Prove that (A') '=A by arbitrary elementary method ​

1 Elementary Set Theory

Notation:

{} enclose a set.

{1, 2, 3} = {3, 2, 2, 1, 3} because a set is not defined by order or multiplicity.

{0, 2, 4, . . .} = {x|x is an even natural number} because two ways of writing

a set are equivalent.

∅ is the empty set.

x ∈ A denotes x is an element of A.

N = {0, 1, 2, . . .} are the natural numbers.

Z = {. . . , −2, −1, 0, 1, 2, . . .} are the integers.

Q = {

m

n

|m, n ∈ Z and n 6= 0} are the rational numbers.

R are the real numbers.

Axiom 1.1. Axiom of Extensionality Let A, B be sets. If (∀x)x ∈ A iff x ∈ B

then A = B.

Definition 1.1 (Subset). Let A, B be sets. Then A is a subset of B, written

A ⊆ B iff (∀x) if x ∈ A then x ∈ B.

Theorem 1.1. If A ⊆ B and B ⊆ A then A = B.

Proof. Let x be arbitrary.

Because A ⊆ B if x ∈ A then x ∈ B

Because B ⊆ A if x ∈ B then x ∈ A

Hence, x ∈ A iff x ∈ B, thus A = B.

Definition 1.2 (Union). Let A, B be sets. The Union A ∪ B of A and B is

defined by x ∈ A ∪ B if x ∈ A or x ∈ B.

Theorem 1.2. A ∪ (B ∪ C) = (A ∪ B) ∪ C

Proof. Let x be arbitrary.

x ∈ A ∪ (B ∪ C) iff x ∈ A or x ∈ B ∪ C

iff x ∈ A or (x ∈ B or x ∈ C)

iff x ∈ A or x ∈ B or x ∈ C

iff (x ∈ A or x ∈ B) or x ∈ C

iff x ∈ A ∪ B or x ∈ C

iff x ∈ (A ∪ B) ∪ C

Definition 1.3 (Intersection). Let A, B be sets. The intersection A ∩ B of A

and B is defined by a ∈ A ∩ B iff x ∈ A and x ∈ B

Theorem 1.3. A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

Proof. Let x be arbitrary. Then x ∈ A ∩ (B ∪ C) iff x ∈ A and x ∈ B ∩ C

iff x ∈ A and (x ∈ B or x ∈ C)

iff (x ∈ A and x ∈ B) or (x ∈ A and x ∈ C)

iff x ∈ A ∩ B or x ∈ A ∩ C

iff x ∈ (A ∩ B) ∪ (A ∩ C)

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