Evolution on Sets

TL;DR

This paper introduces the concept of evolution on sets, exploring its properties and implications for various structured sets, and presents initial theoretical results on this new mathematical notion.

Contribution

It defines the concept of evolution on sets and investigates its properties across different structured sets, providing foundational results.

Findings

Established basic properties of evolution on sets

Analyzed specific classes of sets with this structure

Provided initial theoretical insights into the concept

Abstract

We introduce the notion of evolution on sets and study several sets endowed with this structure and obtain some results about this new notion.

Equations

$$A:E\to \mathbb{N}\text{and}D:E\to \mathbb{N},$$

$$A(x)=\min \{k\in \mathbb{N}\mid x\in E_k\}\text{and}D(x)=\max \{k\in \mathbb{N}\mid x\in E_k\}+1.$$

$$A:E\to \mathbb{N}\text{and}D:E\to \mathbb{N}\text{be surjective functions such that}A(x)+2\le D(x),\forall x\in E.$$

$$E_k=\{x\in E\mid A(x)\le k<D(x)\}$$

$$E_k\cap E_{k+1}\supset \{x\in E\mid A(x)=k\}\ne \varnothing$$

$$E_k\backslash E_{k+1}\supset \{x\in E\mid A(x)\le k,D(x)=k+1\}\ne \varnothing$$

$$E_{k+1}\backslash E_k\supset \{x\in E\mid A(x)=k+1\}\ne \varnothing$$

$${\rho }^{-1}(10)=\{y\in \mathbb{N}\mid 2\times \pi (y)=10\}=\{y\in \mathbb{N}\mid \pi (y)=5\}=\{12,13,14,15,16\}.$$

$$G_{k+1}=\bigcup _{x\in M_k}{\rho }^{-1}(m(x)),M_{k+1}=G_{k+1}\cap M\backslash {\cup }_{i=1}^k{M}_i,F_{k+1}=G_{k+1}\cap F\backslash {\cup }_{i=1}^k{F}_i.$$

$$E_k=M_k\cup F_k\cup M_{k+1}\cup F_{k+1},E_0=E_1\cup E\backslash {\cup }_{i=1}^{\infty }{E}_i.$$

$$E_k\cap E_{k+1}=f^{-1}(F_k)\cap f^{-1}(F_{k+1})\ne \varnothing \text{because}{G}_k\cap G_{k+1}=\varnothing ,$$

$$E_k\backslash E_{k+1}=(f^{-1}(F_k)\backslash f^{-1}(F_{k+1}))\cup G_k\ne \varnothing \text{because}{G}_k\cap G_{k+1}=\varnothing ,$$

$$E_{k+1}\backslash E_k=(f^{-1}(F_{k+1})\backslash f^{-1}(F_k))\cup G_{k+1}\ne \varnothing \text{because}{G}_k\cap G_{k+1}=\varnothing ,$$

$$e\in E_k\cap E_{k+n}⟹e\in f^{-1}(F_k)\cap f^{-1}(F_{k+n})⟹e\in f^{-1}(F_k)\cap f^{-1}(F_{k+j})\subset E_k\cap E_{k+j},\forall j\le n$$

$$E=\bigcup _{k=1}^{\infty }{E}_k\text{and}e\in E⟹\exists i\in \mathbb{N}\text{such that}e\notin E_i.$$

$$\underset{k\to \infty }{\lim }{\int }_{E_k}{\varphi }_{k}d\mu .$$

$$\underset{k\to \infty }{\lim }{\int }_{E_k}\varphi d\mu .$$

$$\underset{k\to \infty }{\lim }\mu (E_k).$$

$$\underset{k\to \infty }{\lim }\mu (E_k)=0.$$

$$\mu\bigg{(}\biguplus_{i=1}^{\infty}E_{k_{j_{i}}}\bigg{)}=\infty.$$

$$\underset{k\to \infty }{\lim }{\int }_{E_k}{\varphi }_{k}d\mu =0.$$

$$0=\underset{k\to \infty }{\lim }{\int }_{E_k}-Cd\mu \le \underset{k\to \infty }{\lim }{\int }_{E_k}{\varphi }_{k}d\mu \le \underset{k\to \infty }{\lim }{\int }_{E_k}Cd\mu =0,$$

$$\underset{k\to \infty }{\lim }{\int }_{E_k}{\varphi }_{k}d\mu =L\ne 0,$$

$$\displaystyle\text{either}\,\,\limsup_{k}\bigg{(}\sup_{E_{k}}\phi_{k}\bigg{)}=+\infty\,\,\text{or}\,\,\liminf_{k}\bigg{(}\inf_{E_{k}}\phi_{k}\bigg{)}=-\infty$$

$$\underset{k\to \infty }{\lim }{\int }_{E_k}{\varphi }_{k}d\mu ={\int }_E\varphi d\mu .$$

$$\underset{k\to \infty }{\lim }{\int }_{E_k}\varphi d\mu ={\int }_E\varphi d\mu .$$

$$E_k=\varnothing ⟹F_k=\varnothing ⟹F_{k+1}=\varnothing ⟹E_{k+1}=\varnothing .$$

$$E_k\cap E_{k+1}=[F_k\cap F_{k+1}]\ne \varnothing \text{because}{F}_k\cap F_{k+1}\ne \varnothing ,$$

$$E_k\backslash E_{k+1}\supset [F_k\backslash F_{k+1}]\ne \varnothing \text{because}{F}_k\backslash F_{k+1}\ne \varnothing ,$$

Full text

1. Introduction

When we observe the population of human beings, we know that it changes when some people are born or die. We can assume that in every moment there are people that are born and also people that die.

If we denote by $E$ the set of human beings and by $E_{n}$ the set of people in the $n-th$ change, we have the following properties for the sequence $E_{1},E_{2},\dots,E_{n},\dots$ of subsets of people:

1. (1)

$E_{k}=\emptyset\implies E_{k+1}=\emptyset$;

it means that if the human beings are extinguished some day, they remain like that 2. (2)

$E_{n},E_{n+1}\neq\emptyset\implies E_{n}\cap E_{n+1},E_{n}\backslash E_{n+1},E_{n+1}\backslash E_{n}\neq\emptyset$;

it means that from the $n-th$ step to the $(n+1)-th$ step some people die, some people are born and some people are in commom between them 3. (3)

$e\in E_{n}\cap E_{n+k}\implies e\in E_{n+j},\forall j\leq k$;

it means that if some people live in two steps, then they live in the intermediate steps. 4. (4)

$e\in E\implies\exists\,n,k\in\mathbb{N}\,\,\text{such that}\,\,e\in E_{n},e\not\in E_{n+k}$.

it means that every human being is born at some step and die at some step.

So, motivated by this observation, we introduce the notion of evolution on sets.

2. Preliminaries

Definition 1**.**

Let $E$ be a set. A sequence $E_{1},E_{2},\dots,E_{n},\dots$ of subsets of $E$ is said to be an evolution on the set $E$ if:

1. (1)

$E_{k}=\emptyset\implies E_{k+1}=\emptyset$; 2. (2)

$E_{n},E_{n+1}\neq\emptyset\implies E_{n}\cap E_{n+1},E_{n}\backslash E_{n+1},E_{n+1}\backslash E_{n}\neq\emptyset$; 3. (3)

$e\in E_{n}\cap E_{n+k}\implies e\in E_{n+j},\forall j\leq k$; 4. (4)

$e\in E\implies\exists\,n,k\in\mathbb{N}\,\,\text{such that}\,\,e\in E_{n},e\not\in E_{n+k}$.

We call $E_{k}$ the $k-th$ stage of the evolution.

It is easy to see that for every $e\in E$ there exists $n,k\in\mathbb{N}$ such that $e\in E_{n+j},\forall j\leq k$ and $e\not\in E_{i},i\not\in\{n,\dots,n+k\}$.

We intend to interpret the set $E_{k}$ as the set of ”focused points” in the set $E$ at the moment $k$. Later on, we give examples where this notion is important.

Now we give simple examples of evolution.

Example 1**.**

$E=\mathbb{N},E_{n}=\{n,\dots,(n+1)^{2}\}$* an evolution on the set $\mathbb{N}$.*

We observe that the conditions $3,4$ implies that for every point in $E$ there exist moments when it appears and when it disappears forever. We can define two functions:

[TABLE]

where

[TABLE]

The following proposition shows that we can obtain an evolution from functions of appearing and disappearing.

Proposition 1**.**

Let $E$ be a set and

[TABLE]

Then, the sequence of subsets of $E$ defined as

[TABLE]

is an evolution on $E$.

Proof.

Since the function $A$ is surjective, we have that $E_{k}\neq\emptyset,\forall k\in\mathbb{N}$. Also, $E=\displaystyle\bigcup_{n=1}^{\infty}E_{n}$ and $x\in E_{k}$ if $k>D(x)$, which shows codition $4$. It remains to show the conditions $2$ and $3$. We have that $x\in E_{k}$ if $A(x)\leq k<D(x)$, which shows the condition $3$. Also,

[TABLE]

[TABLE]

[TABLE]

It shows the condition $2$ and completes the proof that the sequence is an evolution. ∎

We call $A$ the appearing function and $D$ the disappearing function of the evolution. It establishes a chronology on $E$.

Once we have a chronology on a set, we have an evolution. So, it is easy to find several evolutions on a given set. Also, as we can expect, the image of an evolution by a bijection is also an evolution:

Proposition 2**.**

Let $f:F\to E$ be a surjective function. If $E_{1},\dots,E_{k},\dots$ is an evolution on $E$, then $f^{-1}(E_{1}),\dots,f^{-1}(E_{k}),\dots$ is an evolution on $F$ with the same appearing and disappearing functions.

Proof.

It is easy to see that the sequence $f^{-1}(E_{k})$ satisfies the conditions to be an evolution because pre-images preserve intersections and differences of sets. ∎

We say that two sets $E$ and $F$ with evolutions $E_{k}$ and $F_{k}$, respectively, is isoevolved if the exists a bijection $f:E\to F$ such that $F_{k}=f(E_{k})$.

We say that an evolution $E_{k}$ if reducible if there exists a subsequence $E_{k_{j}}$ that is also an evolution. Otherwise, we say that the evolution is irreducible.

3. Reproduction

Let $E_{*}\subset E$ be a set such that $E=M\uplus F$ is a decomposition where $M$ (male) and $F$ (female) and $E_{*}=M_{*}\uplus F_{*}$ with $M_{*}\subset M,F_{*}\subset F$ have same cardinality, and we consider $m:M_{*}\to F_{*}$ a bijection, that we call marriage function and $\rho:E\to F$ an arbitrary function, that we call reproduction function. We call the pair $(x,m(x))$ a couple and $\rho^{-1}(m(x))$ is the set of children of the couple $(x,m(x))$. It is clear that distinct couple have disjoint sets of children.

Remark 1**.**

We observe that if the reproduction function is not surjective, then not every couple has a set of children.

Example 2**.**

$E=E_{*}=\mathbb{N},M=M_{*}=\{\text{odd numbers}\},F=F_{*}=\{\text{even numbers}\},m(x)=x+1,\rho(y)=2\times\pi(y)$, where $\pi(y)$ denotes the number of primes less than $y$. If we take $x=9,m(x)=10$ and the set of children of the couple $(9,10)$ is

[TABLE]

Now, we would like to use the notion of reproduction to construct an evolution.

Proposition 3**.**

Given a set $E$, a decomposition on $E$, a marriage function and a reproduction function, there exists an evolution on $E$ such that the children of couples in $E_{k}$ are in $E_{k+1}\backslash\cup_{i=0}^{k}E_{i}$.

Proof.

Let $\emptyset\neq M_{1}\subset M\backslash\rho^{-1}(F),\emptyset\neq F_{1}\subset F\backslash\rho^{-1}(F),G_{1}=M_{1}\cup F_{1}$ and we denote

[TABLE]

We observe that the sequences $M_{k}$ and $F_{k}$ are composed by elements pairwise disjoints.

We claim that the sequence

[TABLE]

is the desired evolution. In fact, let us verify the four conditions:

1. (1)

If $E_{k}=\emptyset$ then $M_{k},F_{k},M_{k+1},F_{k+1}=\emptyset$ which implies $G_{k+2}=\emptyset$. Also, $M_{k+2},F_{k+2}=\emptyset$ and finally $E_{k+1}=\emptyset$. 2. (2)

We have that $E_{n}\cap E_{n+1}=M_{n+1}\cup F_{n+1}$, $E_{n}\backslash E_{n+1}=M_{n}\cup F_{n}$ and $E_{n+1}\backslash E_{n}=M_{n+2}\cup F_{n+2}$. All these sets are not empty. 3. (3)

If $e\in E_{n}\cap E_{n+k}$, we have that $k=1$ and $E_{n}\cap E_{n+k}\neq\emptyset$ and it implies that $k=1$. So, $e\in E_{n}\cap E_{n+j},j\leq k$. 4. (4)

By definition of $M_{k},F_{k},E_{k}$, it is easy to see that for every $e\in E$ there exists $n,k\in\mathbb{N}$ such that $e\in E_{n},e\not\in E_{n+k}$.

∎

The previous proposition guarantees that there exists an evolution such that the children do not appear in the evolution before their parents. Moreover, we can not have cycles, which means that an element can not be a descendent of itself. Also, the evolution starts with the element $E_{1}$ and is constructed by the reproduction, step by step.

4. Evolution on Probability Spaces

With the above propositions, we know that there are several evolutions on a given set and we may ask what is the ”best” evolution on a set. If the set is a probability space, we could try to find an evolution for which all stages has positive measure.

Proposition 4**.**

If $E$ is a set with a probability $\mu$, such that there exists a sequence of two by two disjoint measurable subsets $G_{k}\subset E$ with $\mu(G_{k})>0,\forall k\in\mathbb{N}$ and a measurable bijection $f:E\backslash\cup_{k=1}^{\infty}G_{k}\to E$, then there exists an evolution $E_{k}$ on $E$ such that $\mu(E_{k})>0,\forall k\in\mathbb{N}$.

Proof.

Let $F_{k}$ be an evolution on $E$. It is enough to show that $E_{k}=f^{-1}(F_{k})\uplus G_{k}$ is an evolution on $E$. It follows from the fact that the sequence $f^{-1}(F_{k})$ is an evolution on $E\backslash\cup_{k=1}^{\infty}G_{k}$ (by the previous proposition). In fact, we have $E_{k}\neq\emptyset,\forall k\in\mathbb{N}$ because $\mu(G_{k})>0$. Also, we have

[TABLE]

[TABLE]

[TABLE]

Moreover,

[TABLE]

Finally,

[TABLE]

It completes the proof that $E_{k}$ is an evolution. Also, $\mu(E_{k})=\mu(f^{-1}(F_{k})\uplus G_{k})\geq\mu(G_{k})>0,\forall k\in\mathbb{N}$. ∎

Let us assume that $E$ is a measurable space, the sets $E_{k}$ are measurable and $\mu$ is a probability on $E$. Let $\phi_{k}:E\to\mathbb{R}$ be a sequence of measurable functions. We intend to investigate the following limit:

[TABLE]

If $\phi_{k}\equiv\phi$, we could look for the value of

[TABLE]

In particular, if $\phi\equiv 1$, then the value of

[TABLE]

Proposition 5**.**

Let $\mu$ be a probability on $E$ and the sequence $E_{k}$ of subsets of $E$ an evolution. If there exists a sequence $d_{k}\in\mathbb{N}$ such that $D(x)-A(x)\leq d_{k},\forall x\in E_{k},\forall k\in\mathbb{N}$, then we have

[TABLE]

Proof.

If this limit is not equal to zero, there exist $\epsilon>0$ and a subsequence $\mu(E_{k_{j}})\geq\epsilon$. We have that $D(x)\leq A(x)+d_{k}\leq k+d_{k},\forall x\in E_{k}$. So, $E_{k_{j}}\cap E_{k_{j}+d_{k_{j}}+1}=\emptyset\implies E_{k_{j}}\cap E_{n}=\emptyset,\forall n\geq k_{j}+d_{k_{j}}+1$. It means that there exists a subsequence $E_{k_{j_{i}}}$ two by two disjoint. This is a contradiction because we would have

[TABLE]

∎

Corollary 1**.**

Let $\mu$ be a probability on $E$, $E_{k}$ an evolution on $E$ and $\phi_{k}:E\to\mathbb{R}$ a sequence of measurable functions. If there exist $C>0$ and a sequence $d_{k}\in\mathbb{N}$ such that $D(x)-A(x)\leq d_{k}$ and $\mid\phi_{k}(x)\mid\leq C,\forall x\in E_{k},\forall k\in\mathbb{N}$, then we have

[TABLE]

Proof.

In fact, we have that

[TABLE]

because $\displaystyle\lim_{k\to\infty}\mu(E_{k})=0$ by the previous proposition. ∎

Corollary 2**.**

Let $\mu$ be a probability on $E$, $E_{k}$ an evolution on $E$ and $\phi_{k}:E\to\mathbb{R}$ a sequence of measurable functions. If there exists a sequence $d_{k}\in\mathbb{N}$ such that $D(x)-A(x)\leq d_{k}$ and we have

[TABLE]

then the following holds

[TABLE]

Proof.

It follows from the previous corollary because, otherwise, there would exist $C>0$ such that $\mid\phi_{k}\mid<C$ on $E_{k}$. ∎

Given an evolution $E_{k}$ on $E$, we say that the sequence of functions $\phi_{k}:E\to\mathbb{R}$ converges to $\phi:E\to\mathbb{R}$ by the evolution $E_{k}$ if

[TABLE]

In particular, we intend to investigate when we have

[TABLE]

or at least to know if there exists an evolution such that it holds. We already know that $\phi$ must have both positive and negative values and be unbounded (unless $\int_{E}\phi d\mu=0$), since $\mu(E_{k})\to 0$ under ”good” conditions.

Question: when $\phi_{k}$ converges to $\phi$ by all evolutions on $E$ with positive measure stages?

Proposition 6**.**

Let $E$ be connected and $\phi:E\to\mathbb{R}$ continuous, unbounded and assuming positive and negative values. There exists an evolution $E_{k}$ such that

[TABLE]

Proof.

∎

5. Evolution on Vector Spaces

Let $E$ be an infinite dimensional vector space and fix a countable linearly independent subset $B=\{e_{1},\dots,e_{k},\dots\}$. We can use evolutions on $\mathbb{N}$ to obtain evolutions on a subset of $E$. Given $F\subset\mathbb{N}$, we denote by $[F]$ the subspace generated by the vectors in $B$ with index in $F$.

Proposition 7**.**

Let $F_{k}$ be an evolution on $\mathbb{N}$. Then the sequence $E_{k}=[F_{k}]$ of subspaces of $E$ is an evolution on $E^{\prime}=\cup_{k=1}^{\infty}E_{k}$ with the same appearing and disappearing functions. If $E^{\prime}$ is a subspace, then $F_{1}=\mathbb{N},F_{k}=\emptyset,k>2$.

Proof.

Firstly,

[TABLE]

Now, we have

[TABLE]

[TABLE]

[TABLE]

Also,

[TABLE]

Finally,

[TABLE]

∎

Proposition 8**.**

Let $E$ be a vector space and $f:E\to\mathbb{R}$ a non trivial linear functional. If $F_{k}$ is an evolution on $\mathbb{R}$, then $E_{k}=f^{-1}(F_{k})$ is an evolution on $E$ with the same appearing and disappearing functions. Each stage $E_{k}$ is a union of hyperplanes.

Proof.

It follows from the fact that $f^{-1}(t)$ is a hyperplane in $E$ and that a non trivial linear functional is always surjective. ∎

Let $M_{n\times n}$ be the space of $n\times n$ matrices. Given $A\in M_{n\times n}$, $\det A$ denotes the determinant of $A$.

Proposition 9**.**

Let $F_{k}$ be an evolution on $\mathbb{R}$. The following sequence is an evolution on $M_{n\times n}$:

[TABLE]

Proof.

It follows from the fact that the function $\det:M_{n\times n}\to\mathbb{R}$ is surjective. ∎

Proposition 10**.**

Let $E$ be a vector space with inner product $\langle\,,\rangle$ and $F_{k}$ be an evolution on $\mathbb{R}$. Fix $e_{0}\in E$. The following sequence is an evolution on $E$:

[TABLE]

Proof.

It follows from the fact that the function $\langle\cdot,e_{0}\rangle:E\to\mathbb{R}$ is surjective. ∎

If $E$ is a normed vector space, we can find an evolution on $E$ from the metric derived from the norm, as we will see in the next section.

6. Evolution on Metric Spaces and Manifolds

If $E$ is endowed with a metric $d$, we can obtain an evolution on $E$ from an evolution on $\mathbb{N}$ or $[0,\infty)$.

Proposition 11**.**

Let $E$ be a metric space, fix $E_{0}\subset E$ and let $F_{k}$ be an evolution on $[0,\infty)$. Let

[TABLE]

Then, the sequence $E_{k}$ is an evolution on $E$ with the same appearing and disappearing functions.

Proof.

We have that $E_{k}=f^{-1}(F_{k})$, where $f(e)=d(e,E_{0})$, which is a surjective function $f:E\to[0,\infty)$. ∎

Corollary 3**.**

Let $E$ be a metric space and fix $e_{0}\in E$ and let $F_{k}$ be an evolution on $[0,\infty)$. Let

[TABLE]

Then, the sequence $E_{k}$ is an evolution on $E$ with the same appearing and disappearing functions.

Proof.

It follows from the previous proposition, by taking $E_{0}=\{e_{0}\}$. ∎

Corollary 4**.**

Fix $e_{0}\in E$ and let $F_{k}$ be an evolution on $\mathbb{N}$. Let

[TABLE]

Then, the sequence $E_{k}$ is an evolution on $E$ with the same appearing and disappearing functions.

Proof.

It follows from the previous corollary, by taking $G_{k}=\cup_{j\in F_{k}}[j,j+1],k\geq 0$. ∎

Given a differentiable manifold $M$ which is Hausdorff and with countable basis, we can endow $M$ with a riemannian metric and $M$ is also a metric space.