學數學 論

學數學

學 學 數學 學 數學 . 學數學

論 . 學 , .

(Logic), (Set) 數 (Function) . , 學

論. 論 學 數學 .

, ,

. , .

, . , 論

, . , .

v

Set

論 數學 論 .

論. 論, .

3.1. Basic Definition

, .

數學 , .

, . ( set)

, , ( element).

set, A, B, S , set . 數學

. : N 數 , Z 數

,Q 數 , 數 R . x

S , x∈ S , x belongs to S ( x S). x

S , x̸∈ S .

數學 , set element element.

S, x, x∈ S . ,

, . S ={1,2,3}

3 , 1, 2 3. S , 1∈ S,

4̸∈ S. , set-builder notation

. {x : P(x)} , : x

x , : P(x) x P(x).

{x : P(x)} P(x) x .

前, , .

Definition 3.1.1. A, B . B element A element, B A

subset ( ), B is contained in ( ) A, B⊆ A. B⊆ A A⊆ B, A, B equal, A = B. B⊆ A B̸= A, B A proper subset, B⊂ A.

29

subset proper set “⊆” “⊂”, , .

B⊆ A, B x, A , 數學

“x∈ B ⇒ x ∈ A”. A = B “x∈ B ⇔ x ∈ A”.

, , ,

. :

Example 3.1.2. A ={1,2,2}, B = {1,2,3}, C = {3,3,1,2}, D = {n ∈ N : 1 ≤ n ≤ 3}, E ={2,4}.

A 1, 2 , 1∈ B 2∈ B, A⊆ B. 3∈ B 3̸∈ A, B

A, A⊂ B. B = C.

x∈ B, x∈ N 1≤ x ≤ 3, x∈ D. B⊆ D. , x∈ D

x∈ N 1≤ x ≤ 3, x = 1, 2, 3 B , D⊆ B, B = D.

1∈ B 1̸∈ E, B E subset. , 4∈ E 4̸∈ B, E

B subset.

A, B sets, B A subset , B* A . B⊆ A

A* B, B⊂ A.

Question 3.1. P(x), Q(x) statement form. P ={x : P(x)} Q ={x : Q(x)}

:

(1) P⊆ Q P(x)⇒ Q(x).

(2) P = Q P(x)⇔ Q(x).

, . ,

subset , universal set (

). 論 數, R universal set.

x∈ R . universal set

, a, b 數 , universal set Q 論 ax + b = 0 . 論

ax²+ b = 0, universal set R 數 C . ,

, universal set .

universal set , 論 set universal set subset.

empty set ( ). ,

/0 . /0 ?

, “ x∈ /0 ” .

operations, /0

. universal set empty set, .

Proposition 3.1.3. X universal set A set. A⊆ X /0⊆ A.

Proof. X universal set, A X subset, A⊆ X. , /0⊆ A

x∈ /0 x∈ A. x∈ /0 , P P⇒ Q

“x∈ /0 ⇒ x ∈ A” /0⊆ A. 

Question 3.2. Question X universal set. universal set ?

empty set ?

數學 , ,

論 subset .

Proposition 3.1.4. A, B,C sets, .

(1) A⊆ A.

(2) A⊆ B B⊆ C, A⊆ C.

Proof. (1) x∈ A, x∈ A, A⊆ A.

(2) x∈ A, A⊆ B x∈ B. B⊆ C x∈ C. 言 , x∈ A x∈ C,

A⊆ C. 

Question 3.3. Proposition 3.1.4 A = B B = C, A = C.

Question 3.4. A, B,C sets, ?

(1) A⊂ B B⊆ C, A⊆ C.

(2) A⊆ B B⊆ C, A⊂ C.

(3) A⊂ B B⊂ C, A⊆ C.

(4) A⊂ B B⊆ C, A⊂ C.

, A = B A⊆ B B⊆ A .

, 數 , 學

. .

Example 3.1.5. A ={(x,y) ∈ R²: x²− x = y = 2} B ={(2,2),(−1,2)}. A = B.

Proof. (x, y)∈ A, x²− x = 2 y = 2, x = 2 x =−1 y = 2.

(x, y)∈ A, (x, y) = (2, 2) (x, y) = (−1,2). x∈ B, A⊆ B. (x, y)∈ B, (x, y) = (2, 2) (x, y) = (−1,2) x²−x = y = 2, (x, y)∈ A. B⊆ A,

A = B 

Question 3.5. A ={x ∈ R :√

x = x− 2}, B = {1}, C = {4}, D = {1,4}. A, B,C, D .

Venn diagrams . ,

universal set, ( )

set. X set A.

..

A .

X

Venn diagrams A, B .

A B.... X

. A

. B

.

X

. A

.

B .

X

A, B , A, B

, A⊆ B.

Venn diagrams ,

. Proposition 3.1.4 (2) A⊆ B B⊆ C, A⊆ C

.

A...

B .

C .

X

Venn diagrams , .

Question 3.6. A, B,C sets. A⊆ B. B C ,

Venn diagrams. A C ? B C ,

Venn diagrams, A C ? , A,C

Venn diagrams, B,C .

“∈” ( ) “⊆” ( ) . “∈”

, “⊆” . A⊆ B B⊆ C A⊆ C

, A∈ B B∈ C A∈ C.

A ={1}, B ={ {1}}

, C ={{

{1}}}

.

A∈ B B∈ C, A̸∈ C.

3.2. Set Operations

set operation, .

operations, intersection, union set difference, set

operations .

3.2.1. Intersection and Union. intersection union.

Definition 3.2.1. A, B sets. A∩B = {x : x ∈ A and x ∈ B} the intersec- tion of A and B (A B ). A∪ B = {x : x ∈ A or x ∈ B} the union of A and

B (A B ).

A∩ B A, B , A∪ B A, B

. .

Example 3.2.2. A ={1,2,3}, B = {2,4,6}. 2 A B,

A∩ B = {2}. 1, 3 B A, A B 1, 3

A∪ B. 4, 6 A∪ B. 2 A B A

B , 2 A∪ B. 數 A B ,

A∪ B = {1,2,3,4,6}.

. A∩ B = {2} ⊆ A =

{1,2,3} B ={2,4,6} ⊆ A ∪ B = {1,2,3,4,6}. , x∈ A ∩ B, x∈ A x∈ B,

x A x B,

(A∩ B) ⊆ A and (A ∩ B) ⊆ B. (3.1)

A∩ B , A, B disjoint. ,

A, B disjoint . x∈ A, x A B, x∈ A∪B

.

A⊆ (A ∪ B) and B ⊆ (A ∪ B) (3.2)

Question 3.7. (A∩ A) = A (A∪ A) = A.

, .

Proposition 3.2.3. A, B,C, D sets A⊆ B C⊆ D.

(A∩C) ⊆ (B ∩ D) and (A ∪C) ⊆ (B ∪ D).

Proof. A⊆ B, x∈ A x∈ B. C⊆ D, x∈ C x∈ D. x∈ A ∩C, x∈ A x∈ C. x∈ B x∈ D. (A∩C) ⊆ (B ∩ D). , x∈ A ∪C,

x∈ A x∈ C. x∈ A x∈ B, x∈ C x∈ D. x∈ A ∪C

x∈ B ∪ D. (A∪C) ⊆ (B ∪ D). 

, A⊆ B A⊆ D , C = A Proposition 3.2.3 (A∩ A) ⊆ (B ∩ D). (A∩ A) = A, A⊆ (B ∩ D). , A⊆ B C⊆ B ,

(A∪C) ⊆ (B ∪ B). (B∪ B) = B, (A∪C) ⊆ B. .

Proposition 3.2.3 導 , corollary ( ) .

Corollary 3.2.4. A, B,C, D, E sets.

(1) A⊆ B A⊆ C, A⊆ (B ∩C).

(2) D⊆ A E⊆ A, (D∪ E) ⊆ A.

Question 3.8. Corollary 3.2.4, 導 Proposition 3.2.3.

. .

Proposition 3.2.5. A, B sets. equivalent.

(1) A⊆ B.

(2) (A∩ B) = A.

(3) (A∪ B) = B.

Proof. (1)⇔ (2) (1)⇔ (3).

(1)⇔ (2): A⊆ B, (A∩ B) = A. (3.1) (A∩ B) ⊆ A,

A⊆ (A ∩ B). A⊆ A A⊆ B, Corollary 3.2.4 A⊆ (A ∩ B).

(1)⇒ (2). , (3.1) (A∩ B) ⊆ B. A = (A∩ B)

A⊆ B, (2)⇒ (1).

(1)⇔ (3): A⊆ B, (A∪ B) = B. (3.2) B⊆ (A ∪ B),

(A∪ B) ⊆ B. A⊆ B B⊆ B, Corollary 3.2.4, (A∪ B) ⊆ B.

(1)⇒ (3). , (3.2) A⊆ (A ∪ B). (A∪ B) = B

A⊆ B, (3)⇒ (1). 

Definition 3.2.1 “ ” “and” , “ ” “or” .

: (1) A∩ B = B ∩ A.

(2) A∪ B = B ∪ A.

(3) (A∩ B) ∩C = A ∩ (B ∩C).

(4) (A∪ B) ∪C = A ∪ (B ∪C).

(3) , ,

A∩ B ∩C. (4), ,

A∪ B ∪C.

∧,∨ ,

((P∧ Q) ∨ R) ∼ ((P ∨ R) ∧ (Q ∨ R)), ((P ∨ Q) ∧ R) ∼ ((P ∧ R) ∨ (Q ∧ R)), .

Proposition 3.2.6. A, B,C sets,

((A∩ B) ∪C) = (A ∪C) ∩ (B ∪C), ((A ∪ B) ∩C) = (A ∩C) ∪ (B ∩C).

Proof. (A∩ B) ⊆ A C⊆ C Proposition 3.2.3 ((A∩ B) ∪C) ⊆ (A ∪C).

((A∩B)∪C) ⊆ (B∪C). Corollary 3.2.4 ((A∩B)∪C) ⊆ (A∪C)∩(B∪C).

x∈ (A∪C)∩(B∪C) x∈ A∪C x∈ B∪C. proof in cases, x∈ C

x̸∈ C . x∈ C, x∈ (A∩B)∪C. x̸∈ C, x∈ A∪C x∈ B∪C,

x∈ A x∈ B, x∈ A ∩ B. x∈ (A ∩ B) ∪C. ,

x∈ (A ∪C) ∩ (B ∪C) x∈ (A ∩ B) ∪C, ((A∪C) ∩ (B ∪C)) ⊆ (A ∩ B) ∪C.

((A∩ B) ∪C) = (A ∪C) ∩ (B ∪C).

((A∪ B) ∩ C) = (A ∩ C) ∪ (B ∩ C) , A⊆ (A ∪ B) C⊆ C Proposition 3.2.3 (A∩ C) ⊆ ((A ∪ B) ∩ C), (B∩ C) ⊆ ((A ∪ B) ∩ C).

Corollary 3.2.4 (A∩C)∪(B∩C) ⊆ ((A∪B)∩C). , x∈ (A∪B)∩C, x∈ A∪B

x∈C. x∈ A∪B, x∈ A x∈ B. x∈ A , x∈ C, x∈ A∩C.

x∈ (A∩C)∪(B∩C). , x∈ B , x∈ B∩C. x∈ (A∩C)∪(B∩C), ((A∪ B) ∩C) ⊆ (A ∩C) ∪ (B ∩C). ((A∪ B) ∩C) = (A ∩C) ∪ (B ∩C)  Question 3.9. Proposition 3.2.5 (1)⇒ (2) Proposition 3.2.6 Proposition 3.2.5 (2)⇒ (3).

3.2.2. Set Difference. set difference.

Definition 3.2.7. A, B sets, A\ B = {x : x ∈ A and x ̸∈ B}, the set difference of B in A (B A ). X universal set, A^c= X\ A = {x : x ̸∈ A}

the complement of A (A ).

A^c {x : x ∈ X and x ̸∈ A}, X universal set,

X , x∈ X x̸∈ A.

, . Q , Q^c= /0, R ,

Q^c 數 .

, A\B = A∩B^c, A\B B\A

. (A\ B)∩ (B \ A) = /0. A∩ A^c= /0 B∩ B^c= /0,

(A\ B) ∩ (B \ A) = (A ∩ B^c)∩ (B ∩ A^c) = (A∩ A^c)∩ (B ∩ B^c) = /0.

Example 3.2.8. X ={1,2,3,4,5,6},A = {1,2,3},B = {2,4,6}. 1, 3∈ A 1̸∈ B 3̸∈ B, 1, 3∈ A \ B. 2∈ A 2∈ B, 2̸∈ A \ B. A\ B = {1,3}.

B\ A = {4,6}. (A\ B) ∩ (B \ A) = {1,3} ∩ {4,6} = /0. 1, 3, 5∈ X 1, 3, 5 B , 1, 3, 5∈ B^c. 2, 4, 6∈ B 2, 4, 6 B^c, B^c={1,3,5}. A∩ B^c

A∩ B^c={1,2,3} ∩ {1,3,5} = {1,3} = A \ B.

set difference . A, B,C sets A⊆ B,

x̸∈ B, x̸∈ A. x∈ A, A⊆ B x∈ B, x̸∈ B .

A⊆ B x∈ C \ B, x∈ C x̸∈ B, x∈ C x̸∈ A, x∈ C \ A.

(C\ B) ⊆ (C \ A). , A⊆ C, .

Proposition 3.2.9. A, B,C sets. A⊆ B (C\ B) ⊆ (C \ A).

A⊆ C, A⊆ B (C\ B) ⊆ (C \ A).

Proof. 前 A⊆ B C\ B ⊆ C \ A ( A⊆ C ).

A⊆ C (C\ B) ⊆ (C \ A), A⊆ B, x∈ A x∈ B. C\ B,C \ A

“not” , contradiction method. x∈ A, x̸∈ B, . A⊆ C

x∈ A x∈ C, x̸∈ B, x∈ C \ B. 前 (C\ B) ⊆ (C \ A),

x∈ C \ A, x∈ C x̸∈ A. x∈ A . x∈ A x̸∈ B,

A⊆ B. 

Question 3.10. A⊆ C A⊆ B (C\ B) ⊆ (C \ A) ?

A⊆ C , ?

C = X , A⊆ C = X, Proposition 3.2.9, A⊆ B

(X\ B) ⊆ (X \ A). .

Corollary 3.2.10. A, B sets. A⊆ B B^c⊆ A^c.

Definition 3.2.7 set difference “not” ,

導 set difference , .

Proposition 3.2.11. A, B,C sets, .

(1) (C\ (C \ A)) = (C ∩ A). , (A^c)^c= A.

(2) C\ (A ∩ B) = (C \ A) ∪ (C \ B). , (A∩ B)^c= (A^c∪ B^c).

(3) C\ (A ∪ B) = (C \ A) ∩ (C \ B). , (A∪ B)^c= (A^c∩ B^c).

Proof. 前 equivalence 導,

.

(1): x∈ C \ (C \ A) x∈ C x̸∈ C \ A. x̸∈ A, x∈ C \ A

( x∈ C), x̸∈ C \ A , x∈ A. x∈ (C \ (C \ A)),

x∈ C X∈ A ( x∈ C ∩ A). (C\ (C \ A)) ⊆ (C ∩ A). , x∈ C ∩ A, x∈ C, x̸∈ (C \ A), x∈ C \ (C \ A). x∈ (C \ A) x∈ C x̸∈ A,

x∈ C ∩ A , x̸∈ (C \ A). (C∩ A) ⊆ (C \ (C \ A)), (C\ (C \ A)) = (C ∩ A).

C = X , X\ A = A^c, X\ (X \ A) = X \ A^c= (A^c)^c. X∩ A = A, (A^c)^c= A.

(2): (A∩B) ⊆ A, Proposition 3.2.9 (C\A) ⊆ (C\(A∩B)). (A∩B) ⊆ B (C\ B) ⊆ (C \ (A ∩ B)). Corollary 3.2.4 (2) (C\ A) ∪ (C \ B) ⊆ C \ (A ∩ B).

, x∈ C \ (A ∩ B), x ∈ C x̸∈ A ∩ B. x̸∈ A, x∈ C \ A, x∈ (C \ A) ∪ (C \ B). x∈ A, x̸∈ B, x∈ B x∈ A x∈ B, x∈ A ∩ B

. x∈ C \ B, x∈ (C \ A) ∪ (C \ B). C\ (A ∩ B) ⊆ (C \ A) ∪ (C \ B).

C = X , X\ A = A^c, X\ B = B^c X\ (A ∩ B) = (A ∩ B)^c (A∩ B)^c= (A^c∪ B^c).

(3): (2) (A^c∩B^c)^c= ((A^c)^c∪(B^c)^c), (1), (A^c∩B^c)^c= (A∪B).

complement (1) (A^c∩ B^c) = (A∪ B)^c.

C\ (A ∪ B) = C ∩ (A ∪ B)^c= C∩ (A^c∩ B^c) and (C\ A) ∩ (C \ B) = (C ∩ A^c)∩ (C ∩ B^c), C∩ (A^c∩ B^c) = (C∩ A^c)∩ (C ∩ B^c),

C\ (A ∪ B) = (C \ A) ∩ (C \ B).

 Question 3.11. Proposition 3.2.11 (2) C\ A = C \ (C ∩ A).

Proposition 3.2.9 (C∩ A) ⊆ (C ∩ B) (C\ B) ⊆ (C \ A).

operations , ,

connectives , .

operations 導. ,

operations 導 , . ,

, 導 , 導.

, 學

導 . .

Example 3.2.12. A, B,C B^c⊆ A, ((C\ A) ∪ B) = B.

: . B⊆ ((C \ A) ∪ B),

((C\ A) ∪ B) ⊆ B. x∈ ((C \ A) ∪ B), x∈ B. x∈ ((C \ A) ∪ B)

x∈ C \ A x∈ B. x∈ B, , 論 x∈ C \ A , x∈ C

x̸∈ A. x∈ B, , x̸∈ B .

x̸∈ B, x∈ B^c, B^c⊆ A x∈ A. x̸∈ A , x∈ B.

((C\ A) ∪ B) ⊆ B.

: 前 . B,

Proposition 3.2.5. (C\ A) ⊆ B, ((C\ A) ∪ B) = B.

(C\ A) ⊆ B ? B^c⊆ A . C\ A = C ∩ A^c, Corollary 3.2.10 B^c⊆ A A^c⊆ (B^c)^c, Proposition 3.2.11 (1) A^c⊆ B.

(C\ A) = (C ∩ A^c)⊆ (C ∩ B) ⊆ B.

3.3. Indexed Family

前 , , operation.

, 論 . ,

, . 前

, . ,

, .

論 , 數 , 5

A, B,C, D, E , A∩ B ∩C ∩ D ∩ E .

, .

, . . 數 ,

前 . , ,

100 , A1, A2, . . . , A100 ( , Ai ).

“summation” ,

100∩

i=1

Ai,

100∪

i=1

Ai . Ai= [i− 1,i] ( i− 1 i 數),

100∩

i=1

A_i= /0,

100∪

i=1

A_i= [0, 100]. ? 前 , 數

i∈ N, Ai , A_i . 學

數 ,

∩∞ i=1

Ai,

∪∞ i=1

Ai. , i

數 , , A5, A6, A7, A8

∩8

i=5

A_i,

∪8

i=5

A_i . i m n A_i ,

∩n

i=m

Ai,

∪n

i=m

Ai . i m Ai ,

∩∞ i=m

A_i,

∪∞ i=m

A_i . , 學

∩∞ i=m

A_i,

∪∞ i=m

A_i

∩n

i=m

A_i,

∪n

i=m

A_i n

∞ . , “ ”.

. , 數 數

數 . 數, 數. 數 , (−r,r)

r > 0, , ? ,

index set . index set “ ” . 前

Ai= [i− 1,i] , i 數 N, N index set.

[−r,r] , 數 R⁺ index set,

A_r= [−r,r]. A_r, r∈ R⁺ , indexed

family . , index set .

index set , indexed family. I index set,

A_i , {Ai, i∈ I} indexed family.

, indexed family . ,

, indexed family. A, B, index

set I ={1,2} indexed family, A₁= A, A₂= B. index family

. A, B ,

; A, B . .

Definition 3.3.1. I index set, {Ai, i∈ I}, I index set indexed family.

indexed family intersection

∩

i∈I

A_i={x : x ∈ Ai,∀i ∈ I}.

indexed family union

∪

i∈I

Ai={x : x ∈ Ai,∃i ∈ I}.

, .

Example 3.3.2. index set I 1 數. i∈ I, Ai={m/i : m ∈ Z}.

∩

i∈I

Ai=Z, ^∪

i∈I

Ai=Q.

, n∈ Z, n = ni/i. ni∈ Z, n∈ Ai,∀i ∈ I.

Z ⊆^∩

i∈I

A_i. , x∈^∩

i∈I

A_i, i∈ I x∈ Ai. x∈ A2 x∈ A3,

x = m/2 x = m^′/3, m, m^′∈ Z. 3m = 2m^′, 3m 數.

m 數 2n, n∈ Z. x = m/2 = n∈ Z. ^∩

i∈I

Ai⊆ Z, ^∩

i∈I

Ai=Z.

x∈ Q, 數 , x m/n, m∈ Z, n ∈ N. n = 1, x = m∈ Z.

x x = 2m/2, x∈ A2. n≥ 2, n∈ I, x∈ An. Q ⊆^∪i∈IAi. , x∈^∪i∈IAi, n∈ I, x∈ An. x = m/n,

m∈ Z. x∈ Q, ^∪

i∈I

Ai⊆ Q, ^∪

i∈I

Ai=Q.

Question 3.12. Ai Example 3.3.2 . m∈ Z , p, q 數

p mq, p m , p, q 數, A_p∩ Aq=Z.

m∈ N, ^∩∞

i=m

A_i=Z.

Question 3.13. Ai Example 3.3.2 , m∈ N, ^∪∞

i=m

Ai=Q.

數 m < n

∩n

i=m

A_i=Q?

, indexed

family . Proposition 3.2.3 .

Proposition 3.3.3. {Ai, i∈ I}, {Bi, i∈ I} I index set indexed family.

i∈ I A_i⊆ Bi,

∩

i∈I

Ai⊆^∩

i∈I

Bi and ^∪

i∈I

Ai⊆^∪

i∈I

Bi.

Proof. x∈^∩

i∈I

A_i, i∈ I, x∈ Ai, A_i⊆ Bi, x∈ Bi.

i∈ I , x∈^∩

i∈I

Bi. ^∩

i∈I

Ai⊆^∩

i∈I

Bi. x∈^∪

i∈I

Ai, i∈ I x∈ Ai, Ai⊆ Bi, x∈ Bi. x∈^∪

i∈I

Bi,

∪

i∈I

Ai⊆^∪

i∈I

Bi. 

, i∈ I, A_i= A, ^∩

i∈I

A_i= A ^∪

i∈I

A_i= A.

Proposition 3.3.3 Corollary 3.2.4 .

Corollary 3.3.4. A, B set {Ai, i∈ I}, {Bi, i∈ I} I index set indexed family.

(1) i∈ I A⊆ Ai, A⊆^∩

i∈I

A_i.

(2) i∈ I Bi⊆ B, ^∪

i∈I

Bi⊆ B.

A_i 數 N index set indexed family A₁⊇ A2⊇ ··· ⊇ Ai⊇ ··· ( Ai+1⊆ Ai,∀i ∈ N). n∈ N, An⊆ Ai,∀i ≤ n, Corollary 3.3.4, An⊆

∩n

i=1

Ai. ,

∩n

i=1

Ai⊆ An,

∩n

i=1

Ai= A_n. , i∈ N,

A_i ,

∩n

i=1

A_i . ,

. .

Example 3.3.5. 數 N index set indexed family

{Ai, i∈ I}, i∈ N Ai+1⊆ Ai Ai ,

∩∞ i=1

Ai= /0.

i∈ N A_i (0, 1/i). A_i A_i̸= /0 A_i+1⊆ Ai.

∩∞ i=1

A_i= /0. x∈^∩∞

i=1

A_i, x > 0. n∈ N n > 1/x.

x > 1/n, x̸∈ (0,1/n) = An. x∈^∩∞

i=1

Ai ,

∩∞ i=1

Ai= /0.

, ,

, .

, Proposition 3.2.6 .

Proposition 3.3.6. B set, {Ai, i∈ I} I index set indexed family.

(^∩

i∈I

A_i)∪ B =^∩

i∈I

(A_i∪ B) and (^∪

i∈I

A_i)∩ B =^∪

i∈I

(A_i∩ B).

Proof. k∈ I (^∩

i∈I

A_i)⊆ (Ak∪ B) B⊆ (Ak∪ B), Corollary 3.2.4 (2) ((^∩

i∈I

Ai)∪ B) ⊆ (Ak∪ B). k∈ I , Corollary 3.3.4 (1) ((^∩

i∈I

A_i)∪ B) ⊆^∩

i∈I

(A_i∪ B).

, x∈^∩

i∈I

(A_i∪B), i∈ I, x∈ Ai x∈ B. x∈ B x̸∈ B

論. x∈ B, x∈ (^∩

i∈I

Ai)∪ B. x̸∈ B, x∈ Ai

i∈ I , x∈^∩

i∈I

Ai, x∈ (^∩

i∈I

Ai)∪ B. ^∩

i∈I

(A_i∪ B) ⊆ ((^∩

i∈I

Ai)∪ B), (^∩

i∈I

Ai)∪ B =^∩

i∈I

(Ai∪ B).

, k∈ I (A_k∩ B) ⊆ (^∪

i∈I

Ai) (A_k∩ B) ⊆ B, Corollary 3.2.4 (1) (A_k∩ B) ⊆ (^∪

i∈I

A_i)∩ B. k∈ I , Corollary 3.3.4 (2)

∪

i∈I

(A_i∩ B) ⊆ (^∪

i∈I

A_i)∩ B.

, x∈ (^∪

i∈I

Ai)∩ B, x∈^∪

i∈I

Ai x∈ B. i∈ I, x∈ Ai x∈ B.

i∈ I x∈ Ai∩ B. x∈^∪

i∈I

(A_i∩ B), ((^∪

i∈I

A_i)∩ B) ⊆^∪

i∈I

(A_i∩ B), (^∪

i∈I

A_i)∩ B =^∪

i∈I

(A_i∩ B).

 Question 3.14. Ai, Bi, i∈ I I index set indexed family.

(^∩

i∈I

Ai)∪ (^∩

i∈I

Bi) =^∩

i∈I

(Ai∪ Bi) and (^∪

i∈I

Ai)∩ (^∪

i∈I

Bi) =^∪

i∈I

(Ai∩ Bi)?

DeMorgan’s laws (Proposition 3.2.11 (2)(3)).

Proposition 3.3.7. C sets {Ai, i∈ I} I index set indexed family, .

(1) C\ (^∩

i∈I

Ai) =^∪

i∈I

(C\ Ai). , (^∩

i∈I

Ai)^c=^∪

i∈I

A^c_i. (2) C\ (^∪

i∈I

Ai) =^∩

i∈I

(C\ Ai). , (^∪

i∈I

Ai)^c=^∩

i∈I

A^c_i.

Proof. (1): k∈ I,^∩

i∈I

Ai⊆ Ak, Proposition 3.2.9 (C\Ak)⊆ (C \^∩

i∈I

Ai).

Corollary 3.3.4 (2) ^∪

i∈I

(C\ Ai)⊆ (C \^∩

i∈I

A_i). , x∈ C \^∩

i∈I

A_i, x∈ C x̸∈^∩

i∈I

A_i, x A_i. i∈ I x̸∈ Ai, x∈ C x∈ C \ Ai.

x∈^∪

i∈I

(C\ Ai), (C\^∩

i∈I

A_i)⊆^∪

i∈I

(C\ Ai).

C = X , X\ Ai= A^c_i X\ (^∩

i∈I

A_i) = (^∩

i∈I

A_i)^c

(^∩

i∈I

Ai)^c=^∪

i∈I

A^c_i.

(2): (1) (^∩

i∈I

A^c_i)^c =^∪

i∈I

(A^c_i)^c, Proposition 3.2.11 (1), (^∩

i∈I

A^c_i)^c=^∪

i∈I

Ai. complement Proposition 3.2.11 (1) (^∪

i∈I

Ai)^c=^∩

i∈I

A^c_i.

C\ (^∪

i∈I

A_i) = C∩ (^∪

i∈I

A_i)^c= C∩ (^∩

i∈I

A^c_i) and ^∩

i∈I

(C\ Ai) =^∩

i∈I

(C∩ A^ci), C∩ (^∩

i∈I

A^c_i) =^∩

i∈I

(C∩ A^c_i),

C\ (^∪

i∈I

A_i) =^∩

i∈I

(C\ Ai).

 Question 3.15. C sets {Ai, i∈ I} I index set indexed family.

(^∩

i∈I

Ai)\C ^∩

i∈I

(Ai\C) ^∪

i∈I

(Ai\C), ? (^∪

i∈I

Ai)\C ?

3.4. Power Set and Cartesian Product

前 ( , ) ,

(

), .

3.4.1. Power Set. power set.

Definition 3.4.1. A set. A power set A subsets ,

P(A) .

P(A) = {S : S ⊆ A}.

set A, /0⊆ A A⊆ A, /0∈ P(A) A∈ P(A).

, /0 , /0⊆ P(A). /0∈ P(A)

/0⊆ P(A) . a∈ A, {a} ⊆ A, {a} ∈ P(A). set

power set “ ” “ ” , . .

Example 3.4.2. /0 , P(/0) = {/0}.

A ={1,2,3}, 前 /0, A, {1}, {2}, {3} P(A) . {1,2}, {1,3}, {2,3}

A,

P(A) ={

/0,{1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3}} .

A , finite set. #(A) A

數. Example 3.4.2 #(A) = 3, #(P(A)) = 2³= 8.

finite set, 數, power set 數.

Proposition 3.4.3. A finite set #(A) = n. #(P(A)) = 2ⁿ.

Proof. #(P(A)) = 2ⁿ. 論

, 數學 . A 數 #(A) 數學 . #(A) = 0 ,

A , A = /0. Example 3.4.2, #(A) = 1 = 2⁰. #(A) = 1 ,

A , a, A ={a}. P(A) = {/0,{a}}, #(A) = 2 = a¹.

n = 0, 1 .

數 k . #(A) = k + 1 , A ={a1, . . . , ak, ak+1}.

A^′= A\ {ak+1}. #(A^′) = k, P(A^′) = 2^k, A^′ 2^k

. A^′⊂ A, A^′ subset A subset. P(A) 2^k .

A subset A^′ , a_k+1 subset. S subset,

A_k+1∈ S. S^′= S\ {ak+1}, S^′⊆ A^′. , S^′⊆ A^′, S = S^′∪ {ak+1},

S A subset, A^′ subset. 言 , A subset, A^′ subset,

A^′ subset {ak+1} . A subset 數 2^k+ 2^k= 2^k+1,

#(P(A)) = 2^k+1. 數學 , #(A) = n, #(P(A)) = 2ⁿ. 

論 power set set .

, power set . power set , A set,

S∈ P(A) S⊆ A. , power set

. , power set .

Proposition 3.4.4. A, B sets. A⊆ B P(A) ⊆ P(B).

Proof. (⇒): A⊆ B. S∈ P(A), S⊆ A. A⊆ B S⊆ B, S∈ P(B).

P(A) ⊆ P(B).

(⇒): P(A) ⊆ P(B). A∈ P(A), P(A) ⊆ P(B), A∈ P(B). power set

, A⊆ B. 

Question 3.16. A, B sets. A⊂ B P(A) ⊂ P(B) ?

Power set , .

Proposition 3.4.5. A, B sets. P(A ∩ B) = P(A) ∩ P(B).

Proof. (A∩ B) ⊆ A (A∩ B) ⊆ B Proposition 3.4.4 P(A ∩ B) ⊆ P(A) P(A ∩ B) ⊆ P(B). Corollary 3.2.4 P(A ∩ B) ⊆ P(A) ∩ P(B).

, S∈ P(A) ∩ P(B) S∈ P(A) S∈ P(B), S⊆ A S⊆ B.

Corollary 3.2.4 S⊆ (A ∩ B), S∈ P(A ∩ B). P(A) ∩ P(B) ⊆ P(A ∩ B),

P(A ∩ B) = P(A) ∩ P(B). 

Question 3.17. {Ai, i∈ I} I index set indexed family. P(^∩

i∈I

A_i)

∩

i∈I

P(Ai)?

Power set . P(A) ∪ P(B) ⊆ P(A ∪ B) (

A⊆ (A ∪ B) B⊆ (A ∪ B) Proposition 3.4.4 P(A) ⊆ P(A ∪ B)

P(B) ⊆ P(A ∪ B)), P(A ∪ B) ⊆ P(A) ∪ P(B) .

S∈ P(A∪B) S⊆ (A ∪B), S⊆ A S⊆ B. A ={1}, B = {2},

S ={1,2} ⊆ A∪B, S* A S* B. P(A) = {/0,{1}},P(B) = {/0,{2}}, P(A)∪P(B) = {/0,{1},{2}}. P(A∪B) = {/0,{1},{2},{1,2}}. P(A∪B) ̸=

P(A) ∪ P(B), P(A) ∪ P(B) ⊂ P(A ∪ B).

P(A) ∪ P(B) = P(A ∪ B), A∪ B

A B . A⊆ B B⊆ A,

P(A) ⊆ P(B) P(B) ⊆ P(A). (A∪ B) = B (A∪ B) = A,

P(A) ∪ P(B) = P(A ∪ B). , A* B B* A ( , A⊆ B B⊆ A ),

a∈ A \ B b∈ B \ A. S ={a,b}, S⊆ A ∪ B S* A S* B, S∈ P(A ∪ B) S̸∈ P(A) S̸∈ P(B), S̸∈ (P(A) ∪ P(B)).

P(A ∪ B) ̸= P(A) ∪ P(B). .

Proposition 3.4.6. A, B sets.

(P(A) ∪ P(B)) ⊆ P(A ∩ B).

(P(A) ∪ P(B)) = P(A ∩ B) A⊆ B B⊆ A.

, Power set . A, B sets ,

/0∈ P(A), /0 ∈ P(B). /0̸∈ P(A) \ P(B). /0∈ P(A \ B),

(P(A) \ P(B)) ̸= P(A \ B). S̸= /0 , S∈ P(A \ B), S⊆ (A \ B).

(A\B) ⊆ A, S⊆ A ( S ∈ P(A)). S* B ( S ̸∈ P(B)). S

, x∈ S, S⊆ B, x∈ B. S⊆ (A \ B), x∈ A \ B,

x̸∈ B, . S∈ P(A) S̸∈ P(B) S∈ P(A) \ P(B).

P(A \ B) , P(A) \ P(B) ,

(P(A \ B) \ {/0}) ⊆ (P(A) \ P(B)).

. S∈ P(A) \ P(B) S⊆ A S* B.

S⊆ (A \ B). A\ B ̸= /0 A∩ B ̸= /0, a∈ A \ B b∈ A ∩ B S ={a,b}. {a,b} ⊆ A {a,b} * B ( S∈ P(A) \ P(B)), {a,b} * (A \ B) ( S̸∈ P(A\B)\{/0}). (P(A)\P(B)) * (P(A\B)\{/0}). , A\B = /0 A∩B = /0,

(P(A) \ P(B)) ⊆ (P(A \ B) \ {/0}), 前 Lemma.

Lemma 3.4.7. A, B sets.

(1) A\ B = /0 (P(A) \ P(B)) = /0.

(2) A∩ B = /0 (P(A) \ P(B)) = (P(A) \ {/0}).

Proof. (1): A\ B = /0, A⊆ B. A* B, a∈ A a̸∈ B, a∈ A \ B . Proposition 3.4.4 ,P(A) ⊆ P(B). (P(A) \ P(B)) = /0.

(2): /0 ∈ P(B), {/0} ⊆ P(B). Proposition 3.2.9 (P(A) \ P(B)) ⊆ (P(A) \ {/0}). , S∈ P(A) \ {/0} S∈ P(A) S̸∈ {/0}, S⊆ A S̸= /0.

A∩ B = /0 , S̸∈ P(B). S∈ P(B) S⊆ B, S⊆ A Corollary

3.2.4(1) S⊆ (A ∩ B) = /0, S̸= /0 . A∩ B = /0 , S∈

P(A) \ {/0} S∈ P(A) \ P(B), (P(A) \ {/0}) ⊆ (P(A) \ P(B)). A∩ B = /0

(P(A) \ P(B)) = (P(A) \ {/0}). 

Question 3.18. A, B sets. A\ B = /0 (P(A) \ P(B)) = /0?

A∩ B = /0 (P(A) \ P(B)) = (P(A) \ {/0})?

Lemma 3.4.7, .

Proposition 3.4.8. A, B sets.

(P(A \ B) \ {/0}) ⊆ (P(A) \ P(B)).

(P(A \ B) \ {/0}) = (P(A) \ P(B)) A\ B = /0 A∩ B = /0.

Proof. 前 (P(A\B)\{/0}) ⊆ (P(A)\P(B)). A\B ̸= /0 A∩B ̸= /0 (P(A)\P(B)) * (P(A\B)\{/0}). (P(A)\P(B)) = (P(A\B)\{/0}) A\B = /0

A∩ B = /0. , A\ B = /0 A∩ B = /0

(P(A \ B) \ {/0}) = (P(A) \ P(B)).

A\ B = /0, Lemma 3.4.7(1) (P(A) \ P(B)) = /0. , P(A \ B) = P(/0) = {/0},

(P(A \ B) \ {/0}) = {/0} \ {/0} = /0 = (P(A) \ P(B)).

A∩B = /0, Lemma 3.4.7(2) (P(A)\P(B)) = (P(A)\{/0}). , A\ B = A \ (A ∩ B) = A \ /0 = A,

(P(A \ B) \ {/0}) = (P(A) \ {/0}) = (P(A) \ P(B)).



3.4.2. Cartesian Product. ,

{1,2} {2,1} . S1={{1},{1,2}} S2{{2},{1,2}},

{1} ∈ S1 {1} ̸∈ S2, S1̸= S2. , 1, 2

. .

Definition 3.4.9. A, B sets. a∈ A,b ∈ B, ordered pair (a, b) ={{a},{a,b}}.

A× B = {(a,b) : a ∈ A,b ∈ B}, the Cartesian product of A and B.

ordered pair, 數 , ,

. 前 , (1, 2) ={{1},{1,2}}, (2, 1) ={{2},{1,2}}, (1, 2)̸= (2,1).

a, a^′∈ A, b,b^′∈ B. a = a^′, b = b^′, , (a, b) ={{a},{a,b}} = {{a^′},{a^′, b^′}} = (a^′, b^′).

(a, b) = (a^′, b^′) {{a},{a,b}} = {{a^′},{a^′, b^′}}. a̸= b, {a,b}

, (a, b) = (a^′, b^′), {a^′, b^′} ( {{a^′},{a^′, b^′}}

, {{a},{a,b}}). , ,

{a} = {a^′} {a,b} = {a^′, b^′}. a = a^′ b = b^′. {a,b}

, a = b. {a,b} = {a},

(a, b) = (a, a) ={{a},{a,b}} = {{a},{a}} = {{a}}.

, (a, b) = (a^′, b^′) b^′= a^′= a, a = a^′ b = b^′. .

Proposition 3.4.10. A, B sets, a, a^′∈ A b, b^′∈ B (a, b) = (a^′, b^′) a = a^′ b = b^′.

, a∈ A, b ∈ B, {{a},{a,b}} . {a,b}

, A∪B . {a}, {a,b} A∪B subset, {a} {a,b}

P(A∪B) . {{a},{a,b}} P(A∪B) , {{a},{a,b}} ∈ P(P(A∪B)).

(a, b) P(P(A∪B)) . (a, b) {{a},{a,b}}

, . Proposition 3.4.10 , (a, b)

. (a, b) A×B , (a, b) = (a^′, b^′)

A× B .

Example 3.4.11. (1) A ={a,b}, B = {1,2,3}. A× B A× B = {(a,1),(a,2),(a,3),(b,1),(b,2),(b,3)}.

A× {/0} = {(a, /0),(b, /0)}.

(2) S ={(x,y) ∈ R × R : x²+ y²≤ 1}. S R × R subset, A⊆ R, B⊆ R S = A× B. , S = A× B, (1, 0)∈ S, 1∈ A. (0, 1)∈ S,

1∈ B. (1, 1)∈ A × B. 1²+ 1²= 2 > 1, (1, 1)̸∈ S. S = A× B

, A⊆ R, B ⊆ R S = A× B.

A× /0 A× {/0} . (x, y)∈ A × {/0} x∈ A y∈ {/0},

{/0} /0 , y = /0. (x, y)∈ A × /0 x∈ A

y∈ /0, /0, y . A× /0 ,

A× /0 = /0. /0× B = /0. .

Proposition 3.4.12. A, B sets, A× B = /0 A = /0 B = /0.

Proof. A× B = /0, A = /0 B = /0. contrapositive method, A̸= /0 B̸= /0. a∈ A b∈ B, (a, b)∈ A × B. A× B ̸= /0. 

A, B finite sets , Example 3.4.11 A× B .

a∈ A , (a, y)∈ A × B. Proposition 3.4.10 , y B

, (a, y) . A× B (a, y) #(B)

. a , A× B

#(A)× #(B) , .

Proposition 3.4.13. A, B finite sets. #(A× B) = #(A) × #(B).

#( /0) = 0, Proposition 3.4.13 #(A× /0) = #(A) × #(/0) = 0. 論 Propo- sition 3.4.12 A× /0 = /0 論 .

Cartesian product . , set

A, B = /0 , A× B = /0 A× B A^′× B A, A^′ .

A× B A, B /0 . .

Proposition 3.4.14. A, B,C, D sets A̸= /0 B̸= /0. A⊆ C B⊆ D (A× B) ⊆ (C × D).

Proof. (⇒) : A⊆ C B⊆ D. (x, y)∈ A × B, x∈ A y∈ B, A⊆ C B⊆ D x∈ C y∈ D. (x, y)∈ C × D, (A× B) ⊆ (C × D).

(⇐) : (A× B) ⊆ (C × D). x∈ A, B̸= /0, b∈ B.

(x, b)∈ A×B. (A×B) ⊆ (C ×D), (x, b)∈ C ×D. x∈ C, A⊆ C.

, y∈ B, A̸= /0, a∈ A. (a, y)∈ A × B. (A× B) ⊆ (C × D),

(a, y)∈ C × D. y∈ D, B⊆ D. 

Question 3.19. Proposition 3.4.14 , A̸= /0 B̸= /0 ?

A⊆ C B⊆ D ?

Cartesian product intersection . Proposition 3.4.15. A, B,C, D sets.

(A∩C) × B = (A × B) ∩ (C × B) and A × (B ∩ D) = (A × B) ∩ (A × D).

Proof. (A∩C) ⊆ A (A∩C) ⊆ C Proposition 3.4.14 ((A∩C) × B) ⊆ (A × B)

((A∩C)×B) ⊆ (C×B) ( Proposition 3.4.14 ). Corollary

3.2.4(1) ((A∩C) × B) ⊆ (A × B) ∩ (C × B).

, (x, y)∈ (A × B) ∩ (C × B), (x, y)∈ A × B (x, y)∈ C × B.

x∈ A x∈ C y∈ B, x∈ A ∩ C y∈ B, (x, y)∈ (A ∩ C) × B.

(A× B) ∩ (C × B) ⊆ (A ∩ C) × B, (A∩ C) × B = (A × B) ∩ (C × B).

A× (B ∩ D) = (A × B) ∩ (A × D). 

Proposition 3.4.15 (A∩C) × (B ∩ D). Proposition 3.4.15 B B∩ D , (A∩C) × (B ∩ D) = (A × (B ∩ D)) ∩ (C × (B ∩ D)). A× (B ∩ D) = (A× B) ∩ (A × D) C× (B ∩ D) = (C × B) ∩ (C × D),

(A∩C) × (B ∩ D) = (A × B) ∩ (A × D) ∩ (C × B) ∩ (C × D). (3.3) (x, y)∈ (A×B)∩(C ×D) (x, y)∈ A ×B ( x∈ A, y ∈ B) (x, y)∈ C ×D ( x∈ C, y∈ D), (x, y)∈ A × D ( x∈ A, y ∈ D) (x, y)∈ C × B ( x∈ C, y ∈ B).

(x, y)∈ (A × D) ∩ (C × B), ((A× B) ∩ (C × D)) ⊆ ((A × D) ∩ (C × B)). Proposition

3.2.5 (3.3)

(A∩C) × (B ∩ D) = ((A × B) ∩ (C × D)) ∩ ((A × D) ∩ (C × B)) = (A × B) ∩ (C × D).

.

Corollary 3.4.16. A, B,C, D sets.

(A∩C) × (B ∩ D) = (A × B) ∩ (C × D).

Question 3.20. Corollary 3.4.16 (A∩C) × (B ∩ D) = (A × D) ∩ (C × B) ?

Corollary 3.4.16 .

, Cartesian products operations

Cartesian product. , Cartesian products

Cartesian product. (A×B)∩(C ×D)

Cartesian product S× T . Corollary 3.4.16 .

S = A∩C, T = B∩D, (A×B)∩(C ×D) = S×T. , Cartesian

products Cartesian product.

Question 3.21. (A× B) ∩ (C × D) = (A × D) ∩ (C × B).

Question 3.22. {Ai, i∈ I}, {Bi, i∈ I} I index set indexed family.

∩

i∈I

(A_i× Bi) = (^∩

i∈I

A_i)× (^∩

i∈I

B_i).

Cartesian product union Proposition 3.4.15 . Proposition 3.4.17. A, B,C, D sets.

(A∪C) × B = (A × B) ∪ (C × B) and A × (B ∪ D) = (A × B) ∪ (A × D).

Proof. A⊆ (A ∪C) C⊆ (A ∪C) Proposition 3.4.14 (A× B) ⊆ ((A ∪C) × B) (C× B) ⊆ ((A ∪C) × B). Corollary 3.2.4(2) (A× B) ∪ (C × B) ⊆ ((A ∪C) × B).

, (x, y)∈ (A ∪C) × B, x∈ A ∪C y∈ B, x∈ A x∈ C y∈ B. x∈ A, y∈ B (x, y)∈ A×B, x∈C, y∈ B (x, y)∈C×B. (x, y)∈ A× B (x, y)∈ C × B, (x, y)∈ (A × B) ∪ (C × B). ((A∪C) × B) ⊆ (A × B) ∪ (C × B), (A∪C) × B = (A × B) ∪ (C × B). A× (B ∪ D) = (A × B) ∪ (A × D)  Proposition 3.4.17 (A∪C) × (B ∪ D). Proposition 3.4.17 B B∪ D , (A∪C) × (B ∪ D) = (A × (B ∪ D)) ∪ (C × (B ∪ D)). A× (B ∪ D) = (A× B) ∪ (A × D) C× (B ∪ D) = (C × B) ∪ (C × D), .

Corollary 3.4.18. A, B,C, D sets.

(A∪C) × (B ∪ D) = (A × B) ∪ (A × D) ∪ (C × B) ∪ (C × D).

(A∪C)×(B∪D) Corollary 3.4.16 .

(A∪C)×(B∪D) (A×B)∪(C ×D) . A×D

(A×B)∪(C ×D) ( A⊆ C D⊆ B). A* C D* B,

. A, D /0 B = C = /0, (A∪C)×(B∪D) = A×D /0 ( Proposition 3.4.12), (A× B) ∪ (C × D) = /0 ∪ /0 = /0. (A∪C) × (B ∪ D) ̸= (A × B) ∪ (C × D).

Question 3.23. (A∪C) × (B ∪ D) ̸= (A × B) ∪ (C × D).

Cartesian product Cartesian product.

(A×B)∪(C ×D) Cartesian product S×T ? A, B,C, D

/0, (A×B) (C×D) /0, . , Corollary

3.4.18 . S, T (A× B) ∪ (C × D) = (S × T) s∈ S,t ∈ T, (s,t)∈ (A × B) ∪ (C × D), s∈ A, t ∈ B s∈ C,

t∈ D. s A C t B D , s∈ A ∪C t∈ B ∪ D.

S⊆ A ∪C T ⊆ B ∪ D. x∈ A ∪C, x∈ A x∈ C

論. x∈ A, b∈ B ( B̸= /0), (x, b)∈ A × B,

(x, b)∈ (A × B) ∪ (C × D) = (S × T) x∈ S. x∈ C, D̸= /0, x∈ S.

A∪C ⊆ S, S = A∪C. T = B∪D. A, B,C, D

/0 , (A×B)∪(C ×D) = (S×T) S = A∪C T = B∪D. Corollary 3.4.18 (A×B)∪(C×D) = (A∪C)×(B∪D) (A×D)∪(C×B) ⊆ (A×B)∪(C×D).

A* C D* B, a∈ A\C d∈ D\B. (a, d)∈ A×D (a, d)̸∈ A×B (a, d)̸∈ C ×D, (a, d)̸∈ (A×B)∪(C ×D). (A×D) ⊆ (A×B)∪(C ×D) .

(A×B)∪(C ×D) = (A∪C)×(B∪D) , A⊆ C D⊆ B. C* A B* D,

(c, d)∈ C × D, c∈ C \ A b∈ B \ D, (C× D) ⊆ (A × B) ∪ (C × D) . (A× B) ∪ (C × D) = (A ∪C) × (B ∪ D) , C⊆ A B⊆ D .

.

Proposition 3.4.19. A, B,C, D sets. S, T sets (A× B) ∪ (C × D) =

S× T A, B,C, D :

(1) A, B,C, D /0.

(2) A = C (3) B = D

(4) A⊆ C B⊆ D (5) C⊆ A D⊆ B

Proof. (⇒): (1) , A, B,C, D /0 , 前 S, T (A×

B)∪(C ×D) = S ×T, S = A∪C T = B∪D. (A×B)∪(C ×D) = (A∪C)×(B∪D),

前 論 A⊆ C D⊆ B C⊆ A B⊆ D .

((A⊆ C) ∨ (D ⊆ B))

∧(

(C⊆ A) ∨ (B ⊆ D))

. ∨,∧ ( 1.3),

(((A⊆ C) ∨ (D ⊆ B)) ∧ (C ⊆ A))

∨(

((A⊆ C) ∨ (D ⊆ B)) ∧ (B ⊆ D))

. ,

((A⊆ C) ∧ (C ⊆ A)) ∨ ((D ⊆ B) ∧ (C ⊆ A)) ∨ ((A ⊆ C) ∧ (B ⊆ D)) ∨ ((D ⊆ B) ∧ (B ⊆ D)).

(A⊆ C)∧(C ⊆ A) A = C, (D⊆ B)∧(B ⊆ D) B = D, (2),

(3), (4), (5) .

(⇐): (1) , (A× B) (C× D) /0, S, T (A× B) ∪

(C× D) = S × T.

(2) Proposition 3.4.17

(A× B) ∪ (C × D) = (A × B) ∪ (A × D) = A × (B ∪ D),

S = A, T = (B∪ D) .

(3) Proposition 3.4.17

(A× B) ∪ (C × D) = (A × B) ∪ (C × B) = (A ∪C) × B, S = (A∪C),T = B .

(4) Proposition 3.4.14 (A× B) ⊆ (C × D), (A× B) ∪ (C × D) = (C × D).

S = C, T = D .

(5) Proposition 3.4.14 (C× D) ⊆ (A × B), (A× B) ∪ (C × D) = (A × B).

S = A, T = B . 

Cartesian product set difference . Proposition 3.4.20. A, B,C, D sets.

(C\ A) × B = (C × B) \ (A × B) and A × (D \ B) = (A × D) \ (A × B).

Proof. (x, y)∈ (C \ A) × B, x∈ C \ A y∈ B. x∈ C x̸∈ A y∈ B. (x, y)∈ C × B (x, y)̸∈ A × B ( (x, y)∈ A × B 導 x∈ A ).

(x, y)∈ (C × B) \ (A × B), (C\ A) × B ⊆ (C × B) \ (A × B).

, (x, y)∈ (C × B) \ (A × B), (x, y)∈ C × B ( x∈ C, y ∈ B) (x, y)̸∈ A×B ( x ̸∈ A y̸∈ B). (x, y)∈ C ×B y∈ B, (x, y)̸∈ A×B x̸∈ A

( x∈ A y∈ B (x, y)∈ A×B ). x∈ C x̸∈ A y∈ B,

(x, y)∈ (C \A)×B, (C×B)\(A×B) ⊆ (C \A)×B. (C\A)×B = (C ×B)\(A×B).

A× (D \ B) = (A × D) \ (A × B). 

Proposition 3.4.20 (C\ A) × (D \ B). Corollary 3.4.16 (C\ A) × (D \ B) = ((C \ A) ∩C) × (D ∩ (D \ B)) = ((C \ A) × D) ∩ (C × (D \ B)).

Proposition 3.4.20 (C\ A) × D = (C × D) \ (A × D) C× (D \ B) = (C × D) \ (C× B).

(C\ A) × (D \ B) = ((C × D) \ (A × D)) ∩ ((C × D) \ (C × B)).

Proposition 3.2.11(3),

((C× D) \ (A × D)) ∩ ((C × D) \ (C × B)) = (C × D) \(

(A× D) ∪ (C × B)) . .

Corollary 3.4.21. A, B,C, D sets.

(C\ A) × (D \ B) = ((C \ A) × D) ∩ (C × (D \ B)) = (C × D) \(

(A× D) ∪ (C × B)) . Cartesian product Cartesian product.

(C× D) \ (A × B) S× T ? ,

. , (C×D)\(A×B) = (C ×D)\((C ×D)∩(A×B)) ( Question 3.11).

Corollary 3.4.16, (C× D) ∩ (A × B) = (C ∩ A) × (D ∩ B) = (C × B) ∩ (A × D).

(C× D) \ (A × B) = (C × D) \ ((C × B) ∩ (A × D)). (3.4) Proposition 3.2.11(2)

(C× D) \ ((C × B) ∩ (A × D)) = ((C × D) \ (C × B)) ∪ ((C × D) \ (A × D)). (3.5) Proposition 3.4.20

(C× D) \ (C × B) = C × (D \ B) and (C × D) \ (A × D) = (C \ A) × D. (3.6)

(3.4), (3.5), (3.6) .

Corollary 3.4.22. A, B,C, D sets.

(C× D) \ (A × B) = (C × (D \ B)) ∪ ((C \ A) × D).

Corollary 3.4.21 Corollary 3.4.22 .

Corollary 3.4.22, Proposition 3.4.19 S, T S×T =

(C× D) \ (A × B) . Proposition 3.4.19 (C× (D \ B)) ∪ ((C \ A) × D)

S× T , : (1) C, D, (D\ B),(C \ A) /0,

D⊆ B C⊆ A ( C D /0 ). (2) C = C\ A, A∩C = /0. (3)

D = D\ B, B∩ D = /0. (4) C ⊆ (C \ A) (D\ B) ⊆ D. (D\ B) D, C⊆ (C \A), C = C\A, (2) . (5) D⊆ (D\B) (C\A) ⊆ C.

(4) . (3) . 論 .

Proposition 3.4.23. A, B,C, D sets. S, T sets (C× D) \ (A × B) =

S× T A, B,C, D :

(1) D⊆ B.

(2) C⊆ A.

(3) A∩C = /0.

(4) B∩ D = /0.

Question 3.24. Proposition 3.4.23 S, T ?

論 Cartesian product complement . ,

Cartesian product .

A X , B Y , 論 A×B. A×B

X×Y. A complement A^c X complement,

A^c= X\ A. B^c B Y complement, B^c= Y\ B. A× B complement (A× B)^c, A× B X×Y complement, (A× B)^c= (X×Y) \ (A × B).

, complement universal sets complement. , Corollary 3.4.22 C = X D = Y

(A× B)^c= (X×Y) \ (A × B) = (X × (Y \ B)) ∪ ((X \ A) ×Y) = (X × B^c)∪ (A^c×Y).

X = A∪ A^c Y = B∪ B^c, Proposition 3.4.17

X× B^c= (A∪ A^c)× B^c= (A× B^c)∪ (A^c× B^c),

A^c×Y = (A^c× B) ∪ (A^c× B^c). .

Proposition 3.4.24. A, B sets. (A× B)^c= (A× B^c)∪ (A^c× B^c)∪ (A^c× B).

, 論 Cartesian product.

, .