∪B = B ↔ A ⊆ B

(→) Assume that A ∪B = B. By 1, A ⊆ A ∪B = B. So A ⊆ B.(←) Assume that A ⊆ B.(A ∪B ⊆ B) Let x ∈ A ∪B.

Then x ∈ A or x ∈ B.If x ∈ A, then x ∈ B because A ⊆ B. Thus, A ∪B ⊆ B.If x ∈ B, it is done. Thus, A ∪B ⊆ B.(B ⊆ A ∪B) By 1, B ⊆ A ∪B.


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Theorem (2.2.9)


1. A ⊆ A ∪B and B ⊆ A ∪B


Proof .


1. A ⊆ A ∪B x ∈ A → (x ∈ A ∨ x ∈ B) → x ∈ A ∪B

B ⊆ A ∪B x ∈ B ↔ (x ∈ A ∨ x ∈ B) → x ∈ A ∪B

2. A ∪B = B ↔ A ⊆ B




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Theorem (2.2.9)


1. A ⊆ A ∪B and B ⊆ A ∪B


Proof .


1. A ⊆ A ∪B x ∈ A → (x ∈ A ∨ x ∈ B) → x ∈ A ∪B

B ⊆ A ∪B x ∈ B ↔ (x ∈ A ∨ x ∈ B) → x ∈ A ∪B

2. A ∪B = B ↔ A ⊆ B


If x ∈ A,

then x ∈ B because A ⊆ B. Thus, A ∪B ⊆ B.If x ∈ B, it is done. Thus, A ∪B ⊆ B.(B ⊆ A ∪B) By 1, B ⊆ A ∪B.


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Theorem (2.2.9)


1. A ⊆ A ∪B and B ⊆ A ∪B


Proof .


1. A ⊆ A ∪B x ∈ A → (x ∈ A ∨ x ∈ B) → x ∈ A ∪B

B ⊆ A ∪B x ∈ B ↔ (x ∈ A ∨ x ∈ B) → x ∈ A ∪B

2. A ∪B = B ↔ A ⊆ B


If x ∈ A, then x ∈ B because A ⊆ B.

Thus, A ∪B ⊆ B.If x ∈ B, it is done. Thus, A ∪B ⊆ B.(B ⊆ A ∪B) By 1, B ⊆ A ∪B.


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Theorem (2.2.9)


1. A ⊆ A ∪B and B ⊆ A ∪B


Proof .


1. A ⊆ A ∪B x ∈ A → (x ∈ A ∨ x ∈ B) → x ∈ A ∪B

B ⊆ A ∪B x ∈ B ↔ (x ∈ A ∨ x ∈ B) → x ∈ A ∪B

2. A ∪B = B ↔ A ⊆ B


If x ∈ A, then x ∈ B because A ⊆ B. Thus, A ∪B ⊆ B.

If x ∈ B, it is done. Thus, A ∪B ⊆ B.(B ⊆ A ∪B) By 1, B ⊆ A ∪B.


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Theorem (2.2.9)


1. A ⊆ A ∪B and B ⊆ A ∪B


Proof .


1. A ⊆ A ∪B x ∈ A → (x ∈ A ∨ x ∈ B) → x ∈ A ∪B

B ⊆ A ∪B x ∈ B ↔ (x ∈ A ∨ x ∈ B) → x ∈ A ∪B

2. A ∪B = B ↔ A ⊆ B


If x ∈ A, then x ∈ B because A ⊆ B. Thus, A ∪B ⊆ B.If x ∈ B, it is done.

Thus, A ∪B ⊆ B.(B ⊆ A ∪B) By 1, B ⊆ A ∪B.


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Theorem (2.2.9)


1. A ⊆ A ∪B and B ⊆ A ∪B


Proof .


1. A ⊆ A ∪B x ∈ A → (x ∈ A ∨ x ∈ B) → x ∈ A ∪B

B ⊆ A ∪B x ∈ B ↔ (x ∈ A ∨ x ∈ B) → x ∈ A ∪B

2. A ∪B = B ↔ A ⊆ B


If x ∈ A, then x ∈ B because A ⊆ B. Thus, A ∪B ⊆ B.If x ∈ B, it is done. Thus, A ∪B ⊆ B.

(B ⊆ A ∪B) By 1, B ⊆ A ∪B.


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Theorem (2.2.9)


1. A ⊆ A ∪B and B ⊆ A ∪B


Proof .


1. A ⊆ A ∪B x ∈ A → (x ∈ A ∨ x ∈ B) → x ∈ A ∪B

B ⊆ A ∪B x ∈ B ↔ (x ∈ A ∨ x ∈ B) → x ∈ A ∪B

2. A ∪B = B ↔ A ⊆ B


If x ∈ A, then x ∈ B because A ⊆ B. Thus, A ∪B ⊆ B.If x ∈ B, it is done. Thus, A ∪B ⊆ B.(B ⊆ A ∪B)

By 1, B ⊆ A ∪B.


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Theorem (2.2.9)


1. A ⊆ A ∪B and B ⊆ A ∪B


Proof .


1. A ⊆ A ∪B x ∈ A → (x ∈ A ∨ x ∈ B) → x ∈ A ∪B

B ⊆ A ∪B x ∈ B ↔ (x ∈ A ∨ x ∈ B) → x ∈ A ∪B

2. A ∪B = B ↔ A ⊆ B




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Theorem (2.2.10)

Let A,B,C and D be sets. Then

1. if A ⊆ B and C ⊆ D, then A ∪ C ⊆ B ∪D

2. if A = B and C = D, then A ∪ C = B ∪D


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Proof .


1. A ⊆ B ∧ C ⊆ D → A ∪ C ⊆ B ∪D


x ∈ A ∪ C → x ∈ A ∨ x ∈ C

→ x ∈ B ∨ x ∈ D (∵ A ⊆ B and C ⊆ D)

→ x ∈ B ∪D

Thus, A ∪ C ⊆ B ∪D.

2. A = B ∧ C = D → A ∪ C = B ∪D

A = B ∧ C = D → A ⊆ B ∨ C ⊆ D

→ A ∪ C ⊆ B ∪D ( By 1 )

A = B ∧ C = D → B ⊆ A ∨D ⊆ C

→ B ∪D ⊆ A ∪ C ( By 1 )

Hence, A ∪ C = B ∪D.


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Proof .


1. A ⊆ B ∧ C ⊆ D → A ∪ C ⊆ B ∪D

Assume that A ⊆ B and C ⊆ D.

Then

x ∈ A ∪ C → x ∈ A ∨ x ∈ C

→ x ∈ B ∨ x ∈ D (∵ A ⊆ B and C ⊆ D)

→ x ∈ B ∪D


2. A = B ∧ C = D → A ∪ C = B ∪D

A = B ∧ C = D → A ⊆ B ∨ C ⊆ D

→ A ∪ C ⊆ B ∪D ( By 1 )

A = B ∧ C = D → B ⊆ A ∨D ⊆ C

→ B ∪D ⊆ A ∪ C ( By 1 )



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Proof .


1. A ⊆ B ∧ C ⊆ D → A ∪ C ⊆ B ∪D


x ∈ A ∪ C → x ∈ A ∨ x ∈ C

→ x ∈ B ∨ x ∈ D (∵ A ⊆ B and C ⊆ D)

→ x ∈ B ∪D


2. A = B ∧ C = D → A ∪ C = B ∪D

A = B ∧ C = D → A ⊆ B ∨ C ⊆ D

→ A ∪ C ⊆ B ∪D ( By 1 )

A = B ∧ C = D → B ⊆ A ∨D ⊆ C

→ B ∪D ⊆ A ∪ C ( By 1 )



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Theorem (2.2.11)


1. A ∩ (B ∪ C) = (A ∩B) ∪ (A ∩ C) (distributive law for intersection)

2. A ∪ (B ∩ C) = (A ∪B) ∩ (A ∪ C) (distributive law for union)


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Proof .


1. A ∩ (B ∪ C) = (A ∩B) ∪ (A ∩ C)

x ∈ A ∩ (B ∪ C) ↔ x ∈ A ∧ x ∈ B ∪ C

↔ x ∈ A ∧ (x ∈ B ∨ x ∈ C)

↔ (x ∈ A ∧ x ∈ B) ∨ (x ∈ A ∧ x ∈ C)

↔ x ∈ A ∩B ∨ x ∈ A ∩ C

↔ x ∈ (A ∩B) ∪ (A ∩ C)

2. A ∪ (B ∩ C) = (A ∪B) ∩ (A ∪ C)

x ∈ A ∪ (B ∩ C) ↔ x ∈ A ∨ x ∈ B ∩ C

↔ x ∈ A ∨ (x ∈ B ∧ x ∈ C)

↔ (x ∈ A ∨ x ∈ B) ∧ (x ∈ A ∨ x ∈ C)

↔ x ∈ A ∪B ∧ x ∈ A ∪ C

↔ x ∈ (A ∪B) ∩ (A ∪ C)


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Theorem (2.2.12)


x ∈ C ↔ (x ∈ A ∧ x /∈ B).

Proof .

Let A and B be sets. Let p(x) be statement x /∈ B. By the axiom of specification , there is a setC such that

x ∈ C ↔ (x ∈ A ∧ x /∈ B).

Let C2 be a set satisfyingx ∈ C2 ↔ (x ∈ A ∧ x /∈ B).


Thus, C = C2.


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Theorem (2.2.12)


x ∈ C ↔ (x ∈ A ∧ x /∈ B).

Proof .


Let p(x) be statement x /∈ B. By the axiom of specification , there is a setC such that

x ∈ C ↔ (x ∈ A ∧ x /∈ B).



Thus, C = C2.


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Theorem (2.2.12)


x ∈ C ↔ (x ∈ A ∧ x /∈ B).

Proof .

Let A and B be sets. Let p(x) be statement x /∈ B.

By the axiom of specification , there is a setC such that

x ∈ C ↔ (x ∈ A ∧ x /∈ B).



Thus, C = C2.


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Theorem (2.2.12)


x ∈ C ↔ (x ∈ A ∧ x /∈ B).

Proof .

Let A and B be sets. Let p(x) be statement x /∈ B. By the axiom of specification , there is a setC such that

x ∈ C ↔ (x ∈ A ∧ x /∈ B).



Thus, C = C2.


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Definition (2.2.3)

The set C in theorem 2.2.12 is different of A and B, denoted A−B , i.e.,

A−B = {x : x ∈ A ∧ x /∈ B} or x ∈ A−B ↔ (x ∈ A ∧ x /∈ B).


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Theorem (2.2.13)


1. A−A = ∅

2. A−∅ = A

3. ∅−A = ∅

4. A−B ⊆ A

Proof.


1. A−A = ∅ Since x ∈ A ∧ x /∈ A is false for any x and

x ∈ A−A ↔ x ∈ A ∧ x /∈ A,

x ∈ A−A is also false. Hence, A−A has no element or A−A = ∅.


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Theorem (2.2.13)


1. A−A = ∅

2. A−∅ = A

3. ∅−A = ∅

4. A−B ⊆ A

Proof.


1. A−A = ∅

Since x ∈ A ∧ x /∈ A is false for any x and

x ∈ A−A ↔ x ∈ A ∧ x /∈ A,



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Theorem (2.2.13)


1. A−A = ∅

2. A−∅ = A

3. ∅−A = ∅

4. A−B ⊆ A

Proof.


1. A−A = ∅ Since x ∈ A ∧ x /∈ A is false for any x

and

x ∈ A−A ↔ x ∈ A ∧ x /∈ A,



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Theorem (2.2.13)


1. A−A = ∅

2. A−∅ = A

3. ∅−A = ∅

4. A−B ⊆ A

Proof.



x ∈ A−A ↔ x ∈ A ∧ x /∈ A,



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Theorem (2.2.13)


1. A−A = ∅

2. A−∅ = A

3. ∅−A = ∅

4. A−B ⊆ A

Proof.



x ∈ A−A ↔ x ∈ A ∧ x /∈ A,

x ∈ A−A is also false.

Hence, A−A has no element or A−A = ∅.


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Theorem (2.2.13)


1. A−A = ∅

2. A−∅ = A

3. ∅−A = ∅

4. A−B ⊆ A

Proof.



x ∈ A−A ↔ x ∈ A ∧ x /∈ A,



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2. A− ∅ = A

For any x

A− ∅ ↔ x ∈ A ∧ x /∈ A

↔ x ∈ A (x /∈ ∅ is true )

So, A− ∅ = A.3. ∅−A = ∅ Since x ∈ ∅ ∧ x ∈ A is false for any x and

∅−A ↔ x ∈ ∅ ∧ x /∈ A

x ∈ ∅−A is also false. Hence, ∅−A has no element or ∅−A = ∅.4. A−B ⊆ A For any x

x ∈ A−B ↔ x ∈ A ∧ x /∈ B

→ x ∈ A

Thus, A−B ⊆ A.


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2. A− ∅ = A For any x

A− ∅ ↔ x ∈ A ∧ x /∈ A

↔ x ∈ A (x /∈ ∅ is true )


∅−A ↔ x ∈ ∅ ∧ x /∈ A


x ∈ A−B ↔ x ∈ A ∧ x /∈ B

→ x ∈ A

Thus, A−B ⊆ A.


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A− ∅ ↔ x ∈ A ∧ x /∈ A

↔ x ∈ A (x /∈ ∅ is true )

So, A− ∅ = A.

3. ∅−A = ∅ Since x ∈ ∅ ∧ x ∈ A is false for any x and

∅−A ↔ x ∈ ∅ ∧ x /∈ A


x ∈ A−B ↔ x ∈ A ∧ x /∈ B

→ x ∈ A

Thus, A−B ⊆ A.


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A− ∅ ↔ x ∈ A ∧ x /∈ A

↔ x ∈ A (x /∈ ∅ is true )

So, A− ∅ = A.3. ∅−A = ∅

Since x ∈ ∅ ∧ x ∈ A is false for any x and

∅−A ↔ x ∈ ∅ ∧ x /∈ A


x ∈ A−B ↔ x ∈ A ∧ x /∈ B

→ x ∈ A

Thus, A−B ⊆ A.


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A− ∅ ↔ x ∈ A ∧ x /∈ A

↔ x ∈ A (x /∈ ∅ is true )

So, A− ∅ = A.3. ∅−A = ∅ Since x ∈ ∅ ∧ x ∈ A is false for any x

and

∅−A ↔ x ∈ ∅ ∧ x /∈ A


x ∈ A−B ↔ x ∈ A ∧ x /∈ B

→ x ∈ A

Thus, A−B ⊆ A.


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A− ∅ ↔ x ∈ A ∧ x /∈ A

↔ x ∈ A (x /∈ ∅ is true )


∅−A ↔ x ∈ ∅ ∧ x /∈ A


x ∈ A−B ↔ x ∈ A ∧ x /∈ B

→ x ∈ A

Thus, A−B ⊆ A.


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A− ∅ ↔ x ∈ A ∧ x /∈ A

↔ x ∈ A (x /∈ ∅ is true )


∅−A ↔ x ∈ ∅ ∧ x /∈ A

x ∈ ∅−A is also false.

Hence, ∅−A has no element or ∅−A = ∅.4. A−B ⊆ A For any x

x ∈ A−B ↔ x ∈ A ∧ x /∈ B

→ x ∈ A

Thus, A−B ⊆ A.


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A− ∅ ↔ x ∈ A ∧ x /∈ A

↔ x ∈ A (x /∈ ∅ is true )


∅−A ↔ x ∈ ∅ ∧ x /∈ A

x ∈ ∅−A is also false. Hence, ∅−A has no element or ∅−A = ∅.

4. A−B ⊆ A For any x

x ∈ A−B ↔ x ∈ A ∧ x /∈ B

→ x ∈ A

Thus, A−B ⊆ A.


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A− ∅ ↔ x ∈ A ∧ x /∈ A

↔ x ∈ A (x /∈ ∅ is true )


∅−A ↔ x ∈ ∅ ∧ x /∈ A

x ∈ ∅−A is also false. Hence, ∅−A has no element or ∅−A = ∅.4. A−B ⊆ A

For any x

x ∈ A−B ↔ x ∈ A ∧ x /∈ B

→ x ∈ A

Thus, A−B ⊆ A.


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A− ∅ ↔ x ∈ A ∧ x /∈ A

↔ x ∈ A (x /∈ ∅ is true )


∅−A ↔ x ∈ ∅ ∧ x /∈ A


x ∈ A−B ↔ x ∈ A ∧ x /∈ B

→ x ∈ A

Thus, A−B ⊆ A.


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Theorem (2.2.14)


1. A ⊆ B if and only if A−B = ∅

2. if A ⊆ B, then (A− C) ⊆ (B − C)

3. if A = B, then (A− C) = (B − C)

Proof.


1. A ⊆ B ↔ A−B = ∅

A ⊆ B ↔ ∀x (x ∈ A→ x ∈ B)

A * B ↔ ¬∀x (x ∈ A→ x ∈ B)

↔ ∃x (x ∈ A ∧ x /∈ B)

↔ ∃x (x ∈ A−B)

↔ A−B ̸= ∅


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2. A ⊆ B → (A− C) ⊆ (B − C)

Assume A ⊆ B. Then, for any x,

x ∈ A− C ↔ x ∈ A ∧ x /∈ C

→ x ∈ B ∧ x /∈ C (∵ A ⊆ B)

↔ x ∈ B − C

Thus, (A− C) ⊆ (B − C).

3. A = B → (A− C) = (B − C) Assume A = B. Then, for any x,

x ∈ A− C ↔ x ∈ A ∧ x /∈ C

↔ x ∈ B ∧ x /∈ C (∵ A = B)

↔ x ∈ B − C

Thus, (A− C) = (B − C).


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2. A ⊆ B → (A− C) ⊆ (B − C) Assume A ⊆ B.

Then, for any x,

x ∈ A− C ↔ x ∈ A ∧ x /∈ C

→ x ∈ B ∧ x /∈ C (∵ A ⊆ B)

↔ x ∈ B − C

Thus, (A− C) ⊆ (B − C).


x ∈ A− C ↔ x ∈ A ∧ x /∈ C

↔ x ∈ B ∧ x /∈ C (∵ A = B)

↔ x ∈ B − C

Thus, (A− C) = (B − C).


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2. A ⊆ B → (A− C) ⊆ (B − C) Assume A ⊆ B. Then, for any x,

x ∈ A− C ↔ x ∈ A ∧ x /∈ C

→ x ∈ B ∧ x /∈ C (∵ A ⊆ B)

↔ x ∈ B − C

Thus, (A− C) ⊆ (B − C).


x ∈ A− C ↔ x ∈ A ∧ x /∈ C

↔ x ∈ B ∧ x /∈ C (∵ A = B)

↔ x ∈ B − C

Thus, (A− C) = (B − C).


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x ∈ A− C ↔ x ∈ A ∧ x /∈ C

→ x ∈ B ∧ x /∈ C (∵ A ⊆ B)

↔ x ∈ B − C

Thus, (A− C) ⊆ (B − C).

3. A = B → (A− C) = (B − C)

Assume A = B. Then, for any x,

x ∈ A− C ↔ x ∈ A ∧ x /∈ C

↔ x ∈ B ∧ x /∈ C (∵ A = B)

↔ x ∈ B − C

Thus, (A− C) = (B − C).


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x ∈ A− C ↔ x ∈ A ∧ x /∈ C

→ x ∈ B ∧ x /∈ C (∵ A ⊆ B)

↔ x ∈ B − C

Thus, (A− C) ⊆ (B − C).

3. A = B → (A− C) = (B − C) Assume A = B.

Then, for any x,

x ∈ A− C ↔ x ∈ A ∧ x /∈ C

↔ x ∈ B ∧ x /∈ C (∵ A = B)

↔ x ∈ B − C

Thus, (A− C) = (B − C).


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x ∈ A− C ↔ x ∈ A ∧ x /∈ C

→ x ∈ B ∧ x /∈ C (∵ A ⊆ B)

↔ x ∈ B − C

Thus, (A− C) ⊆ (B − C).


x ∈ A− C ↔ x ∈ A ∧ x /∈ C

↔ x ∈ B ∧ x /∈ C (∵ A = B)

↔ x ∈ B − C

Thus, (A− C) = (B − C).


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Take a BreakFor 10 Minutes


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Definition (2.2.4)

A universal set is the set of all elements under consideration, denoted by

U .

Definition (2.2.5)

Let A be a set with a universal set U . The complement of A, denoted Ac ,is

Ac = U −A = {x : x ∈ U ∧ x /∈ A} or x ∈ Ac ↔ (x ∈ U ∧ x /∈ A).

Note that

x /∈ Ac ↔ ¬(x ∈ U ∧ x /∈ A)

↔ (x /∈ U ∨ x ∈ A) (∵ x /∈ U is impossible )

↔ x ∈ A


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Theorem (2.2.15)

Let A be a set with a universal set U . Then1 (Ac)c = A

2 ∅c = U3 Uc = ∅4 A ∩Ac = ∅5 A ∪Ac = U


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Proof.

Let A be a set with a universal set U .

1. (Ac)c = A

x ∈ (Ac)c ↔ x ∈ U ∧ x /∈ Ac

↔ x ∈ U ∧ x ∈ A

↔ x ∈ A (∵ A ⊆ U)

Thus, (Ac)c = A.

2. ∅c = U

x ∈ ∅c ↔ x ∈ U ∧ x /∈ ∅↔ x ∈ U (∵ x /∈ ∅ is true )

So, ∅c = U .


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Proof.

Let A be a set with a universal set U .1. (Ac)c = A

x ∈ (Ac)c ↔ x ∈ U ∧ x /∈ Ac

↔ x ∈ U ∧ x ∈ A

↔ x ∈ A (∵ A ⊆ U)

Thus, (Ac)c = A.

2. ∅c = U

x ∈ ∅c ↔ x ∈ U ∧ x /∈ ∅↔ x ∈ U (∵ x /∈ ∅ is true )

So, ∅c = U .


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3. Uc = ∅

Since x ∈ U ∧ x /∈ U is false for any x and

x ∈ Uc ↔ x ∈ U ∧ x /∈ U ,

x ∈ Uc is also false. So, Uc has no element or Uc = ∅.4. A ∩Ac = ∅ Since x ∈ A ∧ x /∈ A is false for any x and

x ∈ A ∩Ac ↔ x ∈ A ∧ x ∈ Ac

↔ x ∈ A ∧ (x ∈ U ∧ x /∈ A)

↔ (x ∈ A ∧ x /∈ A) ∧ x ∈ U

x ∈ A ∩Ac is also false. So, A ∩Ac has no element or A ∩Ac = ∅.


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3. Uc = ∅ Since x ∈ U ∧ x /∈ U is false for any x

and

x ∈ Uc ↔ x ∈ U ∧ x /∈ U ,


x ∈ A ∩Ac ↔ x ∈ A ∧ x ∈ Ac

↔ x ∈ A ∧ (x ∈ U ∧ x /∈ A)

↔ (x ∈ A ∧ x /∈ A) ∧ x ∈ U



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3. Uc = ∅ Since x ∈ U ∧ x /∈ U is false for any x and

x ∈ Uc ↔ x ∈ U ∧ x /∈ U ,


x ∈ A ∩Ac ↔ x ∈ A ∧ x ∈ Ac

↔ x ∈ A ∧ (x ∈ U ∧ x /∈ A)

↔ (x ∈ A ∧ x /∈ A) ∧ x ∈ U



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x ∈ Uc ↔ x ∈ U ∧ x /∈ U ,

x ∈ Uc is also false.

So, Uc has no element or Uc = ∅.4. A ∩Ac = ∅ Since x ∈ A ∧ x /∈ A is false for any x and

x ∈ A ∩Ac ↔ x ∈ A ∧ x ∈ Ac

↔ x ∈ A ∧ (x ∈ U ∧ x /∈ A)

↔ (x ∈ A ∧ x /∈ A) ∧ x ∈ U



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x ∈ Uc ↔ x ∈ U ∧ x /∈ U ,

x ∈ Uc is also false. So, Uc has no element or Uc = ∅.

4. A ∩Ac = ∅ Since x ∈ A ∧ x /∈ A is false for any x and

x ∈ A ∩Ac ↔ x ∈ A ∧ x ∈ Ac

↔ x ∈ A ∧ (x ∈ U ∧ x /∈ A)

↔ (x ∈ A ∧ x /∈ A) ∧ x ∈ U



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x ∈ Uc ↔ x ∈ U ∧ x /∈ U ,

x ∈ Uc is also false. So, Uc has no element or Uc = ∅.4. A ∩Ac = ∅

Since x ∈ A ∧ x /∈ A is false for any x and

x ∈ A ∩Ac ↔ x ∈ A ∧ x ∈ Ac

↔ x ∈ A ∧ (x ∈ U ∧ x /∈ A)

↔ (x ∈ A ∧ x /∈ A) ∧ x ∈ U



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x ∈ Uc ↔ x ∈ U ∧ x /∈ U ,

x ∈ Uc is also false. So, Uc has no element or Uc = ∅.4. A ∩Ac = ∅ Since x ∈ A ∧ x /∈ A is false for any x

and

x ∈ A ∩Ac ↔ x ∈ A ∧ x ∈ Ac

↔ x ∈ A ∧ (x ∈ U ∧ x /∈ A)

↔ (x ∈ A ∧ x /∈ A) ∧ x ∈ U



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x ∈ Uc ↔ x ∈ U ∧ x /∈ U ,


x ∈ A ∩Ac ↔ x ∈ A ∧ x ∈ Ac

↔ x ∈ A ∧ (x ∈ U ∧ x /∈ A)

↔ (x ∈ A ∧ x /∈ A) ∧ x ∈ U



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x ∈ Uc ↔ x ∈ U ∧ x /∈ U ,


x ∈ A ∩Ac ↔ x ∈ A ∧ x ∈ Ac

↔ x ∈ A ∧ (x ∈ U ∧ x /∈ A)

↔ (x ∈ A ∧ x /∈ A) ∧ x ∈ U

x ∈ A ∩Ac is also false.

So, A ∩Ac has no element or A ∩Ac = ∅.


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x ∈ Uc ↔ x ∈ U ∧ x /∈ U ,


x ∈ A ∩Ac ↔ x ∈ A ∧ x ∈ Ac

↔ x ∈ A ∧ (x ∈ U ∧ x /∈ A)

↔ (x ∈ A ∧ x /∈ A) ∧ x ∈ U



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5. A ∪Ac = U

Since x ∈ A ∨ x /∈ A is true for any x and

x ∈ A ∪Ac ↔ x ∈ A ∨ x ∈ Ac

↔ x ∈ A ∨ (x ∈ U ∧ x /∈ A)

↔ (x ∈ A ∨ x ∈ U) ∧ (x ∈ A ∨ x /∈ A)

↔ x ∈ A ∨ x ∈ U↔ x ∈ U (∵ A ⊆ U)

Hence, A ∪Ac = U .


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5. A ∪Ac = U Since x ∈ A ∨ x /∈ A is true for any x

and

x ∈ A ∪Ac ↔ x ∈ A ∨ x ∈ Ac

↔ x ∈ A ∨ (x ∈ U ∧ x /∈ A)

↔ (x ∈ A ∨ x ∈ U) ∧ (x ∈ A ∨ x /∈ A)

↔ x ∈ A ∨ x ∈ U↔ x ∈ U (∵ A ⊆ U)



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5. A ∪Ac = U Since x ∈ A ∨ x /∈ A is true for any x and

x ∈ A ∪Ac ↔ x ∈ A ∨ x ∈ Ac

↔ x ∈ A ∨ (x ∈ U ∧ x /∈ A)

↔ (x ∈ A ∨ x ∈ U) ∧ (x ∈ A ∨ x /∈ A)

↔ x ∈ A ∨ x ∈ U↔ x ∈ U (∵ A ⊆ U)



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Theorem (2.2.16)

Let A and B be sets with a universal set U . Then

1. A−B = A ∩Bc

2. A ⊆ B if and only if Bc ⊆ Ac


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Proof.

Let A and B be a set with a universal set U .

1. A−B = A ∩Bc For any x

x ∈ A ∩Bc ↔ x ∈ A ∧ x ∈ Bc

↔ x ∈ A ∧ (x ∈ U ∧ x /∈ B)

↔ (x ∈ A ∧ x ∈ U) ∧ x /∈ B

↔ x ∈ A ∧ x /∈ B (∵ A ⊆ U)↔ x ∈ A−B

So, A−B = A ∩Bc.2. A ⊆ B ↔ Bc ⊆ Ac

Bc ⊆ Ac ↔ ∀x (x ∈ Bc → x ∈ Ac)

↔ ∀x (x /∈ Ac → x /∈ Bc)

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

↔ ∀x [(x /∈ U ∨ x ∈ A)→ (x /∈ U ∨ x ∈ B)]

↔ ∀x [x ∈ A→ x ∈ B] (∵ x /∈ U is false )

↔ A ⊆ B


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Proof.

Let A and B be a set with a universal set U .1. A−B = A ∩Bc

For any x

x ∈ A ∩Bc ↔ x ∈ A ∧ x ∈ Bc

↔ x ∈ A ∧ (x ∈ U ∧ x /∈ B)

↔ (x ∈ A ∧ x ∈ U) ∧ x /∈ B

↔ x ∈ A ∧ x /∈ B (∵ A ⊆ U)↔ x ∈ A−B

So, A−B = A ∩Bc.2. A ⊆ B ↔ Bc ⊆ Ac

Bc ⊆ Ac ↔ ∀x (x ∈ Bc → x ∈ Ac)

↔ ∀x (x /∈ Ac → x /∈ Bc)

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

↔ ∀x [(x /∈ U ∨ x ∈ A)→ (x /∈ U ∨ x ∈ B)]

↔ ∀x [x ∈ A→ x ∈ B] (∵ x /∈ U is false )

↔ A ⊆ B


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Proof.

Let A and B be a set with a universal set U .1. A−B = A ∩Bc For any x

x ∈ A ∩Bc ↔ x ∈ A ∧ x ∈ Bc

↔ x ∈ A ∧ (x ∈ U ∧ x /∈ B)

↔ (x ∈ A ∧ x ∈ U) ∧ x /∈ B

↔ x ∈ A ∧ x /∈ B (∵ A ⊆ U)↔ x ∈ A−B

So, A−B = A ∩Bc.2. A ⊆ B ↔ Bc ⊆ Ac

Bc ⊆ Ac ↔ ∀x (x ∈ Bc → x ∈ Ac)

↔ ∀x (x /∈ Ac → x /∈ Bc)

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

↔ ∀x [(x /∈ U ∨ x ∈ A)→ (x /∈ U ∨ x ∈ B)]

↔ ∀x [x ∈ A→ x ∈ B] (∵ x /∈ U is false )

↔ A ⊆ B


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Proof.


x ∈ A ∩Bc ↔ x ∈ A ∧ x ∈ Bc

↔ x ∈ A ∧ (x ∈ U ∧ x /∈ B)

↔ (x ∈ A ∧ x ∈ U) ∧ x /∈ B

↔ x ∈ A ∧ x /∈ B (∵ A ⊆ U)↔ x ∈ A−B

So, A−B = A ∩Bc.

2. A ⊆ B ↔ Bc ⊆ Ac

Bc ⊆ Ac ↔ ∀x (x ∈ Bc → x ∈ Ac)

↔ ∀x (x /∈ Ac → x /∈ Bc)

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

↔ ∀x [(x /∈ U ∨ x ∈ A)→ (x /∈ U ∨ x ∈ B)]

↔ ∀x [x ∈ A→ x ∈ B] (∵ x /∈ U is false )

↔ A ⊆ B


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Proof.


x ∈ A ∩Bc ↔ x ∈ A ∧ x ∈ Bc

↔ x ∈ A ∧ (x ∈ U ∧ x /∈ B)

↔ (x ∈ A ∧ x ∈ U) ∧ x /∈ B

↔ x ∈ A ∧ x /∈ B (∵ A ⊆ U)↔ x ∈ A−B

So, A−B = A ∩Bc.2. A ⊆ B ↔ Bc ⊆ Ac

Bc ⊆ Ac ↔ ∀x (x ∈ Bc → x ∈ Ac)

↔ ∀x (x /∈ Ac → x /∈ Bc)

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

↔ ∀x [(x /∈ U ∨ x ∈ A)→ (x /∈ U ∨ x ∈ B)]

↔ ∀x [x ∈ A→ x ∈ B] (∵ x /∈ U is false )

↔ A ⊆ B


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De Morgan’s Law

Theorem (2.2.17)

Let A and B be sets with a universal set U . Then1 (A ∩B)c = Ac ∪Bc

2 (A ∪B)c = Ac ∩Bc


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Proof.

Let A and B be a set with a universal set U .

1. (A ∩B)c = Ac ∪Bc For any x

x ∈ (A ∩B)c ↔ x ∈ U ∧ x /∈ A ∩B

↔ x ∈ U ∧ ¬(x ∈ A ∩B)

↔ x ∈ U ∧ ¬(x ∈ A ∧ x ∈ B)

↔ x ∈ U ∧ (x /∈ A ∨ x /∈ B)

↔ (x ∈ U ∧ x /∈ A) ∨ (x ∈ U ∧ x /∈ B)

↔ x ∈ Ac ∨ x ∈ Bc

↔ x ∈ Ac ∪Bc

So, (A ∩B)c = Ac ∪Bc.


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Proof.

Let A and B be a set with a universal set U .1. (A ∩B)c = Ac ∪Bc

For any x

x ∈ (A ∩B)c ↔ x ∈ U ∧ x /∈ A ∩B

↔ x ∈ U ∧ ¬(x ∈ A ∩B)

↔ x ∈ U ∧ ¬(x ∈ A ∧ x ∈ B)

↔ x ∈ U ∧ (x /∈ A ∨ x /∈ B)

↔ (x ∈ U ∧ x /∈ A) ∨ (x ∈ U ∧ x /∈ B)

↔ x ∈ Ac ∨ x ∈ Bc

↔ x ∈ Ac ∪Bc



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Proof.

Let A and B be a set with a universal set U .1. (A ∩B)c = Ac ∪Bc For any x

x ∈ (A ∩B)c ↔ x ∈ U ∧ x /∈ A ∩B

↔ x ∈ U ∧ ¬(x ∈ A ∩B)

↔ x ∈ U ∧ ¬(x ∈ A ∧ x ∈ B)

↔ x ∈ U ∧ (x /∈ A ∨ x /∈ B)

↔ (x ∈ U ∧ x /∈ A) ∨ (x ∈ U ∧ x /∈ B)

↔ x ∈ Ac ∨ x ∈ Bc

↔ x ∈ Ac ∪Bc



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2. (A ∪B)c = Ac ∩Bc

For any x

x ∈ (A ∪B)c ↔ x ∈ U ∧ x /∈ A ∪B

↔ x ∈ U ∧ ¬(x ∈ A ∪B)

↔ x ∈ U ∧ ¬(x ∈ A ∨ x ∈ B)

↔ x ∈ U ∧ (x /∈ A ∧ x /∈ B)

↔ (x ∈ U ∧ x /∈ A) ∧ (x ∈ U ∧ x /∈ B)

↔ x ∈ Ac ∧ x ∈ Bc

↔ x ∈ Ac ∩Bc

Thus, (A ∪B)c = Ac ∩Bc.


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2. (A ∪B)c = Ac ∩Bc For any x

x ∈ (A ∪B)c ↔ x ∈ U ∧ x /∈ A ∪B

↔ x ∈ U ∧ ¬(x ∈ A ∪B)

↔ x ∈ U ∧ ¬(x ∈ A ∨ x ∈ B)

↔ x ∈ U ∧ (x /∈ A ∧ x /∈ B)

↔ (x ∈ U ∧ x /∈ A) ∧ (x ∈ U ∧ x /∈ B)

↔ x ∈ Ac ∧ x ∈ Bc

↔ x ∈ Ac ∩Bc

Thus, (A ∪B)c = Ac ∩Bc.


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Theorem (2.2.18)


1. A− (B ∩ C) = (A−B) ∪ (A− C)

2. A− (B ∪ C) = (A−B) ∩ (A− C)

Proof. Assignment 2


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The Aixiomatic Set Theory The Power Set Axiom

2.3 The Power Set Axiom

Axiom 2.3.1 (Axiom of Power set)

Let A be a set. There is a set C satisfying

x ∈ C ↔ x ⊆ A.

Theorem (2.3.1)

Let A be a set. There is a unique set C satisfying

x ∈ C ↔ x ⊆ A.

Proof .

Let A be a set and let C1 and C2 be sets satisfying condition of theorem 2.3.1. Then

x ∈ C1 ↔ x ⊆ A

x ∈ C2 ↔ x ⊆ A

∴ x ∈ C1 ↔ x ∈ C2

Therefore, C1 = C2.


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x ∈ C ↔ x ⊆ A.

Theorem (2.3.1)


x ∈ C ↔ x ⊆ A.

Proof .

Let A be a set and let C1 and C2 be sets satisfying condition of theorem 2.3.1.

Then

x ∈ C1 ↔ x ⊆ A

x ∈ C2 ↔ x ⊆ A

∴ x ∈ C1 ↔ x ∈ C2

Therefore, C1 = C2.


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x ∈ C ↔ x ⊆ A.

Theorem (2.3.1)


x ∈ C ↔ x ⊆ A.

Proof .


x ∈ C1 ↔ x ⊆ A

x ∈ C2 ↔ x ⊆ A

∴ x ∈ C1 ↔ x ∈ C2

Therefore, C1 = C2.


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x ∈ C ↔ x ⊆ A.

Theorem (2.3.1)


x ∈ C ↔ x ⊆ A.

Proof .


x ∈ C1 ↔ x ⊆ A

x ∈ C2 ↔ x ⊆ A

∴ x ∈ C1 ↔ x ∈ C2

Therefore, C1 = C2.Thanatyod Jampawai, Ph.D. MAT1202 SET THEORY :Week2 44 / 1

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Definition (2.3.1)

The set C in therorem 2.3.1 is a power set of A, denoted P(A) , i.e.,

P(A) = {x : x ⊆ A} or x ∈ P(A) ↔ x ⊆ A.

Theorem (2.3.2)

Let A be a set. Then

1. A ∈ P(A)

2. ∅ ∈ P(A)

3. P (∅) = {∅}


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Proof.

Let A be a set.

1. A ∈ P(A) Since A ⊆ A, A ∈ P(A).

2. ∅ ∈ P(A) Since ∅ ⊆ A, ∅ ∈ P(A).

3. P(∅) = {∅}

x ∈ P(∅) ↔ x ⊆ ∅↔ x = ∅ ( By theorem 2.1.15 )

↔ x ∈ {∅}


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Proof.

Let A be a set.1. A ∈ P(A)

Since A ⊆ A, A ∈ P(A).

2. ∅ ∈ P(A) Since ∅ ⊆ A, ∅ ∈ P(A).

3. P(∅) = {∅}

x ∈ P(∅) ↔ x ⊆ ∅↔ x = ∅ ( By theorem 2.1.15 )

↔ x ∈ {∅}


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Proof.

Let A be a set.1. A ∈ P(A) Since A ⊆ A, A ∈ P(A).

2. ∅ ∈ P(A) Since ∅ ⊆ A, ∅ ∈ P(A).

3. P(∅) = {∅}

x ∈ P(∅) ↔ x ⊆ ∅↔ x = ∅ ( By theorem 2.1.15 )

↔ x ∈ {∅}


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Proof.


2. ∅ ∈ P(A)

Since ∅ ⊆ A, ∅ ∈ P(A).

3. P(∅) = {∅}

x ∈ P(∅) ↔ x ⊆ ∅↔ x = ∅ ( By theorem 2.1.15 )

↔ x ∈ {∅}


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Proof.


2. ∅ ∈ P(A) Since ∅ ⊆ A, ∅ ∈ P(A).

3. P(∅) = {∅}

x ∈ P(∅) ↔ x ⊆ ∅↔ x = ∅ ( By theorem 2.1.15 )

↔ x ∈ {∅}


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Example (2.3.1)

Draw Hasse diagram of all elements of power set A = {x, y, z}.

Solution

P(A) = {∅, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}

{x, y, z}

{x, y} {y, z} {x, z}

{y} {x} {z}

∅


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Theorem (2.3.3)

If a set has n elements, then its power set will contain 2n elements.

Proof .

Let n be the number of elements of A. Then numbers of subset sets of A which have

0, 1, 2, ..., n elements,

respectively, are (n0

),

(n1

),

(n2

), ...,

(nn

).

By binomial theorem, it implies that(n0

)+

(n1

)+

(n2

)+ · · ·+

(nn

)= 2n.

Thus, the number of elements in the power set A is 2n.


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Theorem (2.3.3)


Proof .

Let n be the number of elements of A.

Then numbers of subset sets of A which have

0, 1, 2, ..., n elements,


),

(n1

),

(n2

), ...,

(nn

).


)+

(n1

)+

(n2

)+ · · ·+

(nn

)= 2n.



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Theorem (2.3.3)


Proof .


0, 1, 2, ..., n elements,

respectively,

are (n0

),

(n1

),

(n2

), ...,

(nn

).


)+

(n1

)+

(n2

)+ · · ·+

(nn

)= 2n.



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Theorem (2.3.3)


Proof .


0, 1, 2, ..., n elements,


),

(n1

),

(n2

), ...,

(nn

).


)+

(n1

)+

(n2

)+ · · ·+

(nn

)= 2n.



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Theorem (2.3.3)


Proof .


0, 1, 2, ..., n elements,


),

(n1

),

(n2

), ...,

(nn

).

By binomial theorem, it implies that

(n0

)+

(n1

)+

(n2

)+ · · ·+

(nn

)= 2n.



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Theorem (2.3.3)


Proof .


0, 1, 2, ..., n elements,


),

(n1

),

(n2

), ...,

(nn

).


)+

(n1

)+

(n2

)+ · · ·+

(nn

)= 2n.



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Example (2.3.2)

Compute number of elements of a power set

A = {x ∈ Z : |x| < 100 and 3 | x}.

Solution

A = {x ∈ Z : |x| < 100 and 3 | x}A = {0,±3,±6, ...,±99}

Hence, A has number of elements to be 33× 2 + 1 = 67.Therefore, number of elements of P(A) is 267.


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Example (2.3.2)


A = {x ∈ Z : |x| < 100 and 3 | x}.

Solution

A = {x ∈ Z : |x| < 100 and 3 | x}A = {0,±3,±6, ...,±99}

Hence, A has number of elements to be 33× 2 + 1 = 67.

Therefore, number of elements of P(A) is 267.


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Example (2.3.2)


A = {x ∈ Z : |x| < 100 and 3 | x}.

Solution

A = {x ∈ Z : |x| < 100 and 3 | x}A = {0,±3,±6, ...,±99}

Hence, A has number of elements to be 33× 2 + 1 = 67.Therefore, number of elements of P(A) is 267.


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Theorem (2.3.4)


1. A ⊆ B if and only if P(A) ⊆ P(B)

2. P(A ∩B) = P(A) ∩ P(B)

3. P(A ∪B) ⊇ P(A) ∪ P(B)


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Proof.


1. A ⊆ B ↔ P(A) ⊆ P(B)

(→) Assume A ⊆ B.

x ∈ P(A) → x ⊆ A

→ x ⊆ B (∵ A ⊆ B)

→ x ∈ P(B)

Thus, P(A) ⊆ P(B).(←) Assume P(A) ⊆ P(B).

x ∈ A → {x} ⊆ A

→ {x} ∈ P(A)

→ {x} ∈ P(B) (∵ P(A) ⊆ P(B))

→ {x} ⊆ B

→ x ∈ B

Thus, A ⊆ B.


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Proof.

Let A and B be sets.1. A ⊆ B ↔ P(A) ⊆ P(B)


x ∈ P(A) → x ⊆ A

→ x ⊆ B (∵ A ⊆ B)

→ x ∈ P(B)


x ∈ A → {x} ⊆ A

→ {x} ∈ P(A)

→ {x} ∈ P(B) (∵ P(A) ⊆ P(B))

→ {x} ⊆ B

→ x ∈ B

Thus, A ⊆ B.


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Proof.



x ∈ P(A) → x ⊆ A

→ x ⊆ B (∵ A ⊆ B)

→ x ∈ P(B)

Thus, P(A) ⊆ P(B).

(←) Assume P(A) ⊆ P(B).

x ∈ A → {x} ⊆ A

→ {x} ∈ P(A)

→ {x} ∈ P(B) (∵ P(A) ⊆ P(B))

→ {x} ⊆ B

→ x ∈ B

Thus, A ⊆ B.


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Proof.



x ∈ P(A) → x ⊆ A

→ x ⊆ B (∵ A ⊆ B)

→ x ∈ P(B)


x ∈ A → {x} ⊆ A

→ {x} ∈ P(A)

→ {x} ∈ P(B) (∵ P(A) ⊆ P(B))

→ {x} ⊆ B

→ x ∈ B

Thus, A ⊆ B.


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2. P(A ∩B) = P(A) ∩ P(B)

For any x

x ∈ P(A) ∩ P(B) ↔ x ∈ P(A) ∧ x ∈ P(B)

↔ x ⊆ A ∧ x ⊆ B

↔ x ⊆ A ∩B ( By theorem 2.2.5)↔ x ∈ P(A ∩B)

3. P(A ∪B) ⊇ P(A) ∪ P(B)

x ∈ P(A) ∪ P(B) → x ∈ P(A) ∨ x ∈ P(B)

→ x ⊆ A ∨ x ⊆ B

→ x ⊆ A ∪B ∨ x ⊆ A ∪B (∵ A ⊆ A ∪B and B ⊆ A ∪B)

→ x ∈ P(A ∪B)


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2. P(A ∩B) = P(A) ∩ P(B) For any x

x ∈ P(A) ∩ P(B) ↔ x ∈ P(A) ∧ x ∈ P(B)

↔ x ⊆ A ∧ x ⊆ B

↔ x ⊆ A ∩B ( By theorem 2.2.5)↔ x ∈ P(A ∩B)

3. P(A ∪B) ⊇ P(A) ∪ P(B)

x ∈ P(A) ∪ P(B) → x ∈ P(A) ∨ x ∈ P(B)

→ x ⊆ A ∨ x ⊆ B

→ x ⊆ A ∪B ∨ x ⊆ A ∪B (∵ A ⊆ A ∪B and B ⊆ A ∪B)

→ x ∈ P(A ∪B)


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Axiom 2.3.2 (The Axiom of Regularity)

Every non-empty set x contains a member y such that x and y are disjoint sets.

∀x [∃a (a ∈ x) → ∃y (y ∈ x ∧ ¬∃z (z ∈ x ∧ z ∈ y))]

In other word, for any A ̸= ∅, there exists C ∈ A such that

∀x [x ∈ C → x /∈ A] or C ∩A = ∅.


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Theorem (2.3.5)

For any set A, A /∈ A.

Proof .

Let A be a set. We will prove by contradiction. Suppose that A ∈ A.Since A ∈ {A}, A ∈ A ∩ {A}. So,

A ∩ {A} ̸= ∅ (1)

Since {A} ̸= ∅ by the axiom of regularity , there is set C ∈ {A} such that C ∩ {A} = ∅.

But C ∈ {A}means that C = A, so A ∩ {A} = ∅.

It contradics (1).


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Theorem (2.3.11)

For set A and B, A /∈ B or B /∈ A.

Proof .

Let A and B be sets. We will prove by contradiction. Suppose that A ∈ B and B ∈ A. SinceA ∈ {A,B} and B ∈ {A,B},

(A ∈ B ∩ {A,B}) ∧ (B ∈ A ∩ {A,B}) (2)

Since {A,B} ̸= ∅ by the axiom of regularity , there is set C ∈ {A,B} such thatC ∩ {A,B} = ∅. But C ∈ {A,B}means that C = A or C = B, so

A ∩ {A,B} = ∅ or B ∩ {A,B} = ∅.

It contradics (2).


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Conclusion

The Existential Axiom There is a set at least one.The Axiom of Extensionality Two sets are equal (are the same set) if they have the sameelements.The Axiom of Specification To every set A and to every condition p(x) therecorresponds a set B whose elements are exactly those elements x of A for which p(x)holds.Axiom of Union Let A and B be sets. There is a set C satisfying

x ∈ C ↔ (x ∈ A ∨ x ∈ B).

Axiom of Power set Let A be a set. There is a set C satisfying

x ∈ C ↔ x ⊆ A.


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Assignment 2 (30 Minutes)

1. (EVEN and ODD) Let A1, A2, ..., An, B1, B2, ..., Bn be sets. Then, for any n ∈ N,

A1 = B1 ∧A2 = B2 ∧ ...∧An = Bn → (A1 ∩A2 ∩ ...∩An) = (B1 ∩B2 ∩ ...∩Bn).

2. Let A,B and C be sets. Prove that

(a) (EVEN) A− (B ∩ C) = (A−B) ∪ (A− C)

(b) (ODD) A− (B ∪ C) = (A−B) ∩ (A− C)

3. (EVEN and ODD) Let A = {1, 2, 3, 4}.

(a) Cumpute numbers of subset sets of A which have 0, 1, 2, 3, 4 elements by binomialtheorem.

(b) Write out elements of P(A).(c) Draw Hasse diagram of elements of P(A).


MAT1202 SET THEORY :Week2 · Thanatyod Jampawai, Ph.D. MAT1202 SET THEORY :Week2 4 / 1..... The...

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Transcript of MAT1202 SET THEORY :Week2 · Thanatyod Jampawai, Ph.D. MAT1202 SET THEORY :Week2 4 / 1..... The...