[set theory] set identities (idempotent law | exchange law | combination law | distribution rate | De Morgan law | absorption rate | zero law | identity | exclusion law | contradiction law | complemen

One 、 Set identity

---

1. Idempotent law :

A

∪

A

=

A

A \cup A = A

A∪A=A ,

A

∩

A

=

A

A \cap A = A

A∩A=A

2. Commutative law :

A

∪

B

=

B

∪

A

A \cup B = B \cup A

A∪B=B∪A ,

A

∩

B

=

B

∩

A

A \cap B = B \cap A

A∩B=B∩A

3. Associative law :

(

A

∪

B

)

∪

C

=

A

∪

(

B

∪

C

)

(A \cup B) \cup C = A \cup ( B \cup C )

(A∪B)∪C=A∪(B∪C) ,

(

A

∩

B

)

∩

C

=

A

∩

(

B

∩

C

)

(A \cap B) \cap C = A \cap ( B \cap C )

(A∩B)∩C=A∩(B∩C)

4. Distribution rate :

A

∪

(

B

∩

C

)

=

(

A

∪

B

)

∩

(

A

∪

C

)

A \cup ( B \cap C ) = ( A \cup B ) \cap ( A \cup C )

A∪(B∩C)=(A∪B)∩(A∪C) ,

A

∩

(

B

∪

C

)

=

(

A

∩

B

)

∪

(

A

∩

C

)

A \cap ( B \cup C ) = ( A \cap B ) \cup ( A \cap C )

A∩(B∪C)=(A∩B)∪(A∩C)

5. De Morgan law :

① absolute form :

∼

(

A

∪

B

)

=

∼

A

∩

∼

B

\sim ( A \cup B ) = \sim A \cap \sim B

∼(A∪B)=∼A∩∼B ,

∼

(

A

∩

B

)

=

∼

A

∪

∼

B

\sim ( A \cap B ) = \sim A \cup \sim B

∼(A∩B)=∼A∪∼B

② Relative form :

A

−

(

B

∪

C

)

=

(

A

−

B

)

∩

(

A

−

C

)

A - (B \cup C) = ( A - B ) \cap (A - C)

A−(B∪C)=(A−B)∩(A−C) ,

A

−

(

B

∩

C

)

=

(

A

−

B

)

∪

(

A

−

C

)

A - (B \cap C) = ( A - B ) \cup (A - C)

A−(B∩C)=(A−B)∪(A−C)

6. absorptivity :

A

∪

(

A

∩

B

)

=

A

A \cup ( A \cap B ) = A

A∪(A∩B)=A ,

A

∩

(

A

∪

B

)

=

A

A \cap (A \cup B) = A

A∩(A∪B)=A

7. Law of zero :

A

∪

E

=

E

A \cup E = E

A∪E=E ,

A

∩

∅

=

∅

A \cap \varnothing = \varnothing

A∩∅=∅

8. The same thing :

A

∪

∅

=

A

A \cup \varnothing = A

A∪∅=A ,

A

∩

E

=

A

A \cap E = A

A∩E=A

( An empty set is the unit element of a union operation , The complete set is the unit element of intersection operation )

9. The law of excluded middle :

A

∪

∼

A

=

E

A \cup \sim A = E

A∪∼A=E

10. Law of contradiction :

A

∩

∼

A

=

∅

A \cap \sim A = \varnothing

A∩∼A=∅

11. Complementary law :

∼

∅

=

E

\sim \varnothing = E

∼∅=E ,

∼

E

=

∅

\sim E= \varnothing

∼E=∅

12. The law of double negation :

∼

(

∼

A

)

=

A

\sim ( \sim A ) = A

∼(∼A)=A

13. Complementary transformation law :

A

−

B

=

A

∩

∼

B

A - B = A \cap \sim B

A−B=A∩∼B

( The difference operation of sets is unnecessary , Intersection and complement operations of sets can replace difference operations )

Two 、 Generalization of set identities to set families

---

{

A

α

}

α

∈

S

\{ A_\alpha \}_{\alpha \in S}

{ Aα​}α∈S​ For the set family ,

S

S

S Is the indicator set ,

α

\alpha

α It is the element in the indicator set , about

S

S

S In the collection

α

\alpha

α Elements , There is a set

A

α

A_\alpha

Aα​ With the corresponding ; be-all

A

α

A_\alpha

Aα​ Put it together , Form a family ;

B

B

B Is an arbitrary set ;

1 . Distributive law

Distributive law ① :

B

∪

(

⋂

{

A

α

}

α

∈

S

)

=

⋂

α

∈

S

(

B

∪

A

α

)

B \cup ( \bigcap \{ A_\alpha \}_{\alpha \in S} ) = \bigcap_{\alpha \in S} ( B \cup A_\alpha )

B∪(⋂{ Aα​}α∈S​)=α∈S⋂​(B∪Aα​)

Find the intersection of each set element in the set family , Then with

B

B

B Perform and calculate ; Equivalent to Set each element in the family with

B

B

B Union , Then find the intersection of each of the above union operation results ;

Distributive law ② :

B

∩

(

⋃

{

A

α

}

α

∈

S

)

=

⋃

α

∈

S

(

B

∩

A

α

)

B \cap ( \bigcup \{ A_\alpha \}_{\alpha \in S} ) = \bigcup_{\alpha \in S} ( B \cap A_\alpha )

B∩(⋃{ Aα​}α∈S​)=α∈S⋃​(B∩Aα​)

Join each set element in the set family , Then with

B

B

B Perform intersection ; Equivalent to Set each element in the family with

B

B

B Intersection , Then find the union of each of the above union operation results ;

2 . De Morgan law

De Morgan law ( absolute form ) ① :

∼

(

⋃

{

A

α

}

α

∈

S

)

=

⋂

α

∈

S

(

∼

A

α

)

\sim ( \bigcup \{ A_\alpha \}_{\alpha \in S} ) = \bigcap_{\alpha \in S} ( \sim A_\alpha )

∼(⋃{ Aα​}α∈S​)=α∈S⋂​(∼Aα​)

Generalized union of set family , Then make up ; be equal to Each set in the set family , Make up first , Then find the generalized intersection ;

De Morgan law ( absolute form ) ② :

∼

(

⋂

{

A

α

}

α

∈

S

)

=

⋃

α

∈

S

(

∼

A

α

)

\sim ( \bigcap \{ A_\alpha \}_{\alpha \in S} ) = \bigcup_{\alpha \in S} ( \sim A_\alpha )

∼(⋂{ Aα​}α∈S​)=α∈S⋃​(∼Aα​)

Generalized intersection of set families , Then make up ; be equal to Each set in the set family , Make up first , Then find the generalized union ;

De Morgan law ( Relative form ) ③ :

B

−

(

⋃

{

A

α

}

α

∈

S

)

=

⋂

α

∈

S

(

B

−

A

α

)

B - ( \bigcup \{ A_\alpha \}_{\alpha \in S} ) = \bigcap_{\alpha \in S} ( B - A_\alpha )

B−(⋃{ Aα​}α∈S​)=α∈S⋂​(B−Aα​)

B

B

B Set subtraction Generalized union of set family ( Set family generalized union be relative to aggregate

B

B

B The complement of ) ; be equal to

B

B

B Set minus each set in the set family , First find the relative complement , Then find the generalized intersection ;

De Morgan law ( Relative form ) ④ :

B

−

(

⋂

{

A

α

}

α

∈

S

)

=

⋃

α

∈

S

(

B

−

A

α

)

B - ( \bigcap \{ A_\alpha \}_{\alpha \in S} ) = \bigcup_{\alpha \in S} ( B - A_\alpha )

B−(⋂{ Aα​}α∈S​)=α∈S⋃​(B−Aα​)

B

B

B Set subtraction Generalized intersection of set families ( Set family generalized intersection be relative to aggregate

B

B

B The complement of ) ; be equal to

B

B

B Set minus each set in the set family , First find the relative complement , Then find the generalized union ;