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[set theory] relationship properties (common relationship properties | relationship properties examples | relationship operation properties)

2022-07-03 04:57:00 【Programmer community】

List of articles

One 、 The nature of common relationships
Two 、 Examples of the nature of relationships
3、 ... and 、 Relation operation properties

One 、 The nature of common relationships

stay Set of natural numbers

{

⋯

}

N=\{ 0, 1,2, \cdots \}

$N = {0, 1, 2, \dots}$ On , The nature of the following relationship :

1. Less than or equal to ：

Less than or equal to :

Symbolic description :

≤

{

∣

∈

∧

∈

∧

≤

}

\leq = \{ <x, y> | x \in N \land y \in N \land x \leq y \}

$\leq = {< x, y > ∣ x \in N \land y \in N \land x \leq y}$

The nature of the relationship : introspect , antisymmetric , Pass on

2. Relationship greater than or equal to ：

Relationship greater than or equal to :

Symbolic description :

≥

{

∣

∈

∧

∈

∧

≥

}

\geq = \{ <x, y> | x \in N \land y \in N \land x \geq y \}

$\geq = {< x, y > ∣ x \in N \land y \in N \land x \geq y}$

The nature of the relationship : introspect , antisymmetric , Pass on

3. Less than relationship ：

Less than relationship :

Symbolic description :

{

∣

∈

∧

∈

∧

}

< = \{ <x, y> | x \in N \land y \in N \land x < y \}

$< = {< x, y > ∣ x \in N \land y \in N \land x < y}$

The nature of the relationship : Reflexion , antisymmetric , Pass on

4. Greater than the relationship ：

Greater than the relationship :

Symbolic description :

{

∣

∈

∧

∈

∧

}

> = \{ <x, y> | x \in N \land y \in N \land x > y \}

$> = {< x, y > ∣ x \in N \land y \in N \land x > y}$

The nature of the relationship : Reflexion , antisymmetric , Pass on

5. Division relations :

Division relations :

Symbolic description :

∣

{

∣

∈

∧

∈

∧

∣

}

| = \{ <x, y> | x \in N \land y \in N \land x | y \}

$∣ = {< x, y > ∣ x \in N \land y \in N \land x ∣ y}$

The nature of the relationship : antisymmetric , Pass on

∣

x|y

$x ∣ y$ Medium

∣

$∣$ The symbol means division ,

$x$ to be divisible by

$y$ ;

x
x
$x$ to be divisible by
y
y
$y$ ,
x
x
$x$ It's a divisor ( molecular ) ,
y
y
$y$ It's a dividend ( The denominator ) ;
y
x
\dfrac{y}{x}
$\frac{y}{x}$
y
y
$y$ Can be
x
x
$x$ to be divisible by ,
x
x
$x$ It's a divisor ( molecular ) ,
y
y
$y$ It's a dividend ( The denominator ) ;
y
x
\dfrac{y}{x}
$\frac{y}{x}$
In the divisible relationship , Be sure to pay attention to , Only non
0
0
$0$ to be divisible by
0
0
$0$ ,
0
0
$0$ Cannot divide non
0
0
$0$ , namely
0
0
$0$ Can only be divisor , Cannot divide ;

6. Identity :

Identity :

Symbolic description :

{

∣

∈

∧

∈

∧

}

I_N = \{ <x, y> | x \in N \land y \in N \land x = y \}

$I_{N} = {< x, y > ∣ x \in N \land y \in N \land x = y}$

The nature of the relationship : introspect , symmetry , antisymmetric , Pass on

7. Global relations :

Global relations :

Symbolic description :

{

∣

∈

∧

∈

}

E_N = \{ <x, y> | x \in N \land y \in N \} = N \times N

$E_{N} = {< x, y > ∣ x \in N \land y \in N} = N \times N$

The nature of the relationship : introspect , symmetry , Pass on

introspect , Antisymmetric relation , It is called partial order relation ;

Two 、 Examples of the nature of relationships

Relationship diagram relationship determination :

① introspect : All vertices in the graph They all have rings ;
② Reflexion : All vertices in the graph There is no ring ;
③ symmetry : Between two vertices Yes
$2$ A directed edge ;
④ antisymmetric : Between two vertices Yes
$1$ A directed edge ;
⑤ Pass on : Premise $\to b , b\to ca→b,b→c Don't set up Default delivery , Premise a → b , b → c a \to b , b\to ca→b,b→c establish Must satisfy a → c a \to ca→c There is ;$

{

}

R_1 = \{ <a, a> , <a, b> , <b , c> , <a,c> \}

$R_{1} = {< a, a >, < a, b >, < b, c >, < a, c >}$ :

Draw a diagram of the above relationship : antisymmetric , Pass on
Insert picture description here

introspect / Reflexion : Some vertices have rings , Some vertices have no rings , Neither reflexivity nor reflexivity is tenable ;

symmetry / antisymmetric : Between the vertices are

$1$ The strip has a directed edge , There is only

0/1

$0 / 1$ side , yes antisymmetric Of ;

Pass on :

→

a\to b, b \to c

$a \to b, b \to c$ establish ,

→

a \to c

$a \to c$ There is , Transitivity establish ;

{

}

R_2 = \{ <a, a> , <a, b> , <b , c> , <c,a> \}

$R_{2} = {< a, a >, < a, b >, < b, c >, < c, a >}$ :

Draw a diagram of the above relationship : antisymmetric

Insert picture description here

introspect / Reflexion : Some vertices have rings , Some vertices have no rings , Neither reflexivity nor reflexivity is tenable ;

symmetry / antisymmetric : Between the vertices are

$1$ The strip has a directed edge , There is only

0/1

$0 / 1$ side , yes antisymmetric Of ;

Pass on :

→

a\to b, b \to c

$a \to b, b \to c$ establish ,

→

a \to c

$a \to c$ non-existent , Transitivity Don't set up ;

{

}

R_3 = \{ <a, a> , <b, b> , <a,b> , <b,a> , <c,c> \}

$R_{3} = {< a, a >, < b, b >, < a, b >, < b, a >, < c, c >}$ :

Draw a diagram of the above relationship : introspect , symmetry , Pass on

Insert picture description here

introspect / Reflexion : All vertices have rings , reflexivity establish ;

symmetry / antisymmetric : Between the vertices are

$0$ or

$2$ The strip has a directed edge , There is only

0/2

$0 / 2$ side , yes symmetry Of ;

Pass on : Transitivity establish ;

Premise $\to b , b\to aa→b,b→a , Correspondence exists a → a a \to aa→a$
Premise $\to a , a\to bb→a,a→b , Correspondence exists b → b b \to bb→b$

{

}

R_4 = \{ <a, a> , <a,b> , <b,a> , <c,c> \}

$R_{4} = {< a, a >, < a, b >, < b, a >, < c, c >}$ :

Draw a diagram of the above relationship : symmetry

Insert picture description here

introspect / Reflexion : Some vertices have rings , Some vertices have no rings , Neither reflexivity nor reflexivity is tenable ;

symmetry / antisymmetric : Between the vertices are

$0$ or

$2$ The strip has a directed edge , There is only

0/2

$0 / 2$ side , yes symmetry Of ;

Pass on : Transitivity Don't set up ;

Premise $\to b , b\to aa→b,b→a , Correspondence exists a → a a \to aa→a$
Premise $\to a , a\to bb→a,a→b , There is no corresponding b → b b \to b$
$b \to b$ , Here transitivity does not hold ;

{

}

R_5 = \{ <a, a> , <a,b> , <b,b> , <c,c> \}

$R_{5} = {< a, a >, < a, b >, < b, b >, < c, c >}$ :

Draw a diagram of the above relationship : introspect , antisymmetric , Pass on

Insert picture description here

introspect / Reflexion : All vertices have rings , reflexivity establish ;

symmetry / antisymmetric : Between the vertices are

$0$ or

$1$ The strip has a directed edge , There is only

0/1

$0 / 1$ side , yes antisymmetric Of ;

Pass on : The premise doesn't hold , Transitivity establish ;

{

}

R_6 = \{ <a, a> , <b,a> , <b,c> , <a,a> \}

$R_{6} = {< a, a >, < b, a >, < b, c >, < a, a >}$ :

Draw a diagram of the above relationship : It doesn't matter

Insert picture description here

introspect / Reflexion : Some vertices have rings , Some vertices have no rings , Neither reflexivity nor reflexivity is tenable ;

symmetry / antisymmetric : Between the vertices are

$1$ or

$2$ The strip has a directed edge , There is only

0/1

$0 / 1$ The edge is antisymmetric , There is only

0/2

$0 / 2$ The edges are symmetrical , The above symmetry / The objection is not tenable ;

Pass on : Premise

→

a \to b , b \to c

$a \to b, b \to c$ , There is no corresponding

→

a \to c

$a \to c$ , Here transitivity does not hold ;

3、 ... and 、 Relation operation properties

discuss a problem : Specify the nature of the relationship Between them , The nature of the result ; Such as Two relations of reflexivity Perform the reverse order synthesis operation , The result is reflexive ;

The meaning of the table in the following figure is : Such as Second column “ introspect ” And The third column “

∪

R_1 \cup R_2

$R_{1} \cup R_{2}$ ” , Cross table position , representative Relationship

R_1

$R_{1}$ And relationships

R_2

$R_{2}$ It's reflexive , Whether the intersection of its ordered pairs is reflexive , If it is

$1$ , The explanation is reflexive , If there is no value , The explanation is not reflexive ;

	introspect	Reflexion	symmetry	antisymmetric	Pass on
$R_1^{-1}, R_2^{-1}R1−1,R2−1$
$R_1 \cup R_2^{-1}R1∪R2−1$
$R_1 \cap R_2R1∩R2$
$R_1 \circ R_2 , R_2 \circ R_1R1∘R2,R2∘R1$
$R_1 - R_2 , R_2 - R_1R1−R2,R2−R1$
$\sim R_1, \sim R_2∼R1,∼R2$