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.

Keywords

Share and Cite:

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:

· $\left(A,f\left(B\right),{I}_{f\circ g}\right)$ forms a partition of F.

· $\left(B,g\left(A\right),{I}_{g\circ f}\right)$ forms a partition of E.

where,

$A=\left\{y\in F|\forall x\in E,f\left(x\right)\ne y\right\}$ and $B=\left\{x\in E|\forall y\in F,g\left(y\right)\ne x\right\}$

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:

$A=\left\{y\in F|\forall x\in E,f\left(x\right)\ne y\right\}$ and $B=\left\{x\in E|\forall y\in F,g\left(y\right)\ne x\right\}$ are non-empty.

Also, obviously $E=B\cup {I}_{g}$ such as $B\cap {I}_{g}=\varnothing$ as follows from the deﬁnition of ${I}_{g}$.

For any such map ƒ from E to F:

${I}_{f}=f\left(E\right)=f\left(B\cup {I}_{g}\right)=f\left(B\right)\cup f\left({I}_{g}\right)=f\left(B\right)\cup {I}_{f\circ g}$

So $F=A\cup f\left(B\right)\cup {I}_{f\circ g}$

Since ƒ is an injective map then:

$f\left(B\cap {I}_{g}\right)=f\left(B\right)\cap {I}_{f\circ g}=f\left(\varnothing \right)=\varnothing$

Therefore, $\left(A,f\left(B\right),{I}_{f\circ g}\right)$ forms a partition of F.

In analogy to the map g from F to E, $\left(B,g\left(A\right),{I}_{g\circ f}\right)$ 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:

· $E=B\cup g\left(A\right)\cup \left(g\circ f\right)\left(B\right)\cup {I}_{g\circ f\circ g}$

· $F=A\cup f\left(B\right)\cup \left(f\circ g\right)\left(A\right)\cup {I}_{f\circ g\circ f}$

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

· $E={A}_{f}\cup f\left({A}_{f}\right)\cup {f}^{2}\left({A}_{f}\right)\cup {I}_{{f}^{3}}$

· $E={A}_{g}\cup g\left({A}_{g}\right)\cup {g}^{2}\left({A}_{g}\right)\cup {I}_{{g}^{3}}$

· $E={A}_{f}\cup f\left({A}_{g}\right)\cup \left(f\circ g\right)\left({A}_{f}\right)\cup {I}_{f\circ g\circ f}$

· $E={A}_{g}\cup g\left({A}_{f}\right)\cup \left(g\circ f\right)\left({A}_{g}\right)\cup {I}_{g\circ f\circ g}$

· $E={A}_{f}\cup f\left({A}_{g}\right)\cup \left(f\circ g\right)\left({A}_{g}\right)\cup {I}_{f\circ {g}^{2}}$

· $E={A}_{g}\cup g\left({A}_{f}\right)\cup \left(g\circ f\right)\left({A}_{f}\right)\cup {I}_{g\circ {f}^{2}}$

where,

${A}_{f}=\left\{y\in F|\forall x\in E,f\left(x\right)\ne y\right\}$ and ${A}_{g}=\left\{x\in E|\forall y\in F,g\left(y\right)\ne x\right\}$

Example 1

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

· $\forall n\in ℕ,f\left(n\right)=2n$

· $\forall n\in ℕ,g\left(n\right)=2n+1$

Knowing that ${A}_{f}=2ℕ+1$ and ${A}_{g}=2ℕ$ then:

· $f\left({A}_{g}\right)=4ℕ$

· $g\left({A}_{f}\right)=4ℕ+3$

· $\left(f\circ g\right)\left({A}_{f}\right)=8ℕ+6$

· $\left(g\circ f\right)\left({A}_{g}\right)=8ℕ+1$

· $\left(f\circ g\circ f\right)\left(ℕ\right)=8ℕ+2$

· $\left(g\circ f\circ g\right)\left(ℕ\right)=8ℕ+5$

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

$ℕ=\left(2ℕ+1\right)\cup \left(4ℕ\right)\cup \left(8ℕ+6\right)\cup \left(8ℕ+2\right)$

$ℕ=\left(2ℕ\right)\cup \left(4ℕ+3\right)\cup \left(8ℕ+1\right)\cup \left(8ℕ+5\right)$

2.1. Remarkable Partition

Proposition 2

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

${A}_{f}=\left\{y\in E|\forall x\in E,f\left(x\right)\ne y\right\}$

Then,

$\forall n\in ℕ,E=\left[\underset{i=0}{\overset{i=n}{\cup }}{f}^{i}\left({A}_{f}\right)\right]\cup {I}_{{f}^{n+1}}$

where, ${f}^{i}\left({A}_{f}\right)=\underset{i\text{\hspace{0.17em}}\text{times}}{\underset{︸}{f\circ f\circ \cdots \circ f\left({A}_{f}\right)}}$

Proof

By induction:

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

${I}_{f}=f\left(E\right)=\left[\underset{i=0}{\overset{i=n}{\cup }}{f}^{i+1}\left({A}_{f}\right)\right]\cup {I}_{{f}^{n+2}}=\left[\underset{i=1}{\overset{i=n+1}{\cup }}{f}^{i}\left({A}_{f}\right)\right]\cup {I}_{{f}^{n+2}}$

Then,

$E={A}_{f}\cup \left[\underset{i=1}{\overset{i=n+1}{\cup }}{f}^{i}\left({A}_{f}\right)\right]\cup {I}_{{f}^{n+2}}=\left[\underset{i=0}{\overset{i=n+1}{\cup }}{f}^{i}\left({A}_{f}\right)\right]\cup {I}_{{f}^{n+2}}$

Therefore,

$\forall n\in ℕ,E=\left[\underset{i=0}{\overset{i=n}{\cup }}{f}^{i}\left({A}_{f}\right)\right]\cup {I}_{{f}^{n+1}}$

Note 2

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

$\forall n\in ℕ,{A}_{{f}^{n+1}}=\underset{i=0}{\overset{i=n}{\cup }}{f}^{i}\left( A f \right)$

Deﬁnition 1

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

· $x\in E$ is a ﬁxed point of ƒ if, $f\left(x\right)=x$.

· $A\subset E$ is a ﬁxed point set of ƒ if, $f\left(A\right)=A$.

Proposition 3

For all non-surjective injective maps f from an inﬁnite set E to itself, $\forall n\in ℕ$, ${f}^{n}\left({A}_{f}\right)$ contains no ﬁxed points of ƒ.

Proof

By induction:

For $n=0$, let $x\in {A}_{f}$, and by deﬁnition $\forall y\in E$, $f\left(y\right)\ne x$, particularly for $y=x$ so $f\left(x\right)\ne x$ then ${A}_{f}$ contains no ﬁxed points of f. Suppose that the aforementioned property is true for n, let $x\in {f}^{n+1}\left({A}_{f}\right)=f\left({f}^{n}\left({A}_{f}\right)\right)$, then $\exists y\in {f}^{n}\left({A}_{f}\right)$ such that $x=f\left(y\right)$, we have $f\left(y\right)\ne y$ (by inductive hypothesis), since ƒ is injective then $f\left(f\left(y\right)\right)\ne f\left(y\right)$ so $f\left(x\right)\ne x$, then $\forall x\in {f}^{n+1}\left({A}_{f}\right)$, $f\left(x\right)\ne x$. Therefore, $\forall n\in ℕ$, ${f}^{n}\left({A}_{f}\right)$ 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:

$Fi{x}_{f}=\left\{x\in E|f\left(x\right)=x\right\}$ and $S{t}_{f}\left(E\right)=\left\{A\subset E|f\left(A\right)=A\right\}$

· $\forall n\in ℕ,Fi{x}_{f}\cap {f}^{n}\left({A}_{f}\right)=\varnothing$

· $\forall n\in {ℕ}^{*},Fi{x}_{f}\subseteq Fi{x}_{{f}^{n}}$

· $\forall n\in {ℕ}^{*},Fi{x}_{{f}^{n}}\subseteq Fi{x}_{{f}^{2n}}$

· $\forall p\in ℕ,\forall n\in {ℕ}^{*},{f}^{p}\left(Fi{x}_{{f}^{n}}\right)=Fi{x}_{{f}^{n}}$

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

$Fi{x}_{f}\in S{t}_{f}\left( E \right)$

Proposition 4

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

$\forall A\in S{t}_{f}\left(E\right),\forall n\in ℕ,{f}^{n}\left({A}_{f}\right)\cap A=\varnothing$

Proof

Let $A\in S{t}_{f}\left(E\right)$, by induction, For $n=0$, we have, by deﬁnition ${A}_{f}\cap {I}_{f}=\varnothing$ then ${A}_{f}\cap A={A}_{f}\cap f\left(A\right)=\varnothing$  Suppose that the said property is true for $n\in ℕ$, let $x\in {f}^{n+1}\left({A}_{f}\right)\cap A$, then, $\exists y\in {f}^{n}\left({A}_{f}\right),x=f\left(y\right)$ and $\exists {y}_{0}\in A,x=f\left({y}_{0}\right)$. Since ƒ is injective so $y={y}_{0}\in {f}^{n}\left({A}_{f}\right)\cap A$, which contradicts the inductive hypothesis. Then,

${f}^{n+1}\left({A}_{f}\right)\cap A=\varnothing$.

Therefore, $\forall A\in S{t}_{f}\left(E\right),\forall n\in ℕ,{f}^{n}\left({A}_{f}\right)\cap A=\varnothing$

Lemma 1 (Generalization)

$\forall p\in {ℕ}^{*},\forall A\in S{t}_{{f}^{p}}\left(E\right),\forall n\in ℕ,{f}^{n}\left({A}_{f}\right)\cap A=\varnothing$

Proof

By complete (strong) induction,

Let $p\in {ℕ}^{*}$, and $A\in S{t}_{{f}^{p}}\left(E\right)=\left\{A\subset E|{f}^{p}\left(A\right)=A\right\}$ : For $n=0$, we have, by deﬁnition ${A}_{f}\cap {I}_{f}=\varnothing$, since $\exists p\in {ℕ}^{*},{I}_{{f}^{p}}\subseteq {I}_{f}$, then: ${A}_{f}\cap A={A}_{f}\cap {f}^{p}\left(A\right)=\varnothing$

Suppose that the said property is true for all $i\in \left\{1,\cdots ,n\right\}$, let $x\in {f}^{n+1}\left({A}_{f}\right)\cap A$, then, $\exists y\in {A}_{f},x={f}^{n+1}\left(y\right)$ and $\exists {y}_{0}\in A,x={f}^{p}\left({y}_{0}\right)$. Since ƒ is injective then,

· ${f}^{n-p+1}\left(y\right)={y}_{0}$, if $n+1>p$, which contradicts the inductive hypothesis, because $i=n-p+1\in \left\{1,\cdots ,n\right\}$

· $y={f}^{p-n-1}\left({y}_{0}\right)$, if $p>n+1$, which contradicts ${A}_{f}\cap {I}_{{f}^{p-n-1}}=\varnothing$

· $y={y}_{0}$, if $n+1=p$, which contradicts ${A}_{f}\cap A=\varnothing$

Then,

${f}^{n+1}\left({A}_{f}\right)\cap A=\varnothing$,

Therefore,

$\forall p\in {ℕ}^{*},\forall A\in S{t}_{{f}^{p}}\left(E\right),\forall n\in ℕ,{f}^{n}\left({A}_{f}\right)\cap A=\varnothing$

QED

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

· ${S}_{{f}^{n}}\left(E\right)=\left\{A\subset E|\exists p\in \left\{1,\cdots ,n+1\right\},{f}^{p}\left(A\right)=A\right\}$

· ${S}_{{f}^{\infty }}\left(E\right)=\left\{A\subset E|\exists p\in {ℕ}^{*},{f}^{p}\left(A\right)=A\right\}$

· ${S}_{{f}^{n}}=\left\{x\in A|A\in {S}_{{f}^{n}}\left(E\right)\right\}$

· ${S}_{{f}^{\infty }}=\left\{x\in A|A\in {S}_{{f}^{\infty }}\left(E\right)\right\}$

Theorem 1

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

$E=\left[\underset{n\in ℕ}{\cup }{f}^{n}\left({A}_{f}\right)\right]\cup {S}_{{f}^{\infty }}$

Proof

Note that for $n=0$, ${S}_{{f}^{0}}\left(E\right)=S{t}_{f}\left(E\right)$, The sequence of subsets of E, ${\left({S}_{{f}^{n}}\right)}_{n\in ℕ}$ is increasing by inclusion, so it is convergent, the limit is: ${\cup }_{n\in ℕ}{S}_{{f}^{n}}={S}_{{f}^{\infty }}$

$\forall n\in ℕ,E=\left[\underset{i=0}{\overset{i=n}{\cup }}{f}^{i}\left({A}_{f}\right)\right]\cup {I}_{{f}^{n+1}}$

Otherly, according to the Lemma 1, $\forall n\in ℕ,E=\left[{\cup }_{i=0}^{i=n}{f}^{i}\left({A}_{f}\right)\right]\cap {S}_{{f}^{n}}=\varnothing$

Let’s define the following, $\forall n\in ℕ,\stackrel{^}{{I}_{{f}^{n+1}}}={I}_{{f}^{n+1}}\{S}_{{f}^{n}}$ the sequence of subsets of ${\left(\stackrel{^}{{I}_{{f}^{n+1}}}\right)}_{n\in ℕ}$ is decreasing by inclusion then, it is convergent , the limit is:

$\underset{n\in ℕ}{\cap }\stackrel{^}{{I}_{{f}^{n+1}}}=\varnothing$

Because, let $x\in {\cap }_{n\in ℕ}\stackrel{^}{{I}_{{f}^{n+1}}}$, then:

$x\in \underset{n\in ℕ}{\cap }{I}_{{f}^{n+1}}\{S}_{{f}^{n}}⇔\forall n\in ℕ,x\in {I}_{{f}^{n+1}}\{S}_{{f}^{n}}⇔\forall n\in ℕ,x\in {I}_{{f}^{n+1}}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}x\in \stackrel{¯}{{S}_{{f}^{n}}}$

$⇔\forall n\in ℕ,x\in {I}_{{f}^{n+1}}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\forall n\in ℕ,x\in \stackrel{¯}{{S}_{{f}^{n}}}⇔x\in \underset{n\in ℕ}{\cap }{I}_{{f}^{n+1}}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}x\in \underset{n\in ℕ}{\cap }\stackrel{¯}{{S}_{{f}^{n}}}$

$⇔x\in \underset{n\in ℕ}{\cap }{I}_{{f}^{n+1}}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}x\in \stackrel{¯}{\underset{n\in ℕ}{\cup }{S}_{{f}^{n}}}⇔x\in \underset{n\in ℕ}{\cap }{I}_{{f}^{n+1}}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}x\in \stackrel{¯}{{S}_{{f}^{\infty }}}$

$⇔x\in \left[\underset{n\in ℕ}{\cap }{I}_{{f}^{n+1}}\right]\{S}_{{f}^{\infty }}$

On the other hand, the sequence of subsets of E, ${\left({I}_{{f}^{n+1}}\right)}_{n\in ℕ}$ is strictly decreasing by inclusion, then it is convergent and the limit is $H={\cap }_{n\in ℕ}{I}_{{f}^{n+1}}$, and since ƒ is injective, so $f\left(H\right)=H$ then ƒ is bijective from H to itself, and $H\in S{t}_{f}\left(E\right)\subset {S}_{{f}^{\infty }}\left(E\right)$.

$\forall n\in ℕ,E=\left[\underset{i=0}{\overset{i=n}{\cup }}{f}^{i}\left({A}_{f}\right)\right]\cup \stackrel{^}{{I}_{{f}^{n+1}}}\cup {S}_{{f}^{n}}$

Therefore,

$E=\left[\underset{n\in ℕ}{\cup }{f}^{n}\left({A}_{f}\right)\right]\cup {S}_{{f}^{\infty }}$

N.B. If ${S}_{{f}^{\infty }}=\varnothing$, then we get:

$E=\underset{n\in ℕ}{\cup }{f}^{n}\left( A f \right)$

QED

Example 2

$ℝ$ is the set of real numbers, and $P\left(ℝ\right)$ is the set of subsets of $ℝ$.

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

$f\left(x\right)=\left\{\begin{array}{l}x+1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}x\ge 0\\ x,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}x<0\end{array}$

Therefore, ƒ is injective non-surjective from $ℝ$ to itself, because ƒ is injective and ${A}_{f}=\left[0,1\left[$.

where,

$\forall n\in {ℕ}^{*},{f}^{n}\left(x\right)=\left\{\begin{array}{l}x+n,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{if}\text{\hspace{0.17em}}x\ge 0\\ x,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}x<0\end{array}$

We have, $f\left({A}_{f}\right)=\left[f\left(0\right),f\left(1\right)\left[=\left[1,2\left[\subset {ℝ}_{+}$ since ƒ is increasing Therefore,

$\forall n\in {ℕ}^{*},{f}^{n}\left({A}_{f}\right)=\left[n,n+1\left[$

· $\forall n\in ℕ,{S}_{{f}^{n}}={S}_{{f}^{\infty }}={ℝ}_{-}^{*}$

· $\forall n\in ℕ,{S}_{{f}^{n}}\left(E\right)={S}_{{f}^{\infty }}\left(E\right)=P\left({ℝ}_{-}^{*}\right)$

According to the theorem 1, we can write:

$ℝ={ℝ}_{-}^{*}\cup \left[\underset{n\in ℕ}{\cup }\left[n,n+1\left[\right]$

Example 3

Remark

If h is an aﬃne map from $ℝ$ to $ℝ$, so that $h\left(x\right)=ax+b/a,b\in ℝ$ and $a\ne 0$, then:

$\forall n\in {ℕ}^{*},\forall x\in ℝ,{h}^{n}\left(x\right)={a}^{n}x+b\left(1+a+\cdots +{a}^{n-1}\right)$

If $a\ne 1$, then:

$\forall n\in {ℕ}^{*},\forall x\in ℝ,{h}^{n}\left(x\right)={a}^{n}x+b\left(\frac{1-{a}^{n}}{1-a}\right)$

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

$\forall x\in \left[0,4\right],f\left(x\right)=\left\{\begin{array}{l}-x+1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}x\in \left[0,1\right]\\ x+1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}x\in \right]1,2\right]\\ \frac{1}{2}x+2,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}x\in \left[2,4\right]\end{array}$

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

· ${A}_{f}=\right]1,2\right]$

· $Fi{x}_{f}=\left\{4\right\}$

· The set $A=\left[0,1\right]\subset E$, fulﬁlls the condition $f\left(A\right)=A$

· We have ${A}_{f}=\right]1,2\right]$, $f\left({A}_{f}\right)=f\left(\right]1,2\right]\right)=\right]2,3\right]\subset \left[2,4\right]$, $f\left(\right]2,3\right]\right)=\right]3,\frac{7}{2}\right]$, $\cdots$

· ${S}_{{f}^{\infty }}=\left\{4\right\}\cup \left[0,1\right]$

The restriction of ƒ to $\left[2,4\right]$ is an aﬃne function such as $a=\frac{1}{2}$, and $b=2$, then:

$\forall n\in {ℕ}^{*},{f}^{n}\left(x\right)=\frac{1}{{2}^{n}}x+2\left(1+\frac{1}{2}+\cdots +\frac{1}{{2}^{n-1}}\right)$

We have: ${f}^{n}\left(2\right)=4-\frac{1}{{2}^{n-1}}$, and ${f}^{n}\left(3\right)=4-\frac{1}{{2}^{n}}$,

According to Theorem 1:

$\left[0,4\right]=\left\{\right]1,2\right]{\cup }_{n\in ℕ}\right]4-\frac{1}{{2}^{n-1}},4-\frac{1}{{2}^{n}}\right]\right\}\cup \left\{4\right\}\cup \left[0,1\right]$

Example 4

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

$\forall x\in \left[0,3\right],f\left(x\right)=\left\{\begin{array}{l}x+2,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{ }\text{ }\text{if}\text{\hspace{0.17em}}x\in \left[0,1\right]\\ \frac{1}{2}x+1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}x\in \right]1,2\left[\\ x-2,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}x\in \left[2,3\right]\end{array}$

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

· ${A}_{f}=\right]1,\frac{3}{2}\right]$

· $Fi{x}_{f}=\varnothing$

· The set $A=\left[0,1\right]\subset E$, fulﬁlls the condition $f\left(A\right)\ne A$, and ${f}^{2}\left(A\right)=A$

· The set $B=\left[2,3\right]\subset E$, fulﬁlls the condition $f\left(B\right)\ne B$, and ${f}^{2}\left(B\right)=B$

· ${S}_{{f}^{\infty }}=\left[0,1\right]\cup \left[2,3\right]$

We have,

· $\forall x\in \right]1,2\left[,\forall n\in {ℕ}^{*},{f}^{n}\left(x\right)=\frac{1}{{2}^{n}}x+2\left(1-\frac{1}{{2}^{n}}\right)$

· ${\mathrm{lim}}_{x\to {1}^{-}}{f}^{n}\left(x\right)=2-\frac{1}{{2}^{n}}$ and ${f}^{n}\left(\frac{3}{2}\right)=2-\frac{1}{{2}^{n+1}}$

Therefore, according to Theorem 1:

$\left[0,3\right]=\left\{{\cup }_{n\in ℕ}\right]2-\frac{1}{{2}^{n}},2-\frac{1}{{2}^{n+1}}\right]\right\}\cup \left[0,1\right]\cup \left[2,3\right]$

Corollary 1

Let $f,g$ be non-surjective injective maps from an inﬁnite set E to itself such as ${S}_{{f}^{\infty }}=\stackrel{¯}{{S}_{{g}^{\infty }}}$, then:

$E=\underset{n\in ℕ}{\cup }{B}_{n}$

where, $\forall n\in ℕ,{B}_{n}={f}^{n}\left({A}_{f}\right)\cup {g}^{n}\left( A g \right)$

Therefore, ${\left({B}_{n}\right)}_{n\in ℕ}$ forms a partition of E.

Deﬁnition 2

· Let E be an inﬁnite set, we write: $Injns\left(E\right)$ as being the set of non-surjective injective maps from E to itself.

· $Injns\left(E,F\right)$ 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 $|E|\le |F|$ ( $|E|$ is the cardinal of E).

Properties

1) $\left(Injns\left(E\right),\circ \right)$ is a semigroup, because the composite of 2 (two) injective maps ƒ and g is injective and ${I}_{f\circ g}\subset {I}_{f}$, so $\forall f,g\in Injns\left(E\right)$, $f\circ g\in Injns\left(E\right)$.

2) $\forall f,g\in Injns\left(E\right)$, ${A}_{f\circ g}={A}_{f}\cup f\left({A}_{g}\right)$, $\left({A}_{f},f\left({A}_{g}\right),{I}_{f\circ g}\right)$ is, indeed, a partition of E and $E={A}_{f\circ g}\cup {I}_{f\circ g}$, because $f\circ g\in Injns\left(E\right)$.

3) There is no idempotent element for the law of composition in $Injns\left(E\right)$ and if such an element exists, then ${f}^{2}=f$ and since ƒ is injective, then $f={I}_{d}$ which is contradictory, because, $f\in Injns\left(E\right)$.

4)Let $f\in Injns\left(E\right)$, and assuming that ƒ is a map from E to itself such as: $\stackrel{˜}{f}\left(x\right)=\left\{\begin{array}{l}x,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}x\in {A}_{f}\\ f\left(x\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}x\in {I}_{f}\end{array}$

5) We have, $\forall f\in Injns\left(E\right),\stackrel{˜}{f}\in Injns\left(E\right)$ knowing that ${I}_{\stackrel{˜}{f}}={A}_{f}\cup {I}_{{f}^{2}}$ and ${A}_{\stackrel{˜}{f}}=f\left( A f \right)$

6) Let $f\in Injns\left(E\right)$, we have, $\stackrel{˜}{f}\left({A}_{\stackrel{˜}{f}}\right)=\stackrel{˜}{f}\left(f\left({A}_{f}\right)\right)={f}^{2}\left({A}_{f}\right)$, and

${I}_{{\stackrel{˜}{f}}^{2}}=\stackrel{˜}{f}\left({I}_{\stackrel{˜}{f}}\right)=\stackrel{˜}{f}\left({A}_{f}\cup {I}_{{f}^{2}}\right)=\stackrel{˜}{f}\left({A}_{f}\right)\cup \stackrel{˜}{f}\left({I}_{{f}^{2}}\right)={A}_{f}\cup {I}_{{f}^{3}}$

Then,

$E={A}_{\stackrel{˜}{f}}\cup \stackrel{˜}{f}\left({A}_{\stackrel{˜}{f}}\right)\cup {I}_{{\stackrel{˜}{f}}^{2}}={A}_{f}\cup f\left({A}_{f}\right)\cup {f}^{2}\left({A}_{f}\right)\cup {I}_{{f}^{3}}$

Therefore, $\stackrel{˜}{f}$ allow us to reduce order of iteration.

7) $\forall f\in Injns\left(E\right)$,

· ${A}_{\stackrel{˜}{f}\circ f}={A}_{\stackrel{˜}{f}}\cup \stackrel{˜}{f}\left({A}_{f}\right)={A}_{f}\cup f\left({A}_{f}\right)={A}_{{f}^{2}}$, then ${I}_{\stackrel{˜}{f}\circ f}={I}_{{f}^{2}}$

· ${A}_{f\circ \stackrel{˜}{f}}={A}_{f}\cup f\left({A}_{\stackrel{˜}{f}}\right)={A}_{f}\cup {f}^{2}\left({A}_{f}\right)$,then, ${I}_{f\circ \stackrel{˜}{f}}=f\left({A}_{f}\right)\cup {I}_{{f}^{3}}$

· ${A}_{\stackrel{˜}{\stackrel{˜}{f}}}=\stackrel{˜}{f}\left({A}_{\stackrel{˜}{f}}\right)={f}^{2}\left({A}_{f}\right)$, then ${I}_{\stackrel{˜}{\stackrel{˜}{f}}}={A}_{f}\cup f\left({A}_{f}\right)\cup {I}_{{f}^{3}}$

8) Let $f\in Injns\left(E,F\right)$ and $g\in Injns\left(F,E\right)$, so:

$f\circ g\in Injns\left(F\right)$ and $g\circ f\in Injns\left( E \right)$

2.2. Equivalence Relations

Example 5

Let $f,g\in Injns\left(E\right)$, we define the binary relation R on $Injns\left(E\right)$, by:

$fRg⇔{A}_{f}={A}_{f}⇔{I}_{f}={I}_{g}$

R is, indeed, an equivalence relation, because:

· R is reﬂexive, $fRf,\forall f\in Injns\left( E \right)$

· R is symmetric, $fRg⇔{A}_{f}={A}_{g}⇔gRf,\forall f,g\in Injns\left( E \right)$

· R is transitive, $\forall f,g$ and $h\in Injns\left(E\right)$, $fRg$ and $gRh⇔{A}_{f}={A}_{g}$ and ${A}_{g}={A}_{h}⇔{A}_{f}={A}_{h}⇔fRh$

Example 6

Let $f,g\in Injns\left(E\right)$, we deﬁne the binary relation R on $Injns\left(E\right)$, by:

$\forall f,g\in Injns\left(E\right):fRg⇔\forall n\in {ℕ}^{*},{I}_{{f}^{n}}={I}_{{g}^{n}}$

R is, indeed, an equivalence relation, because:

· R is reﬂexive, $fRf,\forall f\in Injns\left( E \right)$

· R is symmetric, $fRg⇔\forall n\in {ℕ}^{*},{I}_{{f}^{n}}={I}_{{g}^{n}}⇔gRf,\forall f,g\in Injns\left( E \right)$

· R is transitive, $f,g$ and $h\in Injns\left(E\right)$, $fRg$ and $gRh⇔\forall n\in {ℕ}^{*},{I}_{{f}^{n}}={I}_{{g}^{n}}$ and $\forall n\in {ℕ}^{*},{I}_{{g}^{n}}={I}_{{h}^{n}}⇔\forall n\in {ℕ}^{*},{I}_{{f}^{n}}={I}_{{h}^{n}}⇔fRh$

Example 7

$\forall f,g\in Injns\left(E\right),fRg⇔f\left({A}_{f}\right)=g\left( A g \right)$

Example 8

$\forall f,g\in Injns\left(E\right),fRg⇔{S}_{{f}^{\infty }}={S}_{{g}^{\infty }}⇔\underset{n\in ℕ}{\cup }{f}^{n}\left({A}_{f}\right)=\underset{n\in ℕ}{\cup }{g}^{n}\left( A g \right)$

$\underset{n\in ℕ}{\cup }{f}^{n}\left({A}_{f}\right)$ : as being the ƒ-semicoverage of a set E.

3. Part II

Let E be an infinite set, $f\in Injns\left(E\right)$, and $\stackrel{¯}{f}=\left\{g\in Injns\left(E\right)|{I}_{g}\cap {I}_{f}\ne \varnothing \right\}$

We get the following:

· $\forall f\in Injns\left(E\right),\forall n\in {ℕ}^{*},{f}^{n}\in \stackrel{¯}{f}$

· $\forall f,g\in Injns\left(E\right):{I}_{f}\subset {I}_{g}⇒\stackrel{¯}{f}\subset \stackrel{¯}{g}$

Let $h\in \stackrel{¯}{f}$ i.e. ${I}_{f}\cap {I}_{h}\ne \varnothing$ let $x\in {I}_{h}\cap {I}_{f}$, so $x\in {I}_{h}$ and $x\in {I}_{f}\subset {I}_{g}$, so $x\in {I}_{h}$ and $x\in {I}_{g}$, so $x\in {I}_{h}\cap {I}_{g}$ so ${I}_{h}\cap {I}_{g}\ne \varnothing$, then $h\in \stackrel{¯}{g}$ therefore $\stackrel{¯}{f}\subset \stackrel{¯}{g}$

Theorem 2

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

${I}_{\psi }\cap {I}_{\varphi }\ne \varnothing$

Proof

By contradiction, let’s suppose that for all $\psi$ non-surjective injective maps from E to itself, there exists a map ${\varphi }_{\psi }\in Injns\left(E\right)$ such as ${I}_{\psi }\cap {I}_{{\varphi }_{\psi }}=\varnothing$ so that ${I}_{{\varphi }_{\psi }}\subseteq {A}_{\psi }$. Additionally E is an inﬁnite set, so E is equipotential  to $E\\left\{a\right\},\forall a\in E$. Considering this bijection ƒ as a map from E to itself, then $f\in Injns\left(E\right)$, and ${A}_{f}=\left\{a\right\}$, according to all of the above, there exists a map ${f}_{\varphi }\in Injns\left(E\right)$ such as ${I}_{{f}_{\varphi }}\subseteq Af=\left\{a\right\}$, which contradicts the fact that ${f}_{\varphi }$ injective.

QED

Note 4

· $\exists \psi \in Injns\left(E\right)$, so that: $\forall \phi \in Injns\left(E\right)$, ${\psi }^{-1}\left({I}_{\phi }\right)\ne \varnothing$

· If $\psi$ applies to the former theorem then: $\stackrel{¯}{\psi }=Injns\left(E\right)$.

Proposition 5

$\forall f\in Injns\left(E\right)$, $\exists g\in Injns\left(E\right)\\left\{f\right\}$, so that ${A}_{f}\cap {A}_{g}\ne \varnothing$

Proof

By contradiction, assuming that exists a map $f\in Injns\left(E\right)$, so that $\forall g\in Injns\left(E\right)\\left\{f\right\}$, ${A}_{f}\cap {A}_{g}=\varnothing$, then $\forall g\in Injns\left(E\right)\\left\{f\right\}$, ${I}_{f}\cup {I}_{g}=E$.

Let $g={f}^{2}$, so ${I}_{f}\cup {I}_{g}={I}_{f}⊊E$ because $f\in Injns\left(E\right)$ which is contradictory.

Proposition 6

$\forall f\in Injns\left(E\right)$, $\exists g\in Injns\left(E\right)\\left\{f\right\}$, such as ${A}_{f}\cap {I}_{g}\ne \varnothing$

Proof

By contradiction, assuming that exists a map $f\in Injns\left(E\right)$, so that, $\forall g\in Injns\left(E\right)\\left\{f\right\}$, ${A}_{f}\cap {I}_{g}=\varnothing$ then $\forall g\in Injns\left(E\right)\\left\{f\right\},{I}_{f}\cup {A}_{g}=E$. On the other hand, if $g=\stackrel{˜}{f},{I}_{f}\cup {A}_{g}=E$, additionally ${A}_{g}=f\left({A}_{f}\right)\subset {I}_{f}$ so ${I}_{f}\cup {A}_{g}={I}_{f}=E$ which is contradictory.

Note 5

We can deﬁne another composition law T for $Injns\left(E\right)$ so that,

$\forall f,g\in Injns\left(E\right),\left(fTg\right)\left(x\right)=\left\{\begin{array}{l}g\left(x\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}x\in {f}^{-1}\left({I}_{g}\right)\\ f\left(x\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}x\in {f}^{-1}\left( A g \right)\end{array}$

· $fTf=f,\forall f\in Injns\left(E\right)$ then every element is idempotent under the law T.

· $\forall f,g\in Injns\left(E\right)$, if ${I}_{f}\cap {I}_{g}=\varnothing$, then $fTg=f$.

· $\forall f,g\in Injns\left(E\right)$, if ${I}_{f}\subseteq {I}_{g}$, then $fTg=g$.

· $\forall f\in Injns\left(E\right)$, $fT\stackrel{˜}{f}=f$ and $\stackrel{˜}{f}Tf=\stackrel{˜}{f}$.

· Generally, $\forall f,g\in Injns\left(E\right)$, $fTg\ne gTf$.

4. Part III

4.1. Study of the Quotient Set

Let E be an inﬁnite set, and $f,g\in Injns\left(E\right)$. We define the binary relation R on $Injns\left(E\right)$, by:

$fRg⇔{A}_{f}$ and ${A}_{g}$ have the same cardinal

The binary relation R is reﬂexive, symmetric and transitive, so R is an equivalence relation on $Injns\left(E\right)$.

We deﬁne $Cl\left(f\right)=\left\{g\in Injns\left(E\right)||{A}_{g}|=|{A}_{f}|\right\}$ the equivalence class of a map ƒ.

Note 6

· Let $f\in Injns\left(E\right)$, as ${A}_{\stackrel{˜}{f}}=f\left({A}_{f}\right)$ so ${A}_{f}$ and ${A}_{\stackrel{˜}{f}}$ have the same cardinal, because ƒ is injective then $\stackrel{˜}{f}\in Cl\left(f\right)$, therefore,

$\forall f\in Injns\left(E\right),\stackrel{˜}{f}\in Cl\left( f \right)$

· Let $f,g\in Injns\left(E\right)$, assuming that the cardinals of ${A}_{f}$ and ${A}_{f}$ are ﬁnite, and thus, $|{A}_{f\circ g}|=|{A}_{f}\cup f\left({A}_{g}\right)|=|{A}_{f}|+|{A}_{g}|-|{A}_{f}\cap f\left({A}_{g}\right)|$, and since ${A}_{f}\cap f\left({A}_{g}\right)=\varnothing$, then: $|{A}_{f\circ g}|=|{A}_{f}\cup f\left({A}_{g}\right)|=|{A}_{f}|+|{A}_{g}|$.Since the composition of two (2) maps ƒ and g on $Injns\left(E\right)$ yields a disjoint union, i.e. ${A}_{f\circ g}={A}_{f}\cup f\left({A}_{g}\right)$, with ${A}_{f}\cap f\left({A}_{g}\right)=\varnothing$,then we can extend the sum of the cardinals even for infinite sets, such as:

$|{A}_{f\circ g}|=|{A}_{f}\cup f\left({A}_{g}\right)|=|{A}_{f}|+|{A}_{g}|=\mathrm{sup}\left\{|{A}_{f}|,|{A}_{g}|\right\}$

· For all maps $f,g,h$ and $t\in Injns\left(E\right)$ so that $fRg$ and $hRt$ i.e. $|{A}_{f}|=|{A}_{g}|$ and $|{A}_{h}|=|{A}_{t}|$. Since:

$|{A}_{f\circ h}|=\mathrm{sup}\left\{|{A}_{f}|,|{A}_{g}|\right\}=\mathrm{sup}\left\{|{A}_{g}|,|{A}_{t}|\right\}=|{A}_{g\circ t}|$, therefore,

$f\circ gRh\circ t$

· The map’s composition law is compatible under the equivalence relation R, then we can provide the quotient set $\left(Injns\left(E\right)/R,\ast \right)$ with a structure of a semigroup.

· $\forall Cl\left(f\right)$ and $Cl\left(g\right)\in Injns\left(E\right)/R$ : $Cl\left(f\right)\ast Cl\left(g\right)=Cl\left(f\circ g\right)$

4.2. Partial Order

Let $Cl\left(f\right),Cl\left(g\right)\in Injns\left(E\right)/R$, deﬁne a binary relation on $Injns\left(E\right)/R$ by:

$Cl\left(f\right)\le Cl\left(g\right)⇔|{A}_{f}|\le |{A}_{g}|$

· $\forall Cl\left(f\right)\in Injns\left(E\right)/R,Cl\left(f\right)\le Cl\left( f \right)$

So $\le$ is reﬂexive.

· $\forall Cl\left(f\right),Cl\left(g\right)\in Injns\left(E\right)\R$, $Cl\left(f\right)\le Cl\left(g\right)$ and $Cl\left(g\right)\le Cl\left(f\right)$ $⇔$ $|{A}_{f}|\le |{A}_{g}|$ and $|{A}_{g}|\le |{A}_{f}|$ $⇔$ $|{A}_{f}|=|{A}_{g}|$ so $Cl\left(f\right)=Cl\left( g \right)$

So $\le$ is asymmetric.

· $\forall Cl\left(f\right),Cl\left(g\right)$ and $Cl\left(h\right)\in Injns\left(E\right)/R$ : $Cl\left(f\right)\le Cl\left(g\right)$ and $Cl\left(g\right)\le Cl\left(h\right)$ $⇔$ $|{A}_{f}|\le |{A}_{g}|$ and $|{A}_{g}|\le |{A}_{h}|$ $⇒$ $|{A}_{f}|\le |{A}_{h}|$ $⇒$ $Cl\left(f\right)\le Cl\left( h \right)$

So $\le$ is transitive.

Therefore, the relation $\le$ is a partial order on $Injns\left(E\right)/R$.

Note 7

· Since $\forall Cl\left(f\right),Cl\left(g\right)\in Injns\left(E\right)/R$, $Cl\left(f\right)\le Cl\left(g\right)$ or $Cl\left(g\right)\le Cl\left(f\right)$ then $\le$ is a total partial order on $Injns\left(E\right)/R$.

· The partial order on the semigroup $\left(Injns\left(E\right)/R,\ast \right)$ is, indeed, compatible  with the equivalence class’s composition law of composition *, then:

$\forall Cl\left(f\right),Cl\left(g\right)$ and $Cl\left(h\right)\in Injns\left(E\right)/R$,

If $Cl\left(f\right)\le Cl\left(g\right)$ then $Cl\left(f\right)\ast Cl\left(h\right)=Cl\left(f\circ h\right)\le Cl\left(g\circ h\right)=Cl\left(g\right)\ast Cl\left(h\right)$ and $Cl\left(h\right)\ast Cl\left(f\right)=Cl\left(h\circ f\right)\le Cl\left(h\circ g\right)=Cl\left(h\right)\ast Cl\left( g \right)$

· $Injns\left(E\right)/R$ is well-ordered, because any non-empty subset has a smallest element.

· $Injns\left(E\right)/R$ is a lattice, because it is ordered and every pair of elements has both upper bound and lower bound .

· $Injns\left(E\right)/R$ provided with the order relation has a minimal element $Cl\left(f\right)$, so that $|{A}_{f}|=1$, and also maximal element $Cl\left(g\right)$, so that ${A}_{g}$ has the same cardinality as E.

· If E is an inﬁnite countable set, the map $\phi$ deﬁned by:

$\phi :Injns\left(E\right)\to {ℕ}^{*}\cup \left\{\aleph 0\right\}=M,\forall f\in Injns\left(E\right),\phi \left(f\right)=|{A}_{f}|$,

where, $\aleph 0$ represents the cardinal of $ℕ$.

Considering the convention: $\forall n\in {ℕ}^{*}$, $n+\aleph 0=\aleph 0$, $\phi$ is a morphism of semigroups of $\left(Injns\left(E\right),\circ \right)$ on $\left(M,+\right)$.

$\forall f,g\in Injns\left(E\right),\phi \left(f\circ g\right)=|{A}_{f\circ g}|=|{A}_{f}|+|{A}_{g}|=\phi \left(f\right)+\phi \left( g \right)$

Complement

Let $f,g\in Injns\left(E\right)$ 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 $\psi$ defined from ${I}_{f}$ to ${I}_{g}$ by:

$\forall x\in {I}_{f}$, $\psi \left(x\right)=\left(g\circ {f}^{-1}\right)\left(x\right)$ is bijective, where ${f}^{-1}$ is deﬁned from ${I}_{f}$ to E, and for all $x\in {I}_{f}$, we associated its inputs by ƒ, we have:

· $\forall x,y\in {I}_{f}$, $\psi \left(x\right)=\psi \left(y\right)$ $⇒$ $\left(g\circ {f}^{-1}\right)\left(x\right)=\left(g\circ {f}^{-1}\right)\left(y\right)$ $⇒$ $g\left({f}^{-1}\left(x\right)\right)=g\left({f}^{-1}\left(y\right)\right)$ $⇒$ ${f}^{-1}\left(x\right)={f}^{-1}\left(y\right)⇒x=y$ (because both g and ${f}^{-1}$ are injectives). Therefore $\psi$ is injective.

· On the other hand $\psi \left({I}_{f}\right)=\left(g\circ {f}^{-1}\right)\left({I}_{f}\right)=g\left({f}^{-1}\left({I}_{f}\right)\right)=g\left(E\right)={I}_{g}$ so $\psi$ is surjective.

Let $\phi \in Injns\left({A}_{f},{A}_{g}\right)$ so that the map $\theta$ is defined by:

$\theta \left(x\right)=\left\{\begin{array}{l}\phi \left(x\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}x\in {A}_{f}\\ \psi \left(x\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}x\in {I}_{f}\end{array}$

Belong to $Injns\left(E\right)$ and ${A}_{\theta }={A}_{\phi }$.

Note 8

· If $g={f}^{2}$ so $\forall x\in {I}_{f},\psi \left(x\right)=f\left(x\right)$ so if $\phi =f/{A}_{f}$ then we have, $\theta =f$, and if $\phi ={I}_{d}/{A}_{f}$ so $\theta =\stackrel{˜}{f}$

· We considered, previously, that, ${A}_{f}$ and ${A}_{g}$ are finite. That said, we can build the $\phi$ 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 $\sigma \left(E\right)$ be the permutations group of E, $f\in Injns\left(E\right)$ and $\sigma \in \sigma \left(E\right)$, $f\circ \sigma \in Injns\left(E\right)$, because, since ƒ and $\sigma$ are injective then $f\circ \sigma$ is injective and $\forall \sigma \in \sigma \left(E\right)$, $\left(f\circ \sigma \right)\left(E\right)=f\left(\sigma \left(E\right)\right)=f\left(E\right)={I}_{f}⊊E$ then $f\circ \sigma$ is non-surjective.

where $\forall \sigma \in \sigma \left(E\right),{I}_{f\circ \sigma }={I}_{f}$ then ${A}_{f\circ \sigma }={A}_{f}$.

We have,

·  $\forall {\sigma }_{1},{\sigma }_{2}\in \sigma \left(E\right)$ and $f\in Injns\left(E\right)$ : $\left(f\circ {\sigma }_{1}\right)\circ {\sigma }_{2}=f\circ \left({\sigma }_{1}\circ {\sigma }_{2}\right)$

· $\forall f\in Injns\left(E\right)$, $f\circ {I}_{d}=f$

Let $\theta :\sigma \left(E\right)×Injns\left(E\right)\to Injns\left( E \right)$

so that $\forall \sigma \in \sigma \left(E\right)$, $\forall f\in Injns\left(E\right)$, $\theta \left(\sigma ,f\right)=f\circ \sigma$

Therefore, the E permutations group operates on the rightmost on $Injns\left(E\right)$.

Note 9

The relation R deﬁned on $Injns\left(E\right)$ by: $fRg⇔\exists \sigma \in \sigma \left(E\right)$ such as $g=f\circ \sigma$ is an equivalence relation, that is called Intransitivity relation .

Proposition 7

Let be, $g\in Injns\left(E\right)$ : $\exists \sigma \in \sigma \left(E\right)$ so that $g=f\circ \sigma ⇔{I}_{f}={I}_{g}$.

Proof

$⇒$ ) If there exists $\sigma \in \sigma \left(E\right)$ so that $g=f\circ \sigma ⇒{I}_{g}={I}_{f\circ \sigma }={I}_{f}$.

$⇐$ ) If ${I}_{f}={I}_{g}$, then the map $\sigma :E\to E$, so that $x↦\sigma \left(x\right)={f}^{-1}\left(g\left(x\right)\right)$ is bijective, because:

· $\sigma \left(E\right)={f}^{-1}\left(g\left(E\right)\right)=E$ so $\sigma$ is surjective.

· $\forall x,y\in E$ : $\sigma \left(x\right)=\sigma \left(y\right)$ $⇒$ ${f}^{-1}\left(g\left(x\right)\right)={f}^{-1}\left(g\left(y\right)\right)$ $⇒$ $g\left(x\right)=g\left(y\right)$ $⇒$ $x=y$ so $\sigma$ is injective.

· $\forall x\in E$, $f\circ \sigma \left(x\right)=f\left(\sigma \left(x\right)\right)=g\left( x \right)$

Note 10

· The equivalence class (Intransitivity relation) of the element ƒ is called the orbit of ƒ, $Cl\left(f\right)=\left\{f\circ \sigma |\sigma \in \sigma \left(E\right)\right\}=\left\{g\in Injns\left(E\right)|{I}_{g}={I}_{f}\right\}$.

· The stabilizer of ƒ is: $\Delta f=\left\{\sigma \in \sigma \left(E\right)|f\circ \sigma =f\right\}=\left\{{I}_{d}\right\}$, i.e. the morphism associated with the said action is injective.

6. Part V

Let $f,g\in Injns\left(E\right)$.

Deﬁnition 3

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

${I}_{f}\cap {I}_{g}\ne \varnothing$ and $\forall x,y\in E,f\left(x\right)=g\left(y\right)⇒x=y$

Let $\stackrel{^}{f}=\left\{g\in Injns\left(E\right)|g\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{Co-injective}\text{\hspace{0.17em}}\text{with}\text{\hspace{0.17em}}f\right\}$

We have $\stackrel{^}{f}\ne \varnothing$ because $\forall f\in Injns\left(E\right)$, ${I}_{f}\cap {I}_{f}\ne \varnothing$ and $\forall x,y\in E$, $f\left(x\right)=f\left(y\right)⇒x=y$ so ƒ is Co-injective with itself.

Therefore $\forall f\in Injns\left(E\right)$, $f\in \stackrel{^}{f}$, then, $\forall f\in Injns\left(E\right)$ : $\stackrel{^}{f}\ne \varphi$

Proposition 8

Let $h\in Injns\left(E\right)$, $\forall f,g\in Injns\left(E\right)$ :

$f,g$ are Co-injectives $⇒$ $h\circ f$ and $h\circ g$ are Co-injectives.

Proof:

Let $h\in Injns\left(E\right)$, and $f,g\in Injns\left(E\right)$ such as ƒ and g are Co-injective. ${I}_{f}\cap {I}_{g}\ne \varnothing$, additionally $h\left({I}_{f}\cap {I}_{g}\right)={I}_{h\circ f}\cap {I}_{h\circ g}\ne \varnothing$ (h is injective).

Let $x,y\in E$, $h\circ f\left(x\right)=h\circ g\left(y\right)⇒f\left(x\right)=g\left(y\right)⇒x=y$ (because h is injective and $f,g$ are Co-injectives).

Therefore $h\circ f$ and $h\circ g$ are Co-injectives.

Note 11

For all $f,g\in Injns\left(E\right)$, so that $f,g$ are Co-injectives, so: 

· ${f}^{2}$ and $f\circ g$ are Co-injectives

· $\forall z\in {I}_{f}\cap {I}_{g}$, $\exists !x\in E$, so that, $z=f\left(x\right)=g\left( x \right)$

· ${f}^{-1}\left({I}_{g}\right)={g}^{-1}\left( I f \right)$

· If, ${I}_{f}={I}_{g}$, then $f=g$ 

· Let $h\in Injns\left(E\right)$, if $f\left({I}_{h}\right)\cap g\left({I}_{h}\right)\ne \varnothing$, then $f\circ h$ and $g\circ h$ are Co-injectives.

Proposition 9

$\forall f,g\in Injns\left(E\right)$, if ${I}_{f}⊊{I}_{g}$, then $f,g$ are not Co-injective.

Proof

$\forall f,g\in Injns\left(E\right)$, suppose that ${I}_{f}⊊{I}_{g}$, then we have ${I}_{g}={I}_{f}\cup \left({I}_{g}\{I}_{f}\right)$. If $f,g$ are Co-injectives, then ${f}^{-1}\left({I}_{g}\right)={g}^{-1}\left({I}_{f}\right)=E$, which is contradictory because:

$E={g}^{-1}\left({I}_{f}\right)\cup {g}^{-1}\left({I}_{g}\{I}_{f}\right)$, and ${g}^{-1}\left({I}_{g}\{I}_{f}\right)\ne \varnothing$, so ${g}^{-1}\left({I}_{f}\right)⊊E$.

Proposition 10

Let $f,g\in Injns\left(E\right)$, so that ƒ and g are Co-injectives, $\forall h\in Injns\left(E\right)$, if g and h are Co-injectives and ${I}_{f}\cap {I}_{h}={I}_{f}\cap {I}_{g}$ then ƒ and h are Co-injectives.

Proof

${I}_{f}\cap {I}_{h}\ne \varnothing$, let $x,y\in E$, so that $f\left(x\right)=h\left(y\right)$. We have $f\left(x\right)=h\left(y\right)\in {I}_{f}\cap {I}_{h}={I}_{f}\cap {I}_{g}$, then $f\left(x\right)=g\left(x\right)=h\left(y\right)$ (because $f,g$ are Co-injectives), and since g and h are Co-injectives then $x=y$. So $\forall x,y\in E$, $f\left(x\right)=h\left(y\right)⇒x=y$. 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.

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