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Page 1: Introduction to Set Theory · Introduction to Set Theory February 4, 2018 Chapter 1. Sets 1.4. Elementary Operations on Sets—Proofs of Theorems Introduction to Set Theory February

Introduction to Set Theory

February 4, 2018

Chapter 1. Sets1.4. Elementary Operations on Sets—Proofs of Theorems

() Introduction to Set Theory February 4, 2018 1 / 3

Page 2: Introduction to Set Theory · Introduction to Set Theory February 4, 2018 Chapter 1. Sets 1.4. Elementary Operations on Sets—Proofs of Theorems Introduction to Set Theory February

Table of contents

1 Theorem 1.4.A

() Introduction to Set Theory February 4, 2018 2 / 3

Page 3: Introduction to Set Theory · Introduction to Set Theory February 4, 2018 Chapter 1. Sets 1.4. Elementary Operations on Sets—Proofs of Theorems Introduction to Set Theory February

Theorem 1.4.A

Theorem 1.4.A

Theorem 1.4.A. Let A, B, and C be sets. Then (Distributivity):A ∩ (B ∪ C ) = (A ∩ B) ∪ (A ∩ C ).

Proof. Let x ∈ A ∩ (B ∪ C ). Then x ∈ A and x ∈ B ∪ C . Therefore,x ∈ A and either x ∈ B or x ∈ C .

So either (x ∈ A and x ∈ B) or (x ∈ Aor x ∈ C ); that is, either x ∈ A ∩ B or x ∈ A ∩ C . Hencex ∈ (A ∩ B) ∪ (A ∩ C ). Ergo, every element of A ∩ (B ∪ C ) is in(A ∩ B) ∪ (A ∩ C ) and so A ∩ (B ∪ C ) ⊆ (A ∩ B) ∪ (A ∩ C ).

Let x ∈ (A∩B)∪ (A∩ C ). Then either x ∈ A∩B or x ∈ A∩ C . So either(x ∈ A or x ∈ B) or (x ∈ A and x ∈ C ). If x ∈ A and x ∈ B then x ∈ Aand x ∈ B ∪ C (since B ⊆ B ∪ C ), and so x ∈ A ∩ (B ∪ C ). If x ∈ A andx ∈ C then x ∈ A and x ∈ B ∪ C (since C ⊆ B ∪ C ), and sox ∈ A ∩ (B ∪ C ). In either case of (x ∈ A) and x ∈ B) and (x ∈ A andx ∈ C ) we have x ∈ A ∩ (B ∪ C ). Ergo (A ∩ B) ∪ (A ∩ C ) ⊆ A ∩ (B ∪ C ).As observed above, these imply that A∩ (B ∪C ) = (A∩B)∪ (A∩C ).

() Introduction to Set Theory February 4, 2018 3 / 3

Page 4: Introduction to Set Theory · Introduction to Set Theory February 4, 2018 Chapter 1. Sets 1.4. Elementary Operations on Sets—Proofs of Theorems Introduction to Set Theory February

Theorem 1.4.A

Theorem 1.4.A

Theorem 1.4.A. Let A, B, and C be sets. Then (Distributivity):A ∩ (B ∪ C ) = (A ∩ B) ∪ (A ∩ C ).

Proof. Let x ∈ A ∩ (B ∪ C ). Then x ∈ A and x ∈ B ∪ C . Therefore,x ∈ A and either x ∈ B or x ∈ C . So either (x ∈ A and x ∈ B) or (x ∈ Aor x ∈ C ); that is, either x ∈ A ∩ B or x ∈ A ∩ C . Hencex ∈ (A ∩ B) ∪ (A ∩ C ). Ergo, every element of A ∩ (B ∪ C ) is in(A ∩ B) ∪ (A ∩ C ) and so A ∩ (B ∪ C ) ⊆ (A ∩ B) ∪ (A ∩ C ).

Let x ∈ (A∩B)∪ (A∩ C ). Then either x ∈ A∩B or x ∈ A∩ C . So either(x ∈ A or x ∈ B) or (x ∈ A and x ∈ C ). If x ∈ A and x ∈ B then x ∈ Aand x ∈ B ∪ C (since B ⊆ B ∪ C ), and so x ∈ A ∩ (B ∪ C ). If x ∈ A andx ∈ C then x ∈ A and x ∈ B ∪ C (since C ⊆ B ∪ C ), and sox ∈ A ∩ (B ∪ C ). In either case of (x ∈ A) and x ∈ B) and (x ∈ A andx ∈ C ) we have x ∈ A ∩ (B ∪ C ). Ergo (A ∩ B) ∪ (A ∩ C ) ⊆ A ∩ (B ∪ C ).As observed above, these imply that A∩ (B ∪C ) = (A∩B)∪ (A∩C ).

() Introduction to Set Theory February 4, 2018 3 / 3

Page 5: Introduction to Set Theory · Introduction to Set Theory February 4, 2018 Chapter 1. Sets 1.4. Elementary Operations on Sets—Proofs of Theorems Introduction to Set Theory February

Theorem 1.4.A

Theorem 1.4.A

Theorem 1.4.A. Let A, B, and C be sets. Then (Distributivity):A ∩ (B ∪ C ) = (A ∩ B) ∪ (A ∩ C ).

Proof. Let x ∈ A ∩ (B ∪ C ). Then x ∈ A and x ∈ B ∪ C . Therefore,x ∈ A and either x ∈ B or x ∈ C . So either (x ∈ A and x ∈ B) or (x ∈ Aor x ∈ C ); that is, either x ∈ A ∩ B or x ∈ A ∩ C . Hencex ∈ (A ∩ B) ∪ (A ∩ C ). Ergo, every element of A ∩ (B ∪ C ) is in(A ∩ B) ∪ (A ∩ C ) and so A ∩ (B ∪ C ) ⊆ (A ∩ B) ∪ (A ∩ C ).

Let x ∈ (A∩B)∪ (A∩ C ). Then either x ∈ A∩B or x ∈ A∩ C . So either(x ∈ A or x ∈ B) or (x ∈ A and x ∈ C ). If x ∈ A and x ∈ B then x ∈ Aand x ∈ B ∪ C (since B ⊆ B ∪ C ), and so x ∈ A ∩ (B ∪ C ). If x ∈ A andx ∈ C then x ∈ A and x ∈ B ∪ C (since C ⊆ B ∪ C ), and sox ∈ A ∩ (B ∪ C ).

In either case of (x ∈ A) and x ∈ B) and (x ∈ A andx ∈ C ) we have x ∈ A ∩ (B ∪ C ). Ergo (A ∩ B) ∪ (A ∩ C ) ⊆ A ∩ (B ∪ C ).As observed above, these imply that A∩ (B ∪C ) = (A∩B)∪ (A∩C ).

() Introduction to Set Theory February 4, 2018 3 / 3

Page 6: Introduction to Set Theory · Introduction to Set Theory February 4, 2018 Chapter 1. Sets 1.4. Elementary Operations on Sets—Proofs of Theorems Introduction to Set Theory February

Theorem 1.4.A

Theorem 1.4.A

Theorem 1.4.A. Let A, B, and C be sets. Then (Distributivity):A ∩ (B ∪ C ) = (A ∩ B) ∪ (A ∩ C ).

Proof. Let x ∈ A ∩ (B ∪ C ). Then x ∈ A and x ∈ B ∪ C . Therefore,x ∈ A and either x ∈ B or x ∈ C . So either (x ∈ A and x ∈ B) or (x ∈ Aor x ∈ C ); that is, either x ∈ A ∩ B or x ∈ A ∩ C . Hencex ∈ (A ∩ B) ∪ (A ∩ C ). Ergo, every element of A ∩ (B ∪ C ) is in(A ∩ B) ∪ (A ∩ C ) and so A ∩ (B ∪ C ) ⊆ (A ∩ B) ∪ (A ∩ C ).

Let x ∈ (A∩B)∪ (A∩ C ). Then either x ∈ A∩B or x ∈ A∩ C . So either(x ∈ A or x ∈ B) or (x ∈ A and x ∈ C ). If x ∈ A and x ∈ B then x ∈ Aand x ∈ B ∪ C (since B ⊆ B ∪ C ), and so x ∈ A ∩ (B ∪ C ). If x ∈ A andx ∈ C then x ∈ A and x ∈ B ∪ C (since C ⊆ B ∪ C ), and sox ∈ A ∩ (B ∪ C ). In either case of (x ∈ A) and x ∈ B) and (x ∈ A andx ∈ C ) we have x ∈ A ∩ (B ∪ C ). Ergo (A ∩ B) ∪ (A ∩ C ) ⊆ A ∩ (B ∪ C ).As observed above, these imply that A∩ (B ∪C ) = (A∩B)∪ (A∩C ).

() Introduction to Set Theory February 4, 2018 3 / 3

Page 7: Introduction to Set Theory · Introduction to Set Theory February 4, 2018 Chapter 1. Sets 1.4. Elementary Operations on Sets—Proofs of Theorems Introduction to Set Theory February

Theorem 1.4.A

Theorem 1.4.A

Theorem 1.4.A. Let A, B, and C be sets. Then (Distributivity):A ∩ (B ∪ C ) = (A ∩ B) ∪ (A ∩ C ).

Proof. Let x ∈ A ∩ (B ∪ C ). Then x ∈ A and x ∈ B ∪ C . Therefore,x ∈ A and either x ∈ B or x ∈ C . So either (x ∈ A and x ∈ B) or (x ∈ Aor x ∈ C ); that is, either x ∈ A ∩ B or x ∈ A ∩ C . Hencex ∈ (A ∩ B) ∪ (A ∩ C ). Ergo, every element of A ∩ (B ∪ C ) is in(A ∩ B) ∪ (A ∩ C ) and so A ∩ (B ∪ C ) ⊆ (A ∩ B) ∪ (A ∩ C ).

Let x ∈ (A∩B)∪ (A∩ C ). Then either x ∈ A∩B or x ∈ A∩ C . So either(x ∈ A or x ∈ B) or (x ∈ A and x ∈ C ). If x ∈ A and x ∈ B then x ∈ Aand x ∈ B ∪ C (since B ⊆ B ∪ C ), and so x ∈ A ∩ (B ∪ C ). If x ∈ A andx ∈ C then x ∈ A and x ∈ B ∪ C (since C ⊆ B ∪ C ), and sox ∈ A ∩ (B ∪ C ). In either case of (x ∈ A) and x ∈ B) and (x ∈ A andx ∈ C ) we have x ∈ A ∩ (B ∪ C ). Ergo (A ∩ B) ∪ (A ∩ C ) ⊆ A ∩ (B ∪ C ).As observed above, these imply that A∩ (B ∪C ) = (A∩B)∪ (A∩C ).

() Introduction to Set Theory February 4, 2018 3 / 3


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