Topological Spaces and Phenotype-genotype Spaces
M.E.Abd El-Monsef* ; A.M.Kozae*:M.Shokry**;M.S.Badr*
*Mathematics Department,Faculty of Science,Tanta University
**Department of Physics and Engineering Mathematics,Faculty of Engineering,Tanta
University
Tanta, EGYPT
Abstract: Researchers hope that establishing a notion of proximity using topology
will help to clarify the biological processes underlying the evolution of living
organisms. The simple model presented here, using RNA shapes, can carry over to
more general and complex genotype-phenotype systems. Proximity is an important
component of continuity, in both real-world and topological terms. Consequently,
phenotype spaces provide an appropriate setting for modeling and investigating
continuous and discontinuous evolutionary change.
Key words: RNA shapes, phenotype, topology, genotype
1.1Introduction
According to the Darwinian theory of evolution .adaptation results from
spontaneously generated genetic variation and natural selection. Mathematical models
of this process can be seen as describing a dynamics on an algebraic structure which
in turn is defined by the processes which generate genetic variation (mutation and / or
recombination) .The theory of complex adaptive system has shown that the properties
of the algebraic structure induced by mutation and recombination is more important
for understanding the dynamics than the differential equations themselves. This has
motivated new directions in the mathematical analysis of evolutionary models in
which the algebraic properties induced by mutation and recombination are at the
center of interest [8]. In this chapter we summarize some new results on the algebraic
properties of mutation spaces. It is shown that the algebraic structures induced by
crossover can be represented by a map from the pairs of types to the power set of the
pair of types.
The aim of this paper is to give, first, a relation between mutation (genotype and
phenotype) programming to generated topological spaces. Second, studying applied
the metric space in DNA.
The main goal of 1.2 is to spot light on genotype and phenotype. We studied the
Mutations and net work1.3. The main goal of 1.4 is to define mutation and
programming, genotype and phenotype, 1.5 Metric of DNA .
1.2 Genotype and Phenotype [3]
The concepts of genotype and Phenotype are important in evolutionary biology.
Every living organism is the physical realization (Phenotype) of internally coded,
inheritable information (genotype). Evolutionary change from one Phenotype to
anther occurs by change, called mutations, in their corresponding genotypes. We
would like to have a nation of how close one Phenotype is to another, reflecting how
likely it is that a genotype mutation transforms one Phenotype to other, we describe
how molecular biologists establish a notion of evolutionary proximity by
constructing a Topological Space from a set of Phenotypes.
Recent computational work on a biophysical genotype-phenotype model Based
on the folding on RNA sequences into their secondary structures, surges a rather
different pictures [1, 10]. If phenotypes are organized according to Genetic
accessibility, the resulting space lacks a metric and is formalized by an unfamiliar
structure in the simplest case, a pretopology [1].
Topological Spaces and Phenotype Spaces: the objects that we study in Topology
are called Topological Spaces. These are sets of points on which a notion of proximity
between points is established by specifying a collection of subsets called open sets.
The line, the circle, the plane, the sphere, the torus, and the mobius band are all
examples of Topological Spaces. And we will see how to construct Topological
Spaces in a variety of setting and situations.
Definition1.2.1[3] The set of all genotype sequences that result in a particular
RNA shape s is called the neutral net work of s and is denoted N(s).
1.2.1Genetic mutation[3, 6]
In a general biological setting genetic mutation is a random process by which the
genotype changes. As a result of the many-to-one nature of the genotype-phenotype
mapping, many genetic mutations do not alter the resulting phenotype. On the other
hand, there are occasional mutations that result in new phenotypes, and then process
of natural selection determines whether or not the new phenotype persists.
Now, suppose that we have a specific genotype, and let 𝑟 be its corresponding
RNA shape. In this setting, a genetic mutation is a random change in the entries in the
genotype sequence. As these random changes occur, we many remain within the
neutral network of 𝑟 “drifting” through genotype sequences that all result in RNA
shape 𝑟 at some point, however, a change in one entry in a genotype sequence might
take us out of the neutral network of 𝑟 Into that of another RAN shape 𝑠 We say that 𝑟
has mutated to 𝑠.see figure 1.2.1
FIGURE 1.2.1 Genetic mutations from secondary structure 𝑟 to secondary structure 𝑠.
Given RAN shapes r and s , we would like to know how likely it is for r to
mutate to s as a result of a single-entry change in a genotype sequence in the neutral
net work of r .We can define and quantify a probability to make this precise. We will
use this probability to define phenotype-space topologies on sets of RNA shapes.
Before defining and using the probability, however, we introduce several
quantities that play a role in its definition. By a point mutation we mean a mutation
from one genotype sequence to another obtained by changing a single entry in
sequence. As in Example 1.2.1.
Example 1.2.1[12] Consider the following three genotype sequences:
1. GGGCAGUCUC
CUCCCAUCCA
CCGGCGUUUA AGGGAUCCUG - AACUUCGUCG
AUCAGUCCGC CUCACGGAUG
2. GGGCAGUCUC
CUCCCAUCCA
3. GGGCAGUCUC
AACUUCGUCG
GAGUUG
CCGGCGUUUA AGGAAUCCUG - AACUUCGUCG
AUCAGUCCGC
CUCACGGAUG-
CCGGCCUUUA
GAGUUG
AGGGAUCCUG-
CUCCCAUCCA
AUCAGUCCGC
CUCACGGAUGGAGUUG
sequence 2 is obtained by a point mutation from sequence 1, and vice versa (similarly
for sequences 1 and 3).
1.3 Mutations and net work
For RNA shapes 𝑟𝑎𝑛𝑑 𝑠, 𝑙𝑒𝑡 𝑚𝑟,𝑠 are the numbers of point mutations that change
a sequence in 𝑁(𝑟) to a sequence in 𝑁(𝑠), we calculate the mutation (a point
mutation). A common method of implementing the mutation operator involves
generating a random variable for each bit in a sequence. This random variable tells
whether or not a particular bit will be modified. This mutation procedure, based on the
biological point mutation, is called single point mutation [1,3,7, 10,11] .
And thus result in a mutation from r to s note that 𝑚𝑟,𝑠 = 𝑚𝑠,𝑟 because each point
mutation changing a sequence in 𝑁(𝑟) to one in 𝑁(𝑠) has a corresponding inverse
point mutation that changes a sequence in 𝑁(𝑠) to one in 𝑁(𝑟). Let 𝑚𝑟,∗ be the
number of point mutation that changes a sequence in 𝑁(𝑟) to a sequence in any
another neutral net work. We can think of 𝑚𝑟,∗ as the number of point mutation that
takes us out of 𝑁(𝑟).
DEFINITION1.3.1.[3] The mutation probability, 𝑝𝑟,𝑠 is defined by
𝑝𝑟,𝑠 =
𝑚𝑟,𝑠
𝑚𝑟,∗
Even though 𝑚𝑟,𝑠 = 𝑚𝑠,𝑟 , it need not to be case that 𝑝𝑟,𝑠 = 𝑝𝑠𝑟 since the values
of 𝑚𝑟,∗ and 𝑚𝑠,∗ might be different. For example, if 𝑚𝑟,∗ is greater than 𝑚𝑠,∗ then
there are more point mutation out of 𝑁(𝑟) than there are out of𝑁(𝑠) , and there for
the proportions of mutations
𝑚𝑟,𝑠
𝑚𝑟,∗
is smaller than
𝑚𝑠,𝑟
𝑚𝑠,∗
We will see this asymmetry
reflected in the phenotype-space topologies –while RNA shape s might be close to
RNA shape 𝑟, 𝑟 need not be close to s.
Because of the asymmetry in these probabilities, a distance function cannot
provide a means of determining the proximity of two RNA shapes. A distance
function is necessarily symmetric; the distance between 𝑟 and 𝑠 must equal the
distance between 𝑎𝑛𝑑 𝑟 . Since a distance function will not serve for this purpose, a
topological space is a natural alternative to consider.
EXAMPLE 1.3.1 [3]In order to clarify how the asymmetry arises in the mutation
probabilities, in this example we consider probabilities that are defined like the
mutation probabilities, but in a very different setting. Consider the following scenario
In Figure 1.2.1 we show the eight different RNA shapes. The set of RNA shapes of
genotype sequences of length 10 made up of guanine (G) and cytosine (C) only. There
are 210 = 1024 possible genotype sequences, and, upon folding and bonding.
We consider a topology on GC 10, the set of RNA shapes of genotypes of length 10.
The forthcoming process used in defining a topology from mutation probabilities
carries over to RNA shapes associated with genotype sequences of any fixed length.
FIGURE 1.3.1: The set GC1O
We use the associated mutation probabilities in Figure 1.3, which are derived
from numbers of mutations as previously described
Figure 1.3.2. The mutation probabilities for GC10
Figure 1.3.2. The entry in the 𝑖th row and jth column is the probability that a point
mutation out of the neutral network of Si results in a sequence in the neutral network
of Sj.
The notion of proximity that we define on the set of RNA shapes is based on the
likelihood of a mutation from one RNA shape to another. Since there are eight RNA
shapes in GC10, each can potentially mutate to seven others Hence if Pi, j > 1/7, we
think of Si as having more than the average likelihood of mutating to Sj
For each i = 1,... ,8, define Ri = {Si} U {Sj} I Pi, j > 1/7}. Thus Ri consists of Si
along with all of the RNA shapes to which Si has more than the average likelihood of
mutating. The collection R1/7 = {Ri }8i=1 I is not itself a topology, but we extend it to
one, defining 𝜏1/7 to be the minimal topology on GC1O containing RI/7. The topology
𝜏1/7 is generated by a basis formed by taking finite intersections of the sets in R 1/7.
The resulting topological space is referred to as a phenotype space.
A topology on a finite set has a unique minimal basis that generates the topology.
If for each Si we take all of the sets Rj that contain Si, and let Bi be their intersection,
then the collection B 1/7= {Bi }8i is the minimal basis for 𝜏1/7
From the probability table for GC10, it is easy to determine this minimal basis
B1/7. To begin, as shown in Figure 4.4, we put a check mark in each diagonal table
entry and in each table entry where Pi, j > 1/7
.
FIGURE 1.4.1. The GC10 probability-table entries with Pi,j > 1/7
For each i, the checkmarks in row 𝑖 correspond to the elements in Ri. For
example, R 1 = {s 1 , S3, S8} and R2 = {S2, S3, S6, S8}. To determine the basis element
Bi we take the intersection of the rows that contain a check mark in the ith column.
For example, the S6 column is checked in the second and sixth rows. Intersecting R2
and R6 results in B6 = {S2, S6, S8}.
Researchers hope that establishing a notion of proximity using topology will help
to clarify the biological processes underlying the evolution of living organisms. The
simple model presented here, using RNA shapes, can carry over to more general and
complex genotype-phenotype systems. Proximity is an important component of
continuity, in both real-world and topological terms. Consequently, phenotype spaces
provide an appropriate setting for modeling and investigating continuous and
discontinuous evolutionary change
1.4 mutation and programming
While this method did not show mutation rate is not where you came to these
numbers so we will work program calculate these figures and it shows the proportion
of mutations.
We calculate the mutation in this chapter by program ,The idea of this program
that calculates the change or difference between the element and the element or a
group and that by calculating the change between each element of the first group With
the elements of the second group in the sense we take the first element of the group
with the first elements of the second group + second element with the elements of the
second group and so on for the rest of the group the first and so we have calculated
the change between the first and the second group and so on for groups.
And the account is based on the number of elements unequal any different.
1.4.1The program
Public Class Form1
Dim s As Integer
Dim l = 1
Private Sub Button1_Click(ByVal sender As System.Object, ByVal e
As System.EventArgs) Handles Button1.Click
If l = 1 Then
Me.ListBox1.Items.Add(Me.TextBox1.Text)
Me.TextBox1.Clear()
Me.TextBox1.Focus()
s = Me.ListBox1.Items(0).length
l = 2
Me.Button2.Enabled = True
Else
If Me.TextBox1.Text.Length <> s Then
MsgBox("The length Of the text must be the same")
Else
Me.ListBox1.Items.Add(Me.TextBox1.Text)
Me.TextBox1.Clear()
Me.TextBox1.Focus()
End If
End If
End Sub
Private Sub Button2_Click(ByVal sender As System.Object, ByVal e
As System.EventArgs) Handles Button2.Click
Me.ListBox2.Items.Add(Me.TextBox2.Text)
Me.TextBox2.Text = ""
Me.TextBox2.Focus()
End Sub
Private Sub Button3_Click(ByVal sender As System.Object, ByVal e
As System.EventArgs) Handles Button3.Click
Me.Close()
End Sub
Private Sub Button4_Click(ByVal sender As System.Object, ByVal e
As System.EventArgs) Handles Button4.Click
Dim i, j, k, counter As Integer
Dim c1, c2 As String
For i = 0 To Me.ListBox1.Items.Count - 1
For j = 0 To Me.ListBox2.Items.Count - 1
For k = 0 To Me.ListBox1.Items(i).Length - 1
c1 = Me.ListBox1.Items(i).Substring(k, 1)
c2 = Me.ListBox2.Items(j).Substring(k, 1)
If c1 <> c2 Then
counter = counter + 1
End If
Next
Next
Next
Me.Label1.Text = counter
End Sub
Private Sub Button5_Click(ByVal sender As System.Object, ByVal e
As System.EventArgs) Handles Button5.Click
Me.ListBox1.Items.Remove(Me.ListBox1.SelectedItem)
End Sub
Private Sub Button8_Click(ByVal sender As System.Object, ByVal e
As System.EventArgs) Handles Button8.Click
Me.ListBox2.Items.Remove(Me.ListBox2.SelectedItem)
End Sub
Private Sub Button6_Click(ByVal sender As System.Object, ByVal e
As System.EventArgs) Handles Button6.Click
Me.ListBox1.Items.Clear()
Me.Label1.Text = ""
End Sub
Private Sub Button7_Click(ByVal sender As System.Object, ByVal e
As System.EventArgs) Handles Button7.Click
Me.ListBox2.Items.Clear()
Me.Label1.Text = ""
End Sub
Private Sub TextBox1_KeyDown(ByVal sender As Object, ByVal e As
System.Windows.Forms.KeyEventArgs) Handles TextBox1.KeyDown
If e.KeyCode = Keys.Enter Then
If l = 1 Then
Me.ListBox1.Items.Add(Me.TextBox1.Text)
Me.TextBox1.Clear()
Me.TextBox1.Focus()
s = Me.ListBox1.Items(0).length
l = 2
Me.Button2.Enabled = True
Else
If Me.TextBox1.Text.Length <> s Then
MsgBox("The length Of the text must be the same")
Else
Me.ListBox1.Items.Add(Me.TextBox1.Text)
Me.TextBox1.Clear()
Me.TextBox1.Focus()
End If
End If
End If
End Sub
Private Sub TextBox2_KeyDown(ByVal sender As Object, ByVal e As
System.Windows.Forms.KeyEventArgs) Handles TextBox2.KeyDown
If e.KeyCode = Keys.Enter Then
Me.ListBox2.Items.Add(Me.TextBox2.Text)
Me.TextBox2.Clear()
Me.TextBox2.Focus()
End If
End Sub
Private Sub Form1_Load(ByVal sender As System.Object, ByVal e As
System.EventArgs) Handles MyBase.Load
End Sub
End Class
Example 1.4.1
Suppose our genotypes are four-element sequences made up of the letters N and
S, and we have four phenotypes with neutral networks
𝑆1 =(N, N, N, N), 𝑆2 = (N, N, N, S), 𝑆3 =(N, N, S, N) , 𝑆4 = (N, N, S, S)
𝑚1,2 = 1 , 𝑚1,3 = 1 , 𝑚1,4 = 2 , 𝑚2,3 = 2 , 𝑚2,4 = 1 , 𝑚3,4 =1
𝑚1,∗ =𝑚2,∗ =𝑚2,∗ =𝑚4,∗ = 4
𝑝1,2 = .25 , 𝑝1,3 = .25, 𝑝1,4 = .5,𝑝2,1 = .25, 𝑝2,3 = .25 ,
.25 ,𝑝4,1 = .25 ,𝑝4,2 = .25
𝑝4,2 = .25, 𝑝4,3 = .25
𝑝2,4 = .25 ,𝑝3,1 =
𝑆1
𝑆2
–
𝑆1
.25
–
𝑆2
.25
𝑆3
.25
.25
𝑆4
.5
.25
If 𝑃𝑖,𝑗 = .25
𝑆3
𝑆4
.25
.5
.25
.25
–
.25
.25
–
𝑆2
𝑆3
there for
𝑆1
𝑆1



𝑆4
𝑆2




𝑆3






𝑆4

Example 1.4.2
Suppose our genotypes are four-element sequences made up of the letters N and
S, and we have three phenotypes with neutral networks :
N1 = {(N, N, N, N), (N, N, N, S), (N, N, S, N), (N, N, S, S)}
N2 = {(N, S, N, N), (N, S, N, S), (N, S, S, N), (N, S, S, S), (S, N, N, N), (S, N, N, S)}
N3 = {(S, N, S, N), (S, N, S, S), (S, S, N, N), (S, S, N, S), (S, S, S, N), (S, S, S, S)}
𝑆1 =(N, N, N, N), 𝑆2 = (N, N, N, S), 𝑆3 =(N, N, S, N) , 𝑆4 = (N, N, S, S)
𝑚1,2 = 48 , 𝑚1,3 = 64, 𝑚1,∗ = 112 , 𝑚2,1 = 48, 𝑚2,3 = 78, 𝑚3,∗ =126
𝑚3,1 =64 , 𝑚12 =78,𝑚3,∗ = 142
𝑆1
𝑆2
𝑆3
𝑆1
–
.423
.571
𝑆2
.378
–
.622
𝑆3
.448
.557
–
If 𝑃𝑖,𝑗 > .4 there for
𝑆1
𝑆1

𝑆2
𝑆3

𝑆2
𝑆3






R1 = {S1 , S2 , S3 }
𝑅 2 = {𝑆2 , 𝑆3 }
𝑅 3 = {𝑆1 , 𝑆2 , 𝑆3 }
𝐵1 = 𝑅1 ∩ 𝑅3 = {𝑆1 , 𝑆2 , 𝑆3 }
𝐵2 = 𝑅1 ∩ 𝑅2 ∩ 𝑅3 = 𝐵3 = {𝑆2 , 𝑆3 }
𝜏1 = {{𝑆1 , 𝑆2 , 𝑆3 }, {𝑆2 , 𝑆3 }, ∅}
If Pi,j > .5 there for
𝑆1
𝑆2

𝑆1
𝑆3

𝑆2


𝑆3


𝑆2
𝑆3


𝑅 1 = {𝑆1 , 𝑆3 }
𝑅 2 = {𝑆2 , 𝑆3 }
𝑅 3 = {𝑆2 , 𝑆3 }
𝐵1 = 𝑅1 = {𝑆1 , 𝑆3 }
𝐵2 = 𝑅2 ∩ 𝑅3 = {𝑆2 , 𝑆3 }
𝐵3 = 𝑅1 ∩ 𝑅2 ∩ 𝑅3 = {𝑆3 }
𝜏2 = {{𝑆1 , 𝑆2 , 𝑆3 }, {𝑆2 , 𝑆3 }, {𝑆3 }, {𝑆1 , 𝑆3 }, ∅} .
𝑇ℎ𝑒𝑛 𝜏1 ⊆ 𝜏2
If 𝑃𝑖,𝑗 ≥ .6
there for
𝑆1
𝑆1
𝑆2
𝑆3
𝑅 1 = {𝑆1 }
𝑅 2 = {𝑆2 , 𝑆3 }
𝑅 3 = {𝑆3 }
𝐵1 = 𝑅1 = {𝑆1 }
𝐵2 = 𝑅2 ∩ 𝑅3 = {𝑆3 }


𝐵3 = {𝑆3 }
𝜏3 = {{𝑆1 }, {𝑆3 }, {𝑆1 , 𝑆3 }, ∅}
If 𝑃𝑖,𝑗 ≥ .7
there for
𝑆1
𝑆2
𝑆3

𝑆1

𝑆2

𝑆3
𝑅 1 = {𝑆1 }
𝑅 2 = {𝑆2 }
𝑅 3 = {𝑆3 }
𝐵1 = {𝑆1 }
𝐵2 = {𝑆2 }
𝐵3 = {𝑆3 }
𝜏4 = {{𝑆1 , 𝑆2 , 𝑆3 }, {𝑆2 , 𝑆3 }, {𝑆1 } , {𝑆2 } , {𝑆3 }, {𝑆1 , 𝑆2 }, {𝑆1 , 𝑆3 }, ∅}
Proposition1.4.1
𝛼1
If α1≤α2 then 𝑅1
𝛼
⊇ 𝑅2 2 .
Proof
Let N1 ∈ 𝑅2𝛼1 , α1 ≤ α2 .
When we provide the value of ∝ less than the number of times the mutation
reduces the number of groups involved with the seasons, and found the resulting class
must have a number of mutations to achieve minimum and maximum for that.
Then N1 ∈ 𝑅1𝛼1 , implies 𝑅1𝛼1 ⊇ 𝑅2𝛼2
Proposition1.2.2
If α1≤α2thenThen 𝜏 𝛼1 ⊆ 𝜏 𝛼2
Proof. Obvious from Proposition4.4.1
Remark1.4.1
If ∝= 0 the result topology is indiscrete topology.
Remark 1.4.1
If ∝= 1 the result topology is discrete topology.
1.5 Metric of DNA[3]
One of the most important problems in DNA research is how to compare distinct
DNA sequences. How different is one sequence of DNA from another in some sense,
this is a measure of the evolutionary distance between the two sequences (and, by
extension, between the organisms from which they were derived). When a species
splits into two new species, resulting in fork in the evolutionary tree, the species'
initially identical DNA sequences begin to accumulate unique changes. Measuring the
distance between the two sequences as a function of these differences provides insight
into the nature of the evolutionary history of each species.
One of the most common and useful types of topological space is the so-called
metric space. Metric spaces are topological spaces that result from having a means for
measuring distance between points in the underlying set. This notion of measuring
distance goes beyond stretching out a measuring tape to see how far apart two objects
are. For example, as we will see, we can measure the distance between two functions
by considering the area bounded between their graphs, and we can measure the
distance between two words by considering how many letter changes take us from
one to the other. The ability to measure and compare distances between elements of a
set is often crucial, and it provides more structure than a general topological space
possesses Metric spaces play a major role in the mathematical field of analysis, and
they appear in a variety of interesting applications.
1.5.1 Multiplying DNA
D(x, y) =
𝑑(𝑥,𝑦)
𝑁
Crossings occur during the process of multiplying DNA There are
3 enzymes are the Officials from multiplying DNA. And so could crossings occur for
the same chromosome and occurs changes in the properties of cryptographic rules
(switch base, add a rule, delete a rule, additions to large), but with all this, the
incidence of processing very large There is this 20 to the enzyme.
In the absence of a repair call this case the mutation and thus may be a change in the
length and characteristics of the chromosome.
And this change to be possible also in the order of the rules (genotypes).
To determine the proportion of mutation there are several ways.
To determine the change in the chromosome and the difference and similarities
between a set of chromosomes must define the function.
To know the percentage of change between the groups of chromosomes different
and the change in the length of the chromosome and the percentage of the boom and
the similarity between the chromosomes and differences will know two functions
measure of how differences and similarities between genotypes and dependence and
lack of dependence between phenotypes. And also the ratio of the occurrence of this
character in future generations and the expectation of the future generations and to
know the contrary, if this was the status of the parents or Grandparents.
Theorem 1.5.1
Let X be an non empty set of "types" and A, B be non- empty subsets of X and the
distance between A,B is
|𝑐𝐿(𝐴 )∩𝑐𝑙(𝐵)|
µ (A, B) =1-max{|𝑐𝑙(𝐴)|,|cl(𝐵 )|} Then (P(X), µ) is a metric space.
Proof
1- Since A⊆ CL(A) 𝑎𝑛𝑑 CL(A) ∩ CL(B) ⊆ CL(A) 𝑜𝑟 CL(B)
|𝐶𝐿(𝐴 )∩𝑐𝑙(𝐵)|
𝐴 = 𝐵 𝑡ℎ𝑒𝑛 𝐶𝐿(𝐴) = 𝐶𝐿(𝐵) 𝑇ℎ𝑒𝑛 𝑚𝑎𝑥{|𝑐𝑙(𝐴)|,|𝑐𝑙(𝐵 )|} ≤ 1𝑡ℎ𝑒𝑛 µ(A, B) ≥0 If
and max{|𝑐𝑙(𝐴)|, |cl(𝐵)|} = |𝑐𝑙(𝐴)| There for |𝐶𝐿(𝐴 ) ∩ 𝑐𝑙(𝐵)| = |𝑐𝑙(𝐴)|
|𝐶𝐿(𝐴 )∩𝑐𝑙(𝐵)|
𝑇ℎ𝑒𝑛 𝑚𝑎𝑥{|𝑐𝑙(𝐴)|,|𝑐𝑙(𝐵 )|} = 1𝑡ℎ𝑒𝑛 µ(A, B) =0
|𝐶𝐿(𝐴 )∩𝑐𝑙(𝐵)|
|𝐶𝐿(𝐵 )∩𝑐𝑙(𝐴)|
2-since µ (A, B)= 1-max{|𝑐𝑙(𝐴)|,|cl(𝐵 )|}=1 − max{|𝑐𝑙(𝐵)|,|cl(𝐴 )|}= µ(B, A)
Hence µ(A, B)= µ(B, A)
3-since (A ∩ 𝐵) ∩ 𝐶 = (𝐴 ∩ 𝐶) ∩ (𝐵 ∩ 𝐶) , and take closure for two side
⊆
CL ((A ∩ 𝐵)
𝐶𝐿(𝐴 ∩ 𝐶) ∩ 𝐶𝐿(𝐵 ∩ 𝐶)
Since if A⊆ B implise CL(A) ⊆ CL(B).
CL ((A ∩ 𝐵) ∩ 𝐶)) = 𝐶𝐿((𝐴 ∩ 𝐶) ∩ (𝐵 ∩ 𝐶))
⊆ 𝐶𝐿(𝐴 ∩ 𝐶) ∩ 𝐶𝐿(𝐵 ∩ 𝐶), And take cardinality for two sides
|CL ((A ∩ 𝐵) ∩ 𝐶))| ≤ |𝐶𝐿(𝐴 ∩ 𝐶)| + |𝐶𝐿(𝐵 ∩ 𝐶)|
≤ |𝐶𝐿(𝐴 ) ∩ 𝑐𝑙(𝐶)| + |𝐶𝐿(𝐵) ∩ 𝑐𝑙(𝐶)|
Dividing two sides the max{cl|(𝐶)|, cl|(𝐵 )|} + max{cl|(𝐶)|, cl|𝐴|}
|CL ((A ∩𝐵) ∩𝐶))|
There for max{|𝑐𝑙(𝐶)|,|𝑐𝑙(𝐵 )|}+max{cl|(𝐶)|,|𝑐𝑙(𝐴)|} ≤
|𝐶𝐿(𝐴 )∩𝑐𝑙(𝐶)|+|𝐶𝐿(𝐵) ∩𝑐𝑙(𝐶)|
max{|𝑐𝑙(𝐶)|,|𝑐𝑙(𝐵 )|}+max{cl|(𝐶)|,|𝑐𝑙(𝐴)|}
And
Max{|cl(A) |, |cl(B ) | }≤max{|𝑐𝑙(𝐶)|, cl|(𝐵 )|} + max{cl|(𝐶)|, cl|𝐴|}
|CL ((A ∩𝐵) ∩𝐶))|
|𝐶𝐿(𝐴 )∩𝑐𝑙(𝐶)|+|𝐶𝐿(𝐵) ∩𝑐𝑙(𝐶)|
Since max{|𝑐𝑙(𝐶)|,|𝑐𝑙(𝐵 )|}+max{|𝑐𝑙(𝐶)|,|𝑐𝑙(𝐴)|}≤max{|𝑐𝑙(𝐶)|,|𝑐𝑙(𝐵 )|}+max{|𝑐𝑙(𝐶)|,|𝑐𝑙(𝐴)|}
There for
Hence
|CL ((A ∩𝐵) ∩𝐶))|
≤
max{|𝑐𝑙(𝐶)|,|𝑐𝑙(𝐵 )|}+max{|𝑐𝑙(𝐶)|,|𝑐𝑙(𝐴)|}
|𝐶𝐿(𝐴 )∩𝑐𝑙(𝐶)|
max{|𝑐𝑙(𝐶)|,|𝑐𝑙(𝐵 )|}+max{|𝑐𝑙(𝐶)|,|𝑐𝑙(𝐴)|}
|𝐶𝐿(𝐵 )∩𝑐𝑙(𝐶)|
max{|𝑐𝑙(𝐶)|,|𝑐𝑙(𝐵 )|}+max{|𝑐𝑙(𝐶)|,|𝑐𝑙(𝐴)|}
|CL ((A ∩𝐵) ∩𝐶))|
≤
max{|𝑐𝑙(𝐶)|+|𝑐𝑙(𝐵 )|}+max{|𝑐𝑙(𝐶)|+|𝑐𝑙(𝐴)|}
|𝐶𝐿(𝐴 )∩𝑐𝑙(𝐶)|
+
max{|𝑐𝑙(𝐶)|+|𝑐𝑙(𝐴)|}
|𝐶𝐿(𝐵 )∩𝑐𝑙(𝐶)|
max{|𝑐𝑙(𝐶)|+|𝑐𝑙(𝐵 )|}
Since
CL ((A ∩ 𝐵) ∩ 𝐶)) ⊆ 𝐶𝐿(𝐴 ∩ 𝐵) ⊆ 𝐶𝐿(𝐴) ∩ 𝐶𝐿(𝐵) ) there for
+
|CL ((A ∩ 𝐵) ∩ 𝐶))| ≤ |𝐶𝐿(𝐴 ) ∩ 𝑐𝑙(𝐵)|
And
Hence
Max{|𝑐𝑙(𝐴)|, |𝑐𝑙(𝐵 )|} ≤ max{|𝑐𝑙(𝐶)| + cl|(𝐵 )|} + max{cl|(𝐶)| + cl|𝐴|}
|CL ((A ∩𝐵) ∩𝐶))|
max{|𝑐𝑙(𝐶)|+|𝑐𝑙(𝐵 )|}+max{|𝑐𝑙(𝐶)|+|𝑐𝑙(𝐴)|}
≤
|CL ((A ∩𝐵) ∩𝐶))|
Max{|𝑐𝑙(𝐴)|,|𝑐𝑙(𝐵 )|}
Multiply by (-1)
−
1−
|CL ((A ∩ 𝐵) ∩ 𝐶))|
max{|𝑐𝑙(𝐶)| + |𝑐𝑙(𝐵 )|} + max{|𝑐𝑙(𝐶)| + |𝑐𝑙(𝐴)|}
|𝐶𝐿(𝐴 ) ∩ 𝑐𝑙(𝐶)|
|𝐶𝐿(𝐵 ) ∩ 𝑐𝑙(𝐶)|
≥ −{
+
}
max{|𝑐𝑙(𝐶)| + |𝑐𝑙(𝐴)|}
max{|𝑐𝑙(𝐶)| + |𝑐𝑙(𝐵 )|}
|CL ((A ∩𝐵) ∩𝐶))|
max{|𝑐𝑙(𝐶)|+|𝑐𝑙(𝐵 )|}+max{|𝑐𝑙(𝐶)|+|𝑐𝑙(𝐴)|}
1 −{
1−{
≥
|𝐶𝐿(𝐴 ) ∩ 𝑐𝑙(𝐶)|
|𝐶𝐿(𝐵 ) ∩ 𝑐𝑙(𝐶)|
+
}
max{|𝑐𝑙(𝐶)| + |𝑐𝑙(𝐴)|}
max{|𝑐𝑙(𝐶)| + |𝑐𝑙(𝐵 )|}
|𝐶𝐿(𝐴 ) ∩ 𝑐𝑙(𝐶)|
|𝐶𝐿(𝐵 ) ∩ 𝑐𝑙(𝐶)|
+
} < 1𝑎𝑛𝑑
max{|(𝐶)| + |𝐴|}
max{|(𝐶)| + |(𝐵 )|}
|CL ((A ∩ 𝐵) ∩ 𝐶))| ≤ |𝐶𝐿(𝐴 ) ∩ 𝑐𝑙(𝐶)| 𝑎𝑛𝑑 |𝐶𝐿(𝐵 ) ∩ 𝑐𝑙(𝐶)|
1−
|CL ((A ∩𝐵) ∩𝐶))|
max{|(𝐶)|+|(𝐵 )|}+max{|(𝐶)|+|𝐴|}
|𝐶𝐿(𝐴 )∩𝑐𝑙(𝐶)|
2 − { max{|(𝐶)|+|𝐴|} +
1−
|𝐶𝐿(𝐵 )∩𝑐𝑙(𝐶)|
max{|(𝐶)|+|(𝐵 )|}
< 1 𝑇ℎ𝑒𝑛
|𝐶𝐿(𝐵 )∩𝑐𝑙(𝐶)|
max{|(𝐶)|+|(𝐵 )|}
+1−
} ≥ 1−
|𝐶𝐿(𝐴 )∩𝑐𝑙(𝐶)|
max{|(𝐶)|+|𝐴|}
|CL ((A ∩𝐵) ∩𝐶))|
max{|(𝐶)|+|(𝐵 )|}+max{|(𝐶)|+|𝐴|}
≥ 1−
|CL ((A ∩𝐵) ∩𝐶))|
max{|(𝐶)|+|(𝐵 )|}+max{|(𝐶)|+|𝐴|}
Since (A ∩ 𝐵) ∩ 𝐶 ⊆ (𝐴 ∩ 𝐵)implies𝑐𝑙{(A ∩ 𝐵) ∩ 𝐶} ⊆ 𝑐𝑙(𝐴 ∩ 𝐵)
|𝑐𝑙{(A ∩ 𝐵) ∩ 𝐶}| ≤ |𝑐𝑙(𝐴 ∩ 𝐵)| ≤ |𝐶𝐿(𝐴 ) ∩ 𝑐𝑙(𝐵)|
|𝑐𝑙{(A ∩ 𝐵) ∩ 𝐶}|
max{|(𝐶)| + |(𝐵 )|} + max{|(𝐶)| + |𝐴|}
|𝐶𝐿(𝐴 ) ∩ 𝑐𝑙(𝐵)|
≤
max{|𝑐𝑙(𝐶)| + |cl(𝐵 )|} + max{cl|(𝐶)| + |cl𝐴|}
|𝐶𝐿(𝐴 )∩𝑐𝑙(𝐵)|
≤ max{|𝑐𝑙(𝐴)|,cl|(𝐵 )|}
|𝑐𝑙{(A ∩𝐵) ∩𝐶}|
≥ 1−
max{|(𝐶)|+|(𝐵 )|}+max{|(𝐶)|+|𝐴|}
There for 1 −
|𝐶𝐿(𝐵 )∩𝑐𝑙(𝐶)|
max{|(𝐶)|,|(𝐵 )|}
+1−
|𝐶𝐿(𝐴 )∩𝑐𝑙(𝐵)|
max{|𝑐𝑙(𝐴)|+|cl(𝐵 )|}
|𝐶𝐿(𝐴 )∩𝑐𝑙(𝐶)|
max{|(𝐶)|,|𝐴|}
≥ 1−
≥1−
|CL ((A ∩𝐵) ∩𝐶))|
max{|(𝐶)|,|(𝐵 )|}+max{|(𝐶)|,|𝐴|}
|𝐶𝐿(𝐴 ) ∩ 𝑐𝑙(𝐵)|
max{|𝑐𝑙(𝐴)| + cl|(𝐵 )|}
Hence 1 −
|𝐶𝐿(𝐵 )∩𝑐𝑙(𝐶)|
max{|(𝐶)|+|(𝐵
+1−
)|}
|𝐶𝐿(𝐴 )∩𝑐𝑙(𝐶)|
max{|(𝐶)|+|𝐴|}
≥ 1−
|𝐶𝐿(𝐴 )∩𝑐𝑙(𝐵)|
max{|𝑐𝑙(𝐴)|+cl|(𝐵 )|}
There for µ(B, C)+ µ(A, C) ≥ µ(B, A)
Proposition1.5.1
Let X be an non empty set of "types" and A, B be non- empty subsets of X and
|𝐶𝐿(𝐴 )∩𝑐𝑙(𝐵)|
µ∗ (𝐴, 𝐵) = max{|𝑐𝑙(𝐴)|,|cl(𝐵 )|} Then (P(X) µ∗ ) is a metric space?
Proof. Obvious
Proposition1.5.2
Let A, B, C be non- empty subsets of X. Then
(i)µ∗ (𝐴 ∩B, C) ≤ µ∗ (A, C) + µ∗ (B, C)
(ii)µ (𝐴 ∩B, C) ≤ µ (A, C) + µ (B, C)
Proof
(i) Since (A ∩ 𝐵) ∩ 𝐶 = (𝐴 ∩ 𝐶) ∩ (𝐵 ∩ 𝐶)
CL ((A ∩ 𝐵) ∩ 𝐶) = 𝐶𝐿((𝐴 ∩ 𝐶) ∩ (𝐵 ∩ 𝐶))
⊆ 𝐶𝐿(𝐴 ∩ 𝐶) ∩ 𝐶𝐿(𝐵 ∩ 𝐶)
Since if A⊆ B IMPLIES CL(A) ⊆ CL(B).
CL ((A ∩ 𝐵) ∩ 𝐶)) = 𝐶𝐿((𝐴 ∩ 𝐶) ∩ (𝐵 ∩ 𝐶))
⊆ 𝐶𝐿(𝐴 ∩ 𝐶) ∩ 𝐶𝐿(𝐵 ∩ 𝐶), and take cardinality for two sides
|CL ((A ∩ 𝐵) ∩ 𝐶))| ≤ |𝐶𝐿(𝐴 ∩ 𝐶)| + |𝐶𝐿(𝐵 ∩ 𝐶)|
≤ |𝐶𝐿(𝐴 ) ∩ 𝑐𝑙(𝐶)| + |𝐶𝐿(𝐵) ∩ 𝑐𝑙(𝐶)|
Dividing two sides the max{cl|(𝐴 ∩ 𝐵)|, cl|(𝑐)|}
|CL ((A ∩𝐵) ∩𝐶))|
There for max{cl|(𝐴∩𝐵)|,cl|(𝑐)|} ≤
|𝐶𝐿(𝐴 )∩𝑐𝑙(𝐶)|+|𝐶𝐿(𝐵) ∩𝑐𝑙(𝐶)|
max{cl|(𝐴∩𝐵)|,cl|(𝑐)|}
So
that |CL ((A ∩ 𝐵) ∩ 𝐶))|
≤ |𝐶𝐿(𝐴 ) ∩ 𝑐𝑙(𝐶)| + |𝐶𝐿(𝐵) ∩ 𝑐𝑙(𝐶)|
Then max |CL ((A ∩ 𝐵) ∩ 𝐶))|≤max|𝐶𝐿(𝐴 ) ∩ 𝑐𝑙(𝐶)| +max|𝐶𝐿(𝐵) ∩ 𝑐𝑙(𝐶)|
Max{|(𝐶)|, |(𝐵 ∩ 𝐴)|} ≤ max{|(𝐶)|, |(𝐵 )| + |𝐴|}
= max{|(𝐶)|, |(𝐵 )|} + max{|(𝐶)|, |𝐴|}
Then
Max{|𝑐𝑙(𝐶)|, |𝑐𝑙(𝐵 ∩ 𝐴)|}≤
max{|cl(C) |, | cl(B ) | }+max{|cl(𝐶)|, lc|𝐴|}
Max{|𝑐𝑙(𝐴)|, |𝑐𝑙(𝐵 )|}≤max{|𝑐𝑙(𝐶)|, cl|(𝐵 )|} + max{cl|(𝐶)|, cl|𝐴|}
|CL ((A ∩𝐵) ∩𝐶))|
Since max{cl|(𝐴∩𝐵)|,cl|(𝑐)|} ≤
There for
|𝐶𝐿(𝐴 )∩𝑐𝑙(𝐶)|+|𝐶𝐿(𝐵) ∩𝑐𝑙(𝐶)|
|CL ((A ∩𝐵) ∩𝐶))|
max{cl|(𝐴∩𝐵)|,cl|(𝑐)|}
|CL ((A ∩𝐵) ∩𝐶))|
Hence max{cl|(𝐴∩𝐵)|,cl|(𝑐)|} ≤
max{cl|(𝐴∩𝐵)|,cl|(𝑐)|}
|𝐶𝐿(𝐴 )∩𝑐𝑙(𝐶)|+|𝐶𝐿(𝐵) ∩𝑐𝑙(𝐶)|
≤ max{|𝑐𝑙(𝐶)|,cl|(𝐵 )|}+max{cl|(𝐶)|,cl|𝐴|} +
|𝐶𝐿(𝐴 )∩𝑐𝑙(𝐶)|
+
max{|𝑐𝑙(𝐶)|,|𝑐𝑙(𝐴)|}
|𝐶𝐿(𝐵 )∩𝑐𝑙(𝐶)|
max{|𝑐𝑙(𝐶)|,|𝑐𝑙(𝐵 )|}
There for µ∗ (𝐴 ∩B, C) ≤ µ∗ (A, C) + µ∗ (B, C)
(ii) is obvious from therm4.7.1
Proposition1.5.3
Let (P(X), µ) ,( P(X), µ∗ ) be two metric spaces on a nonempty set X, A⊆
C, A, B and C subsers of X, Then
(i) µ(A,B)≥ µ(𝐵, 𝐶)
ii) µ∗ (A,B)≤ µ∗ (𝐵, 𝐶)where 𝐵 ⊄ 𝐴.
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