Partitioning of Any Infinite Set with the Aid of Non-Surjective Injective Maps and the Study of a Remarkable Semigroup

Abstract

In this article, we will present a particularly remarkable partitioning method of any infinite set with the aid of non-surjective injective maps. The non-surjective injective maps from an infinite set to itself constitute a semigroup for the law of composition bundled with certain properties allowing us to prove the existence of remarkable elements. Not to mention a compatible equivalence relation that allows transferring the said law to the quotient set, which can be provided with a lattice structure. Finally, we will present the concept of Co-injectivity and some of its properties.

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Harrafa, C. (2020) Partitioning of Any Infinite Set with the Aid of Non-Surjective Injective Maps and the Study of a Remarkable Semigroup. Open Journal of Discrete Mathematics, 10, 74-88. doi: 10.4236/ojdm.2020.103008.

1. Introduction

The concept of map in mathematics has a primordial role in understanding the links that exist between the different mathematical fields and structures. A map is binary relation over two sets that associates to every element of the first set exactly one element of the second set, sometimes with a specific property. For instance, a “map” is a “linear transformation” in linear algebra, a “continuous function” in topology, operators in analysis and representations in group theory, etc. In this article we will show how non-surjective injective maps allow to partitioning an infinite set in several ways.

2. Part I

Proposition 1

Let E, and F be non-empty sets. If f , g are two non-surjective injective maps from E to F and from F to E respectively, then:

· ( A , f ( B ) , I f g ) forms a partition of F.

· ( B , g ( A ) , I g f ) forms a partition of E.

where,

A = { y F | x E , f ( x ) y } and B = { x E | y F , g ( y ) x }

and I f , I g representing the image sets under f , g respectively.

Proof

Let E and F be two non-empty sets, and let ƒ and g be two non-surjective injective maps from E to F, and from F to E respectively. Since ƒ and g are non-surjective, then the following sets:

A = { y F | x E , f ( x ) y } and B = { x E | y F , g ( y ) x } are non-empty.

Also, obviously E = B I g such as B I g = as follows from the definition of I g .

For any such map ƒ from E to F:

I f = f ( E ) = f ( B I g ) = f ( B ) f ( I g ) = f ( B ) I f g

So F = A f ( B ) I f g

Since ƒ is an injective map then:

f ( B I g ) = f ( B ) I f g = f ( ) =

Therefore, ( A , f ( B ) , I f g ) forms a partition of F.

In analogy to the map g from F to E, ( B , g ( A ) , I g f ) forms a partition of E.

Note 1

This process could be applied repeatedly, and for each iteration, finer partitions of the sets E and F respectively will be obtained, e.g. after 2nd iteration we have:

· E = B g ( A ) ( g f ) ( B ) I g f g

· F = A f ( B ) ( f g ) ( A ) I f g f

In case that ƒ and g are (two) (2) different non-surjective injective maps from an infinite set E to itself, we can compose either by ƒ or by g, or by both indefinitely. Thus several partition classes of E can obtained, for example after a second (2nd) iteration

· E = A f f ( A f ) f 2 ( A f ) I f 3

· E = A g g ( A g ) g 2 ( A g ) I g 3

· E = A f f ( A g ) ( f g ) ( A f ) I f g f

· E = A g g ( A f ) ( g f ) ( A g ) I g f g

· E = A f f ( A g ) ( f g ) ( A g ) I f g 2

· E = A g g ( A f ) ( g f ) ( A f ) I g f 2

where,

A f = { y F | x E , f ( x ) y } and A g = { x E | y F , g ( y ) x }

Example 1

If E = F = i.e. the set of natural numbers, ƒ and g are two maps from to itself (i.e. ƒ and g are non-surjective injective) and defined by: ˆ

· n , f ( n ) = 2 n

· n , g ( n ) = 2 n + 1

Knowing that A f = 2 + 1 and A g = 2 then:

· f ( A g ) = 4

· g ( A f ) = 4 + 3

· ( f g ) ( A f ) = 8 + 6

· ( g f ) ( A g ) = 8 + 1

· ( f g f ) ( ) = 8 + 2

· ( g f g ) ( ) = 8 + 5

Therefore, we can partition the set to the second (2nd) order by ƒ and g such as:

= ( 2 + 1 ) ( 4 ) ( 8 + 6 ) ( 8 + 2 )

= ( 2 ) ( 4 + 3 ) ( 8 + 1 ) ( 8 + 5 )

2.1. Remarkable Partition

Proposition 2

Let E be an infinite set, be the set of natural numbers, and ƒ be a non-surjective injective map from E to itself, such that:

A f = { y E | x E , f ( x ) y }

Then,

n , E = [ i = 0 i = n f i ( A f ) ] I f n + 1

where, f i ( A f ) = f f f ( A f ) i times

Proof

By induction:

For n = 0 , we have, by definition E = A f I f , and for n = 1 the proposition 1 states that, if E = F and f = g then, E = A f f ( A f ) I f 2 Suppose that the said property is true for n. Then, by composing by ƒ, we get:

I f = f ( E ) = [ i = 0 i = n f i + 1 ( A f ) ] I f n + 2 = [ i = 1 i = n + 1 f i ( A f ) ] I f n + 2

Then,

E = A f [ i = 1 i = n + 1 f i ( A f ) ] I f n + 2 = [ i = 0 i = n + 1 f i ( A f ) ] I f n + 2

Therefore,

n , E = [ i = 0 i = n f i ( A f ) ] I f n + 1

Note 2

For all non-surjective injective maps ƒ from an infinite set E to itself,

n , A f n + 1 = i = 0 i = n f i ( A f )

Definition 1

Let ƒ be a map from no-empty set E to itself,

· x E is a fixed point of ƒ if, f ( x ) = x .

· A E is a fixed point set of ƒ if, f ( A ) = A .

Proposition 3

For all non-surjective injective maps f from an infinite set E to itself, n , f n ( A f ) contains no fixed points of ƒ.

Proof

By induction:

For n = 0 , let x A f , and by definition y E , f ( y ) x , particularly for y = x so f ( x ) x then A f contains no fixed points of f. Suppose that the aforementioned property is true for n, let x f n + 1 ( A f ) = f ( f n ( A f ) ) , then y f n ( A f ) such that x = f ( y ) , we have f ( y ) y (by inductive hypothesis), since ƒ is injective then f ( f ( y ) ) f ( y ) so f ( x ) x , then x f n + 1 ( A f ) , f ( x ) x . Therefore, n , f n ( A f ) contains no fixed points of ƒ.

Note 3

Let E be an infinite set, and ƒ be a non-surjective injective map from E to itself, we define the following:

F i x f = { x E | f ( x ) = x } and S t f ( E ) = { A E | f ( A ) = A }

· n , F i x f f n ( A f ) =

· n * , F i x f F i x f n

· n * , F i x f n F i x f 2 n

· p , n * , f p ( F i x f n ) = F i x f n

· For all ƒ non-surjective injective maps from an infinite set E to itself,

F i x f S t f ( E )

Proposition 4

For all ƒ non-surjective injective maps from E to itself,

A S t f ( E ) , n , f n ( A f ) A =

Proof

Let A S t f ( E ) , by induction, For n = 0 , we have, by definition A f I f = then A f A = A f f ( A ) = ˆ Suppose that the said property is true for n , let x f n + 1 ( A f ) A , then, y f n ( A f ) , x = f ( y ) and y 0 A , x = f ( y 0 ) . Since ƒ is injective so y = y 0 f n ( A f ) A , which contradicts the inductive hypothesis. Then,

f n + 1 ( A f ) A = .

Therefore, A S t f ( E ) , n , f n ( A f ) A =

Lemma 1 (Generalization)

p * , A S t f p ( E ) , n , f n ( A f ) A =

Proof

By complete (strong) induction,

Let p * , and A S t f p ( E ) = { A E | f p ( A ) = A } : For n = 0 , we have, by definition A f I f = , since p * , I f p I f , then: A f A = A f f p ( A ) =

Suppose that the said property is true for all i { 1 , , n } , let x f n + 1 ( A f ) A , then, y A f , x = f n + 1 ( y ) and y 0 A , x = f p ( y 0 ) . Since ƒ is injective then,

· f n p + 1 ( y ) = y 0 , if n + 1 > p , which contradicts the inductive hypothesis, because i = n p + 1 { 1 , , n }

· y = f p n 1 ( y 0 ) , if p > n + 1 , which contradicts A f I f p n 1 =

· y = y 0 , if n + 1 = p , which contradicts A f A =

Then,

f n + 1 ( A f ) A = ,

Therefore,

p * , A S t f p ( E ) , n , f n ( A f ) A =

QED

For all n , and for all ƒ non-surjective injective maps from an infinite set E to itself, we define the following:

· S f n ( E ) = { A E | p { 1 , , n + 1 } , f p ( A ) = A }

· S f ( E ) = { A E | p * , f p ( A ) = A }

· S f n = { x A | A S f n ( E ) }

· S f = { x A | A S f ( E ) }

Theorem 1

Let E be an infinite set, and ƒ be a non-surjective injective map from E to itself, then:

E = [ n f n ( A f ) ] S f

Proof

Note that for n = 0 , S f 0 ( E ) = S t f ( E ) , The sequence of subsets of E, ( S f n ) n is increasing by inclusion, so it is convergent, the limit is: n S f n = S f

We have already established that,

n , E = [ i = 0 i = n f i ( A f ) ] I f n + 1

Otherly, according to the Lemma 1, n , E = [ i = 0 i = n f i ( A f ) ] S f n =

Let’s define the following, n , I f n + 1 ^ = I f n + 1 \ S f n the sequence of subsets of ( I f n + 1 ^ ) n is decreasing by inclusion then, it is convergent [1], the limit is:

n I f n + 1 ^ =

Because, let x n I f n + 1 ^ , then:

x n I f n + 1 \ S f n n , x I f n + 1 \ S f n n , x I f n + 1 and x S f n ¯

n , x I f n + 1 and n , x S f n ¯ x n I f n + 1 and x n S f n ¯

x n I f n + 1 and x n S f n ¯ x n I f n + 1 and x S f ¯

x [ n I f n + 1 ] \ S f

On the other hand, the sequence of subsets of E, ( I f n + 1 ) n is strictly decreasing by inclusion, then it is convergent and the limit is H = n I f n + 1 , and since ƒ is injective, so f ( H ) = H then ƒ is bijective from H to itself, and H S t f ( E ) S f ( E ) .

n , E = [ i = 0 i = n f i ( A f ) ] I f n + 1 ^ S f n

Therefore,

E = [ n f n ( A f ) ] S f

N.B. If S f = , then we get:

E = n f n ( A f )

QED

Example 2

is the set of real numbers, and P ( ) is the set of subsets of .

Let ƒ be a map from to itself, defined by:

f ( x ) = { x + 1 , if x 0 x , if x < 0

Therefore, ƒ is injective non-surjective from to itself, because ƒ is injective and A f = [ 0 , 1 [ .

where,

n * , f n ( x ) = { x + n , if x 0 x , if x < 0

We have, f ( A f ) = [ f ( 0 ) , f ( 1 ) [ = [ 1 , 2 [ + since ƒ is increasing Therefore,

n * , f n ( A f ) = [ n , n + 1 [

· n , S f n = S f = *

· n , S f n ( E ) = S f ( E ) = P ( * )

According to the theorem 1, we can write:

= * [ n [ n , n + 1 [ ]

Example 3

Remark

If h is an affine map from to , so that h ( x ) = a x + b / a , b and a 0 , then:

n * , x , h n ( x ) = a n x + b ( 1 + a + + a n 1 )

If a 1 , then:

n * , x , h n ( x ) = a n x + b ( 1 a n 1 a )

Let ƒ be a map defined from E = [ 0 , 4 ] to itself by:

x [ 0 , 4 ] , f ( x ) = { x + 1 , if x [ 0 , 1 ] x + 1 , if x ] 1 , 2 ] 1 2 x + 2 , if x [ 2 , 4 ]

ƒ is non-surjective injective map from E to itself, so that:

· A f = ] 1 , 2 ]

· F i x f = { 4 }

· The set A = [ 0 , 1 ] E , fulfills the condition f ( A ) = A

· We have A f = ] 1 , 2 ] , f ( A f ) = f ( ] 1 , 2 ] ) = ] 2 , 3 ] [ 2 , 4 ] , f ( ] 2 , 3 ] ) = ] 3 , 7 2 ] ,

· S f = { 4 } [ 0 , 1 ]

The restriction of ƒ to [ 2 , 4 ] is an affine function such as a = 1 2 , and b = 2 , then:

n * , f n ( x ) = 1 2 n x + 2 ( 1 + 1 2 + + 1 2 n 1 )

We have: f n ( 2 ) = 4 1 2 n 1 , and f n ( 3 ) = 4 1 2 n ,

According to Theorem 1:

[ 0 , 4 ] = { ] 1 , 2 ] n ] 4 1 2 n 1 , 4 1 2 n ] } { 4 } [ 0 , 1 ]

Example 4

Let ƒ be a map defined from E = [ 0 , 3 ] to itself by:

x [ 0 , 3 ] , f ( x ) = { x + 2 , if x [ 0 , 1 ] 1 2 x + 1 , if x ] 1 , 2 [ x 2 , if x [ 2 , 3 ]

ƒ is non-surjective injective from E to itself, so that:

· A f = ] 1 , 3 2 ]

· F i x f =

· The set A = [ 0 , 1 ] E , fulfills the condition f ( A ) A , and f 2 ( A ) = A

· The set B = [ 2 , 3 ] E , fulfills the condition f ( B ) B , and f 2 ( B ) = B

· S f = [ 0 , 1 ] [ 2 , 3 ]

We have,

· x ] 1 , 2 [ , n * , f n ( x ) = 1 2 n x + 2 ( 1 1 2 n )

· lim x 1 f n ( x ) = 2 1 2 n and f n ( 3 2 ) = 2 1 2 n + 1

Therefore, according to Theorem 1:

[ 0 , 3 ] = { n ] 2 1 2 n , 2 1 2 n + 1 ] } [ 0 , 1 ] [ 2 , 3 ]

Corollary 1

Let f , g be non-surjective injective maps from an infinite set E to itself such as S f = S g ¯ , then:

E = n B n

where, n , B n = f n ( A f ) g n ( A g )

Therefore, ( B n ) n forms a partition of E.

Definition 2

· Let E be an infinite set, we write: I n j n s ( E ) as being the set of non-surjective injective maps from E to itself.

· I n j n s ( E , F ) as the set of non-surjective injective maps from a set E to a set F, that being said, E and F are supposed to be non-empty and | E | | F | ( | E | is the cardinal of E).

Properties

1) ( I n j n s ( E ) , ) is a semigroup, because the composite of 2 (two) injective maps ƒ and g is injective and I f g I f , so f , g I n j n s ( E ) , f g I n j n s ( E ) .

2) f , g I n j n s ( E ) , A f g = A f f ( A g ) , ( A f , f ( A g ) , I f g ) is, indeed, a partition of E and E = A f g I f g , because f g I n j n s ( E ) .

3) There is no idempotent element for the law of composition in I n j n s ( E ) and if such an element exists, then f 2 = f and since ƒ is injective, then f = I d which is contradictory, because, f I n j n s ( E ) .

4)Let f I n j n s ( E ) , and assuming that ƒ is a map from E to itself such as: f ˜ ( x ) = { x , if x A f f ( x ) , if x I f

5) We have, f I n j n s ( E ) , f ˜ I n j n s ( E ) knowing that I f ˜ = A f I f 2 and A f ˜ = f ( A f )

6) Let f I n j n s ( E ) , we have, f ˜ ( A f ˜ ) = f ˜ ( f ( A f ) ) = f 2 ( A f ) , and

I f ˜ 2 = f ˜ ( I f ˜ ) = f ˜ ( A f I f 2 ) = f ˜ ( A f ) f ˜ ( I f 2 ) = A f I f 3

Then,

E = A f ˜ f ˜ ( A f ˜ ) I f ˜ 2 = A f f ( A f ) f 2 ( A f ) I f 3

Therefore, f ˜ allow us to reduce order of iteration.

7) f I n j n s ( E ) ,

· A f ˜ f = A f ˜ f ˜ ( A f ) = A f f ( A f ) = A f 2 , then I f ˜ f = I f 2

· A f f ˜ = A f f ( A f ˜ ) = A f f 2 ( A f ) ,then, I f f ˜ = f ( A f ) I f 3

· A f ˜ ˜ = f ˜ ( A f ˜ ) = f 2 ( A f ) , then I f ˜ ˜ = A f f ( A f ) I f 3

8) Let f I n j n s ( E , F ) and g I n j n s ( F , E ) , so:

f g I n j n s ( F ) and g f I n j n s ( E )

2.2. Equivalence Relations

Example 5

Let f , g I n j n s ( E ) , we define the binary relation R on I n j n s ( E ) , by:

f R g A f = A f I f = I g

R is, indeed, an equivalence relation, because:

· R is reflexive, f R f , f I n j n s ( E )

· R is symmetric, f R g A f = A g g R f , f , g I n j n s ( E )

· R is transitive, f , g and h I n j n s ( E ) , f R g and g R h A f = A g and A g = A h A f = A h f R h

Example 6

Let f , g I n j n s ( E ) , we define the binary relation R on I n j n s ( E ) , by:

f , g I n j n s ( E ) : f R g n * , I f n = I g n

R is, indeed, an equivalence relation, because:

· R is reflexive, f R f , f I n j n s ( E )

· R is symmetric, f R g n * , I f n = I g n g R f , f , g I n j n s ( E )

· R is transitive, f , g and h I n j n s ( E ) , f R g and g R h n * , I f n = I g n and n * , I g n = I h n n * , I f n = I h n f R h

Example 7

f , g I n j n s ( E ) , f R g f ( A f ) = g ( A g )

Example 8

f , g I n j n s ( E ) , f R g S f = S g n f n ( A f ) = n g n ( A g )

n f n ( A f ) : as being the ƒ-semicoverage of a set E.

3. Part II

Let E be an infinite set, f I n j n s ( E ) , and f ¯ = { g I n j n s ( E ) | I g I f }

We get the following:

· f I n j n s ( E ) , n * , f n f ¯

· f , g I n j n s ( E ) : I f I g f ¯ g ¯

Let h f ¯ i.e. I f I h let x I h I f , so x I h and x I f I g , so x I h and x I g , so x I h I g so I h I g , then h g ¯ therefore f ¯ g ¯

Theorem 2

Let E be an infinite set, then there exists a non-surjective injective map ψ from E to itself, so that for any such non-surjective injective map ϕ from E to itself, we have:

I ψ I ϕ

Proof

By contradiction, let’s suppose that for all ψ non-surjective injective maps from E to itself, there exists a map ϕ ψ I n j n s ( E ) such as I ψ I ϕ ψ = so that I ϕ ψ A ψ . Additionally E is an infinite set, so E is equipotential [2] to E \ { a } , a E . Considering this bijection ƒ as a map from E to itself, then f I n j n s ( E ) , and A f = { a } , according to all of the above, there exists a map f ϕ I n j n s ( E ) such as I f ϕ A f = { a } , which contradicts the fact that f ϕ injective.

QED

Note 4

· ψ I n j n s ( E ) , so that: φ I n j n s ( E ) , ψ 1 ( I φ )

· If ψ applies to the former theorem then: ψ ¯ = I n j n s ( E ) .

Proposition 5

f I n j n s ( E ) , g I n j n s ( E ) \ { f } , so that A f A g

Proof

By contradiction, assuming that exists a map f I n j n s ( E ) , so that g I n j n s ( E ) \ { f } , A f A g = , then g I n j n s ( E ) \ { f } , I f I g = E .

Let g = f 2 , so I f I g = I f E because f I n j n s ( E ) which is contradictory.

Proposition 6

f I n j n s ( E ) , g I n j n s ( E ) \ { f } , such as A f I g

Proof

By contradiction, assuming that exists a map f I n j n s ( E ) , so that, g I n j n s ( E ) \ { f } , A f I g = then g I n j n s ( E ) \ { f } , I f A g = E . On the other hand, if g = f ˜ , I f A g = E , additionally A g = f ( A f ) I f so I f A g = I f = E which is contradictory.

Note 5

We can define another composition law T for I n j n s ( E ) so that,

f , g I n j n s ( E ) , ( f T g ) ( x ) = { g ( x ) , if x f 1 ( I g ) f ( x ) , if x f 1 ( A g )

· f T f = f , f I n j n s ( E ) then every element is idempotent under the law T.

· f , g I n j n s ( E ) , if I f I g = , then f T g = f .

· f , g I n j n s ( E ) , if I f I g , then f T g = g .

· f I n j n s ( E ) , f T f ˜ = f and f ˜ T f = f ˜ .

· Generally, f , g I n j n s ( E ) , f T g g T f .

4. Part III

4.1. Study of the Quotient Set

Let E be an infinite set, and f , g I n j n s ( E ) . We define the binary relation R on I n j n s ( E ) , by:

f R g A f and A g have the same cardinal

The binary relation R is reflexive, symmetric and transitive, so R is an equivalence relation on I n j n s ( E ) .

We define C l ( f ) = { g I n j n s ( E ) | | A g | = | A f | } the equivalence class of a map ƒ.

Note 6

· Let f I n j n s ( E ) , as A f ˜ = f ( A f ) so A f and A f ˜ have the same cardinal, because ƒ is injective then f ˜ C l ( f ) , therefore,

f I n j n s ( E ) , f ˜ C l ( f )

· Let f , g I n j n s ( E ) , assuming that the cardinals of A f and A f are finite, and thus, | A f g | = | A f f ( A g ) | = | A f | + | A g | | A f f ( A g ) | , and since A f f ( A g ) = , then: | A f g | = | A f f ( A g ) | = | A f | + | A g | .Since the composition of two (2) maps ƒ and g on I n j n s ( E ) yields a disjoint union, i.e. A f g = A f f ( A g ) , with A f f ( A g ) = ,then we can extend the sum of the cardinals even for infinite sets, such as:

| A f g | = | A f f ( A g ) | = | A f | + | A g | = sup { | A f | , | A g | }

· For all maps f , g , h and t I n j n s ( E ) so that f R g and h R t i.e. | A f | = | A g | and | A h | = | A t | . Since:

| A f h | = sup { | A f | , | A g | } = sup { | A g | , | A t | } = | A g t | , therefore,

f g R h t

· The map’s composition law is compatible under the equivalence relation R, then we can provide the quotient set ( I n j n s ( E ) / R , ) with a structure of a semigroup.

· C l ( f ) and C l ( g ) I n j n s ( E ) / R : C l ( f ) C l ( g ) = C l ( f g )

4.2. Partial Order

Let C l ( f ) , C l ( g ) I n j n s ( E ) / R , define a binary relation on I n j n s ( E ) / R by:

C l ( f ) C l ( g ) | A f | | A g |

· C l ( f ) I n j n s ( E ) / R , C l ( f ) C l ( f )

So is reflexive.

· C l ( f ) , C l ( g ) I n j n s ( E ) \ R , C l ( f ) C l ( g ) and C l ( g ) C l ( f ) | A f | | A g | and | A g | | A f | | A f | = | A g | so C l ( f ) = C l ( g )

So is asymmetric.

· C l ( f ) , C l ( g ) and C l ( h ) I n j n s ( E ) / R : C l ( f ) C l ( g ) and C l ( g ) C l ( h ) | A f | | A g | and | A g | | A h | | A f | | A h | C l ( f ) C l ( h )

So is transitive.

Therefore, the relation is a partial order on I n j n s ( E ) / R .

Note 7

· Since C l ( f ) , C l ( g ) I n j n s ( E ) / R , C l ( f ) C l ( g ) or C l ( g ) C l ( f ) then is a total partial order on I n j n s ( E ) / R .

· The partial order on the semigroup ( I n j n s ( E ) / R , ) is, indeed, compatible [3] with the equivalence class’s composition law of composition *, then:

C l ( f ) , C l ( g ) and C l ( h ) I n j n s ( E ) / R ,

If C l ( f ) C l ( g ) then C l ( f ) C l ( h ) = C l ( f h ) C l ( g h ) = C l ( g ) C l ( h ) and C l ( h ) C l ( f ) = C l ( h f ) C l ( h g ) = C l ( h ) C l ( g )

· I n j n s ( E ) / R is well-ordered, because any non-empty subset has a smallest element.

· I n j n s ( E ) / R is a lattice, because it is ordered and every pair of elements has both upper bound and lower bound [4].

· I n j n s ( E ) / R provided with the order relation has a minimal element C l ( f ) , so that | A f | = 1 , and also maximal element C l ( g ) , so that A g has the same cardinality as E.

· If E is an infinite countable set, the map φ defined by:

φ : I n j n s ( E ) * { 0 } = M , f I n j n s ( E ) , φ ( f ) = | A f | ,

where, 0 represents the cardinal of .

Considering the convention: n * , n + 0 = 0 , φ is a morphism of semigroups of ( I n j n s ( E ) , ) on ( M , + ) .

f , g I n j n s ( E ) , φ ( f g ) = | A f g | = | A f | + | A g | = φ ( f ) + φ ( g )

Complement

Let f , g I n j n s ( E ) so that | A f | < | A g | , with the assumption that A f and A g are considered finite, I f and I g are equipotential because, the map ψ defined from I f to I g by:

x I f , ψ ( x ) = ( g f 1 ) ( x ) is bijective, where f 1 is defined from I f to E, and for all x I f , we associated its inputs by ƒ, we have:

· x , y I f , ψ ( x ) = ψ ( y ) ( g f 1 ) ( x ) = ( g f 1 ) ( y ) g ( f 1 ( x ) ) = g ( f 1 ( y ) ) f 1 ( x ) = f 1 ( y ) x = y (because both g and f 1 are injectives). Therefore ψ is injective.

· On the other hand ψ ( I f ) = ( g f 1 ) ( I f ) = g ( f 1 ( I f ) ) = g ( E ) = I g so ψ is surjective.

Let φ I n j n s ( A f , A g ) so that the map θ is defined by:

θ ( x ) = { φ ( x ) , if x A f ψ ( x ) , if x I f

Belong to I n j n s ( E ) and A θ = A φ .

Note 8

· If g = f 2 so x I f , ψ ( x ) = f ( x ) so if φ = f / A f then we have, θ = f , and if φ = I d / A f so θ = f ˜

· We considered, previously, that, A f and A g are finite. That said, we can build the φ map even in case that A f and A g are infinite and that | A f | = | A g | .

5. Part IV

E Permutations Group Action

Let σ ( E ) be the permutations group of E, f I n j n s ( E ) and σ σ ( E ) , f σ I n j n s ( E ) , because, since ƒ and σ are injective then f σ is injective and σ σ ( E ) , ( f σ ) ( E ) = f ( σ ( E ) ) = f ( E ) = I f E then f σ is non-surjective.

where σ σ ( E ) , I f σ = I f then A f σ = A f .

We have,

· ˆ σ 1 , σ 2 σ ( E ) and f I n j n s ( E ) : ( f σ 1 ) σ 2 = f ( σ 1 σ 2 )

· f I n j n s ( E ) , f I d = f

Let θ : σ ( E ) × I n j n s ( E ) I n j n s ( E )

so that σ σ ( E ) , f I n j n s ( E ) , θ ( σ , f ) = f σ

Therefore, the E permutations group operates on the rightmost on I n j n s ( E ) .

Note 9

The relation R defined on I n j n s ( E ) by: f R g σ σ ( E ) such as g = f σ is an equivalence relation, that is called Intransitivity relation [5].

Proposition 7

Let be, g I n j n s ( E ) : σ σ ( E ) so that g = f σ I f = I g .

Proof

) If there exists σ σ ( E ) so that g = f σ I g = I f σ = I f .

) If I f = I g , then the map σ : E E , so that x σ ( x ) = f 1 ( g ( x ) ) is bijective, because:

· σ ( E ) = f 1 ( g ( E ) ) = E so σ is surjective.

· x , y E : σ ( x ) = σ ( y ) f 1 ( g ( x ) ) = f 1 ( g ( y ) ) g ( x ) = g ( y ) x = y so σ is injective.

· x E , f σ ( x ) = f ( σ ( x ) ) = g ( x )

Note 10

· The equivalence class (Intransitivity relation) of the element ƒ is called the orbit of ƒ, C l ( f ) = { f σ | σ σ ( E ) } = { g I n j n s ( E ) | I g = I f } .

· The stabilizer of ƒ is: Δ f = { σ σ ( E ) | f σ = f } = { I d } , i.e. the morphism associated with the said action is injective.

6. Part V

Let f , g I n j n s ( E ) .

Definition 3

ƒ and g are said to be Co-injectives, if,

I f I g and x , y E , f ( x ) = g ( y ) x = y

Let f ^ = { g I n j n s ( E ) | g is Co-injective with f }

We have f ^ because f I n j n s ( E ) , I f I f and x , y E , f ( x ) = f ( y ) x = y so ƒ is Co-injective with itself.

Therefore f I n j n s ( E ) , f f ^ , then, f I n j n s ( E ) : f ^ ϕ

Proposition 8

Let h I n j n s ( E ) , f , g I n j n s ( E ) :

f , g are Co-injectives h f and h g are Co-injectives.

Proof:

Let h I n j n s ( E ) , and f , g I n j n s ( E ) such as ƒ and g are Co-injective. I f I g , additionally h ( I f I g ) = I h f I h g (h is injective).

Let x , y E , h f ( x ) = h g ( y ) f ( x ) = g ( y ) x = y (because h is injective and f , g are Co-injectives).

Therefore h f and h g are Co-injectives.

Note 11

For all f , g I n j n s ( E ) , so that f , g are Co-injectives, so: ˆ

· f 2 and f g are Co-injectives

· z I f I g , ! x E , so that, z = f ( x ) = g ( x )

· f 1 ( I g ) = g 1 ( I f )

· If, I f = I g , then f = g ˆ

· Let h I n j n s ( E ) , if f ( I h ) g ( I h ) , then f h and g h are Co-injectives.

Proposition 9

f , g I n j n s ( E ) , if I f I g , then f , g are not Co-injective.

Proof

f , g I n j n s ( E ) , suppose that I f I g , then we have I g = I f ( I g \ I f ) . If f , g are Co-injectives, then f 1 ( I g ) = g 1 ( I f ) = E , which is contradictory because:

E = g 1 ( I f ) g 1 ( I g \ I f ) , and g 1 ( I g \ I f ) , so g 1 ( I f ) E .

Proposition 10

Let f , g I n j n s ( E ) , so that ƒ and g are Co-injectives, h I n j n s ( E ) , if g and h are Co-injectives and I f I h = I f I g then ƒ and h are Co-injectives.

Proof

I f I h , let x , y E , so that f ( x ) = h ( y ) . We have f ( x ) = h ( y ) I f I h = I f I g , then f ( x ) = g ( x ) = h ( y ) (because f , g are Co-injectives), and since g and h are Co-injectives then x = y . So x , y E , f ( x ) = h ( y ) x = y . Therefore f , h are Co-injectives.

Acknowledgements

I would like to thank my dear friends, especially Chafik Bouazzaoui for us discussing the Cantor-Bernstein theorem and Mai Mohammed mainly who for his help with translating this document.

Conflicts of Interest

The author declares no conflicts of interest regarding the publication of this paper.

References

[1] Berhuy, G. (2002) Algébre: le grand combat. Clavage et Mounet, Montrouge.
[2] Dehornoy, P. (2017) La théorie des ensembles. Clavage et Mounet, Montrouge.
[3] Gourdon, X. (1994) Les maths en tête Algèbre. Ellipses, Paris.
[4] Jacobson, N. (2009) Basic Algebra I. Dover Publication Inc., New York.
[5] Guégand, T.D.J. (1999) Algèbre en classe préparatoires MP*. Ellipses, Paris.

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