A NEW SET THEORY FOR ANALYSIS

We present the real number system as a generalization of the natural numbers. First, we prove the co-finite topology, Co f (N), is isomorphic to the natural numbers. Then we prove the power set 2Z contains a subset isomorphic to the non-negative real numbers, with all its defining structure of operations and order. Finally, we provide two different constructions of the entire real number line, and we see that the power set 2N can be given the defining structure of R. We give simple rules for calculating addition, multiplication, subtraction, division, powers and rational powers of real numbers, and logarithms. The supremum and infimum are explicitly constructed by means of a well defined algorithm that ends in denumerable steps. In section 5 we give evidence our construction of N and R are canonical; these constructions are as natural as possible. In the same section, we propose a new axiomatic basis for analysis. This answers Benacerraf’s identification problem by giving a canonical representation of numbers as sets. In the last section we provide a series of graphic representations and physical models that can be used to represent the real number system. We conclude that the system of real numbers is completely defined by the order structure of N.


Introduction
In building the continuum we make use of properties of integers and sets.Apart from this, we assume the basic concepts of category theory.Mainly, the concept of isomorphism between categories.Background knowledge on previous axiomatic constructions of the real numbers will be of help.The more modern constructions of the real number system can be found in the references [1][2][3][4][5].It is notable that the real number system has been studied in detail through the generations, and still new insights and more useful constructions are sought.The mathematical objects that have previously been denominated as real numbers, are objects of complex and illusive structure.The mathematician has always had to recur to advanced tools and objects in building the real number structure.In the words of Dr. K. Knopp [6](p.4) "...these preliminary investigations are tedious and troublesome, and have actually, it must be confessed, not yet reached any entirely satisfactory conclusion at all." The construction provided here, stands out in the fact that we do not have to build a new structure.We prove the real number system is isomorphic in terms of structure (not only cardinality) to 2 N .One can find many bijective functions from the set of real numbers to the power set of integers, but it is completely different to prove they have the same structure.Of course, the structure of 2 N can be defined artificially in terms of any bijection, but that does not give us any new information or computational advantages.Our work is to define a simple structure for 2 N and prove this structure is equivalent to the real number system as a totally ordered field with the property of the supremum element (completeness).In other wrods, we provide a set of rules that define order and operations <, ⊕, ⊗ in 2 N , in such a way that F a < F b if and only if a < b, and F (a + b) = F a ⊕ F b and F (a • b) = F a F b, for a bijection F : R → 2 N .
Real numbers are usually presented axiomatically as a set that satisfies certain rules.In undergraduate school, constructions of this system are rarely taught, even in advanced courses of analysis.This leads to a certain gap in the learning of the student.This problem is one of major motivations of the present work.We first provide an isomorphic structure to N, by defining order and operations for the closed sets of the cofinite topology Co f (N).A similar construction is given for the discrete topology 2 N ; the continuum [0, 1] is isomorphic in structure to the collection of all subsets of N. We generalize these constructions to obtain a structure isomorphic to R + .We define order and operation for the closed sets of Z, where Z is the topology of Z whose closed sets are the subsets of Z that are bounded above.In other words, non-negative real numbers are represented by subsets of Z that are bounded above.We define order, addition and product for these objects.

Motivation
Every positive integer has unique representation as sum of natural powers of 2. This expression is usually treated as a sequence of 0s and 1s.A 0 in the n-th place indicates that 2 n is not a summand in the expression.A digit 1 in that same place, would indicate that power is indeed a summand.This is the usual binary expression of natural numbers.We will refer to finite subsets of N as set numbers.Given the binary representation of a positive integer N, we can naturally assign a set number to it.The elements of the set number are the natural numbers that correspond to spaces in the sequence with a digit 1.Where a 0 would appear we say that integer does not belong to the set number.For example, the number 5 = (. . ., 0, 0, 0, 1, 0, 1) is assigned the set number {0, 2} because the 0-th and 2-nd space are occupied by the digit 1.The number 13 = (. . ., 0, 0, 0, 1, 1, 0, 1) will be mapped to the set number {0, 2, 3} and the number 17 = (. . ., 0, 0, 0, 1, 0, 0, 0, 1) would be given the set number {0, 4}.Given set numbers a, b, we say a < b if max(a b) ∈ b, where a b = (a ∪ b) − (a ∩ b) is the symmetric difference of the two sets.This order is isomorphic to the order of natural numbers N < .
Next, we define a sum operation for set numbers, in the form of a recursive formula that ends in finite steps.We express the sum of two set numbers a, b as the sum of two new sets: the symmetric difference and the intersection, with a slight change.We add 1 to the elements of the intersection to form a new set, s(a ∩ b).In other words, to add two numbers, we add the powers of 2 that are not repeated, then we add the powers of 2 that appear repeated.In terms of set numbers, this means we carry out the set number addition between the symmetric difference, and the intersection (with the elements of the intersection increased by 1).We add 1 to the elements of a ∩ b, because in terms of natural numbers it is equivalent to multiplication by 2. Thus, our formula a ⊕ b = (a b) ⊕ s(a ∩ b).The function s : 2 N → 2 N adds 1 to the elements of the argument; in specific s∅ = ∅ and s{0} = {1}.The set number sum 13 In general, the process ends in finite steps.The sum of two sets, a, b is equal to the sum of the sets a = a b and b = s(a ∩ b).The sum of these two is in turn equal to the sum of a = a b and b = s(a ∩ b ), etc....This process ends when b (n) becomes the empty set (after a finite number of iterations).We give another example, 15 This process can be modeled with particles occupying energy levels.A set number will be represented by a column with numbered energy levels, and any given arrangement of particles occupying levels (with at most one particle on each level).To perform addition of two columns, we give one rule: two particles in the same level are replaced by a single particle, one level up.Let us describe this in detail.Given two columns A, B, we call the ordered pair S = (A, B) a state.Thus, a state is determined by two columns with occupied levels.Call the levels of the left column A i and the levels of the right column B j , so that A 4 means the fourth level of the left column and B 0 means the lowest (or 0-th) level of the right column.
We give a system that evolves discretely in time, with our rule for simplifying columns.To go from a state t n to t n+1 , we form two new columns.The left column of state t n+1 is occupied in the energy levels that were not repeated in the previous state t n .The right column of state t n+1 is occupied in the energy level B j+1 if the energy levels A j and B j where both occupied during the state t n .To find the set number sum of two set numbers, we merge the two columns into a single column, by applying our sum formula.The observation is that the right column becomes the empty set in finite steps.We define x ⊕ ∅ = ∅ ⊕ x = x.See Figure 1.
It can be proven the state stabilizes in a finite number of steps.That is to say, there exists a state S(t n ) such that S(t k ) = S(t n ) for all k ≥ n.Say A n i represents the level A i at time t n .Then stability means for all i, j ≥ 0 and for all n ≥ k, it is true A n i = A k i and B n j = 0. Notice, the same diagram is valid under vertical displacements.In figure 2 we illustrate the fact that the same system of states from Figure 1 holds under displacement of the energy levels (Figure 2).
Real numbers in the unit interval can be expressed as the sum of negative powers of 2. We use the same rules, now with arbitrary sets of negative integers.For example, 1  2 3).
If we rearrange the levels so that a = {0} and b = {0, 1}, then Figure 3 is a graphic representation of 1 + 3 = 4.We have denumerable representations of sums, actually.Observe that ∑ i<n 2 i = 2 n , for every integer n.This is equivalent to defining {i} n−1 i=−∞ = {n}.It comes from an iteration of Figure 3, and we will see why this is consistent (Figure 4).
A subset A ⊆ Z, with maximum in Z, is a positive real number where the integer part is given by the (finite) intersection N ∩ A and the fractional part is given by −N ∩ A.

N ∼ = Co f (N)
Let A ⊂ N be a closed set of the cofinite topology of N. Equivalently, A has a maximum and we say that A is a set number.

Order
Given two set numbers A = B, we define an order relation < between them.We say A < B if max(A B) ∈ B. In the contrary case, max(A B) ∈ A and we define B < A. Every two set numbers are comparable because the symmetric difference is non-empty and has maximum.It is not difficult to see that we have a well ordering on the family of set numbers.We give three examples in Figure 5.

Addition
We define the sum of two set numbers as a recursive formula that ends in finite steps.When adding powers of two, we have the following simple rule 2 n+1 = 2 n + 2 n .So, we ask the sum obey {x} ⊕ {x} = {x + 1}, for every x ∈ N. Given any set number A = {x} x∈A ⊂ N, we define sA = {x + 1} x∈A .Then, define the operation as Before complicating things more, we stop here to see what is happening.
Then the previous equalities can be rewritten as We are calculating C k ⊕ D k as a sum C k+1 ⊕ D k+1 , where the term D k+1 is dependent on the intersection.We call the term D k+1 the remainder, and although it is not getting smaller in value (as natural number), the cardinality (as set number) does go to 0. However, it can be the case that the cardinality of the remainder does not get smaller.It may stay constant for finite iterations.It is guaranteed that in a finite number of steps, the remainder becomes the empty set, and our result is now obvious because we have iterated the formula until we reach A ⊕ B = C k+1 ⊕ D k+1 , where D k+1 = ∅.The system stabilizes when It is left to the reader to prove 1) , for any set numbers X, Y where the y i are the elements of Y.The proofs of the algebraic properties are not all trivial.It is not trivial to prove associativity for ⊕.The best thing to do to avoid direct proofs of these properties, is to prove , where F is the isomorphism between natural numbers and set numbers.

Product
In defining the product, consider that multiplying by 2 is equivalent to adding +1 to the elements of the set number.For example, 2 * 14 = {1} {1, 2, 3} = {1 + 1, 2 + 1, 3 + 1} = {2, 3, 4} = 28.In general, 2 • x = s(X), where X is the set number corresponding to x ∈ N. Recall s(X) = {a + 1} a∈X .Multiplication by 4, is equivalent to adding +2 to all the elements of X.In general, multiplication by 2 n is equivalent to adding n, to the elements of X.That is to say, 2 n • x = {a + n} a∈X .The unit of this operation is the set {0}.To multiply numbers that are not powers of 2, we define using distributivity.
To define the product of two set numbers A B, we will refer to A as the pivot and B as the base.The product is the set number sum, of the set numbers obtained by displacing the base; the displacements of the base correspond to elements of the pivot.So to give another example, if we wish to multiply 32 and 12, and take 32 as pivot and 12 as base, we proceed as follows.We know 32 = {5} has one element so that we add 5 to the elements of {2, 3} to obtain {7, 8} which is indeed 384.If we chose 32 to be the base and 12 to be the pivot, then we would have to add the set numbers {5 + 2} ⊕ {5 + 3} = {7} ⊕ {8} = {7, 8}.The product is then expressed by: Then the formula is the set number sum {a 1 + b} b∈B ⊕ {a 2 + b} b∈B ⊕ • • • ⊕ {a n + b} b∈B , where the set {a i + b} b∈B = {a i + b 1 , a i + b 2 , . . ., a i + b m }.Developing the expression we get Figure 7 illustrates that the product formula is also valid for negative integers.
The reader can prove F (a • b) = F a F b. which will be useful in proving commutativity and associativity of product A B = B A. Commutativity can be rewritten as Direct proofs (in terms of sets) of the set sum are difficult, but direct proofs of the product for set numbers seems to be almost impossible.For instance, in proving commutativity, we have to prove that the set number sum of n sets (each with cardinal number m), is equal to a set number sum of m sets (each with cardinal number n).Again, the only easy way out of this is to use the isomorphism F .

Subtraction
We wish to give forms of inverse operating elements, namely we wish to provide well defined algorithms for finding the difference between two object A B and their quotient A ⊗ B in such a way that the corresponding operation in real numbers is equal.In this section we will concentrate on finding an algorithm (and, ultimately, a well defined formula) for subtraction.
Before defining the general case, let us begin with two set numbers subject to the relation A ⊂ B. Then the subtraction is defined by B A = B − A, where B − A is the usual set difference.So that if we put the two columns side by side, the result B − A is obtained by taking away the particles in the column of B that also appear in A. For example, 45 = {0, 2, 3, 5} and 9 = {0, 3} is a subset of 45, so that we can easily find 45 Next step to generality is to consider two set numbers A < B such that max A < max B in the strict sense.We must use our basic rule of addition, now in reverse order, so that we can take away particles from any level of the column B. We have described our basic rule as {x for any set number X and any k ∈ X.Now, we wish to find a set number X such that A ⊕ X = B, and to do this we rewrite B: Let A < B two set numbers, and let

This is represented in
This defines subtraction in the most general case (Figures 11-13).

Continuum
We extend the family of set numbers to include sets that represent reciprocals of natural numbers.The unit interval continuum [0, 1] is the power set 2 −N .First, we prove that every subset of 2 −N has supremum and infimum.In particular sup(2 −N ) = −N < {0}, and {0} is the smallest set number larger than −N.We define {0} = −N without fear of contradiction.This translates to ∑ i∈N 2 −i = 2 0 = 1.In general, ∑ i<n 2 i = 2 n , or equivalently, {n} = {i} i<n , for any integer n.This means a bounded (above and below) subset of Z has two representations.The inverse of After describing the unit interval, we prove Z ∼ = R + .The closed sets of the topology Z are A ⊂ Z that are bounded above.The integer part of A ∈ R + is given by A ∩ N while its fractional part is A ∩ −N.
Supremum and Infimum.We define a limiting process that defines a supremum function in the continuum [0, 1].The order is defined as before, two sets relate A < B if and only if max(A B) ∈ B. The only change is that A, B are arbitrary subsets of −N, instead of finite sets of N. We must prove that the supremum of any family of subsets of −N exists.
First, we give the intuitive idea for the more general case of finding the supremum of a bounded above set X, of positive real numbers A i ⊂ Z.The family of sets X is bounded above in the order of set numbers and every set number is also bounded above in the order of integers.Therefore x 1 = max X exists.The supremum of X is among the sets A j that contain the integer x 1 .We look to see which of those set numbers contains the second largest integer appearing in the family.Let x 2 = max j A j where the index j include only the elements of X such that x 1 ∈ A j .Now we know the supremum is among the sets that contain both x 1 and x 2 .This procedure continues for denumerable steps and generates a denumerable set of integers The set of these integers is sup X.Now we formally discuss this procedure for an arbitrary set of real numbers in the unit interval.Let X ⊆ 2 −N .This means A ⊆ −N for every A ∈ X.Because of the well ordering principle, we know x 1 := max X exists.Define the sub family Y 1 := {A ∈ X|x 1 ∈ A}, of set numbers containing x 1 , and x 2 }, and x 3 := max X 2 .Continue in this manner, so that x n+1 := max X n , where The supremum is defined as sup X = {x i } i , which is by construction greater than every set in X, and it is the least subset of Z that is greater than all the sets of X.In a countable number of steps we have determined a unique set number (Figure 14). 2 2 −N represents the power set, of the power set of −N.The supremum function is a function of the form sup : 2 2 −N → 2 −N .If sup(X) ∈ X, we say sup(X) = max(X).In particular, this is true if X is finite.There is also an infimum function of the form inf : 2 2 −N → 2 −N , where the image of the family is its greatest lower bound.Given a family X ⊆ 2 −N , define X * as the family of set numbers that are smaller than all the elements of X.In other words a set number A ⊆ −N is in X * if and only if A < Y, for all Y ∈ X. Define inf(X) = sup(X * ).If X * = ∅, then ∅ ∈ X, and inf(X) = ∅.In particular, inf[0, 1] = 0. To calculate the infimum, look for the largest integer x 1 such that x 1 < A, for all A ∈ X.If there is no such integer (the elements of X are arbitrarily small) define inf(X) = ∅.If x 1 exists and {x 1 } ∈ X, then inf(X) = {x 1 }.If x 1 exists and {x 1 } / ∈ X, then we compare {x 1 , ∈ inf(X), and we would then have to verify if x 1 − 2 ∈ inf(X).The process is ends in denumerable steps, with a set inf(X) = {x i } i ⊆ −N.
Reciprocals.If we want to multiply A ⊂ N by B ⊆ −N, we use distributivity.Define the product as a natural generalization of the last definition of product.The index of the operation is now a subset of −N.Refer to Figure 7.The unit of product, {0}, is bounded and therefore has a second representation, −N.This is the representation used in providing a well-defined algorithm for finding the reciprocal of a natural number.We illustrate our motivation with an example.To find the inverse (under set product) of the set number 137 = {0, 3, 7}.We seek out a set number 1 If we propose 1 N = {−7} we find we "go over" because: Let us describe the procedure for finding the reciprocal of a set number N = {x 1 , x 2 , . . ., We wish to find a set number 1  N ⊂ −N such that N 1 N = −N.The reciprocal 1  N = {y 1 , y 2 , . . .} ⊂ −N will be found in denumerable steps, as follows.First, make y 1 = −(x 1 + 1) so that Next find the largest negative integer y 2 such that We know there is at least one negative integer that satisifes this inequality.By the well ordering principle, we can find the maximum of such a set of solutions.This maximum is our integer y 2 .The finitude of #(N) plays a fundamental role in allowing us to guarantee such y 2 exists.Now, to find y 3 , we look for the largest negative integer such that If we continue in this manner we have well defined 1 N .

R + ∼ = Z
Topology of Bounded Subsets of Z.A positive real number is a subset of Z, bounded above.The intersection with the negative integers represents the fractional part, and the finite intersection with the positive integers represents the integer part.We use the continuum [0, 1], built in the last section, to produce a series of copies, squeezed in between consecutive natural numbers.We do not need to make any modifications to the basic rules and relations of order and operation already used.We extend the same relations to the closed sets of the topology Z.For example, given a binary representation 110101.00101 1 , the set number is {5, 4, 2, 0, −3, −5}.We have a natural bijection between binary sequences and closed sets of Z.The bijection is A → ∑ i∈A 2 i for every closed set A of Z.The positive real line is constructed by piecing together copies of [0, 1].Let A ⊂ −N be the corresponding set number to x ∈ [0, 1), so that x = ∑ i∈A 2 i .Then A ∪ {0} is the set number corresponding to 1 + x = 2 0 + ∑ i∈A 2 i .More generally, let N ⊂ N be the set number corresponding to a natural number n ∈ N; this means n = ∑ i∈N 2 i .Now we have n We can summarize our work as follows: i. N ∼ = Co f (N) where the cofinite topology uses the order and operations of set numbers.
iii.The continuum of non-negative real numbers is built as a natural generalization of both Co f (N) and 2 −N .We piece together [0, 1], [1,2], [2,3], . . ., into a single continuum [0, ∞).This is done by considering the upper bounded sets of Z and a proper extension of the set number relations.
We generalize the previous methods into a single structure isomorphic to R + .The main difference between our construction of the real numbers is that we do not build a complex space which we will have great difficulty in understanding.Our approach here is to prove the simplest set with cardinality of the continuum can be given the structure of positive real numbers, in a natural manner.In fact, gives reason to speculate it is a canonical construction of R.
Supremum.In the previous section we provided a well defined algorithm for finding the supremum of a family X ⊆ [0, 1], where the elements of X are arbitrary subsets of −N.Now, we generalize this process to define the supremum of a bounded set of positive real numbers.
Observe max X exists if and only if X is a bounded family, of bounded above subsets of Z.That is to say, sup(X) exists if and only if there exists n ∈ Z such that for all A ∈ X it is true A < {n}.Notice it is not the same as saying "X is a set of bounded above subsets of Z"; this would be written as "...for all A ∈ X there exists n ∈ Z such that A < {n}".We need to guarantee the existence of max X in order to find the supremum of X.This is done by asking that X be a bounded above subset of Z (each of the elements of the set are bounded above subsets of Z).For example, there is no max X for X = {{1, 0, −1, −2, . . .}, {2, 1, 0, −1, . . .}, {3, 2, 1, 0, . . .}, {4, 3, 2, 1, . . .}, . . .}, although every element of X has a maximum.
Division.We now describe division and rational numbers.Let us find the rational representation of the set number {5, 0, −2, −4, −5} = 33.34375.Add 5 to the elements; the result is {10, 5, 3, 1, 0} = 1067.The action of adding 5 to the elements of a set number is actually multiplication by 2 5 , so that {5, 0, −2, −4, −5} = 1067 32 .This fraction is said to be irreducible because 32 = {5} is the smallest natural number that we can multiply with {5, 0, −2, −4, −5} and obtain a subset of N, which is the numerator 1067 = {10, 5, 3, 1, 0}.Given a finite subset of integers, we give an infinite collection of ordered pairs of natural numbers, which will be called fractions.In other words, given a finite set number A ⊂ Z, we can give infinite, but equivalent, representations of a single irreducible fraction m n .The set numbers N, M, corresponding to n, m are easy to find; N ⊂ N is the smallest set number such that N A ⊂ N.Then, M is the result of N A. More generally, any set number B ⊂ N such that B A ⊂ N, allows the construction of a fraction α β , where α = ∑ i∈(A B) 2 i and β = ∑ i∈B 2 i .Consider the inverse problem of finding the set number corresponding to an ordered pair m n .We use our algorithm for finding the inverse, 1/N, and multiply that with M. We will approximate 51 137 , by multiplying 1  137 ≈ {−8, −9, −10, −12} with 51 = {5, 4, 1, 0}, to obtain 0.361083984375 = {−2, −4, −5, −6, −10, −11, −12}.For more precision, we must give a better approximation to the number 1 137 ; for example 1 137 ≈ {−8, −9, −10, −12, −13, −14, −17}.Now, we know how to provide an irreducible fraction, given a set number.We can also approximate the set number corresponding to any given fraction of natural numbers.Now let us make our first consideration as far as infinite subsets of Z are concerned.Let A ⊂ Z be a set number with A ∩ −N infinite.Then we are unable to give a set number B ⊂ N such that A B ⊂ N, with one crucial exception.If the set A ∩ −N is infinite and periodic, then we can find a finite set B ⊂ N such that A B ⊂ N. In other words, infinite but periodic set numbers have rational representation as an ordered pair of natural numbers.Any set number that can be identified with two natural numbers, as just described, is called a rational number.Approximate 137 51 by two methods.First, finding 1/51 and multiplying that by {7, , and in a similar manner B m .The sum is defined as A ⊕ B := sup n,m (A n ⊕ B m ) for all n, m ∈ N. The reader can define the multiplication of two infinite set numbers.
Powers.To take powers of set numbers A B , we start by defining A B , where A, B ⊂ N. In this case, X = A B is the result of carrying out successive products of set numbers.The empty set is representing the integer 0, so we define the power A ∅ = {0}.We define the power A {0} = A because {0} = 1.With this, we are able to give a recursive formula A B = A A B {0} .To find power of 127 4 = {0, 1, 2, 3, 4, 5, 6} {2} we first reduce the expression.We know A {2} = A A {2} {0} .We know, from the subtraction of set numbers, that {2} {0} = {0, 1}, then .For now, taking negative powers is only notation for symbolizing the reciprocal of another number, so we have nothing to do in that respect because A −3 is simply 1/A 3 = 1/A {0,1} .When we define negative real numbers, this will make sense as we will extend our definition.When taking powers, we notice two different behavior: 1) if A < {0} = 1 then A X is a decreasing function 2) if A > {0} = 1 then A X is an increasing function.Proving A X is decreasing (or increasing, depending on the case) is not difficult but requires some labor.Before proving it for arbitrary set numbers X, it has to be proven for X ⊂ N and 1/X.
Roots.The process of finding a root is to take a reciprocal power, that is to say A 1 B , for some finite set B ⊂ N. To find X = A 1 B we must find a set X ⊂ Z such that X B = A. Consider the set P(A, B) of all set numbers X such that X B < A, then this set is bounded above.We define A 1 B = sup P. Notice we have three cases: 1) A > {0}, 2) A = {0}, 3) A < {0}.In the first case, A We illustrate this with an example to find 3 1/4 = {0, 1} 1/{2} .If we elevate {0, −1} to the fourth power, the result is greater than 3 = {0, 1}.But, we find that {0, −2} 4 < {0, 1}.In the next step we find {0, −2, −3} > {0, 1}.Later we find {0, −2, −4} < {0, 1}.We continue in this manner, with trial and error to elevate sets to fourth power all the time using larger numbers but making sure the result does not go over 3.If we are to take a rational power A B , in the sense that B is a periodic set as mentioned above, we find its irreducible fraction B = m n .Now we can say A B = (A m ) 1 n is well defined, since it can be proven (A m ) Consider next the more general case where B is not rational.Consider A, B ⊂ Z to be arbitrary set numbers, and the elements of course, for this definition to be justified, we have to prove the set {A Q k } ∞ k=1 is bounded above.Hint: prove the power function A X is increasing with X, then it suffices to show {B k } ∞ k=1 is bounded above by {b 1 + 1} > B k for all k.
Logarithms.In the last section, we only extended the definition of powers.Now, we explore the inverse function.To find log B A we find a set number X such that B X = A. It is not difficult to show we have two defining cases for the logarithm.Consider B > {0} = 1, and A > {0}.Then X > 0 because there is a positive real number X such that B X = A. If A < {0}, then X is negative, because the negative powers of B map to numbers < {0} = 1.For the case B < {0} we have the contrary arrangement; if A > {0} then X < 0, and if A < {0} then X > 0. Let us calculate log 2.5 3.125 which is the logarithm base 2.5 of 3.125; the numerical value is ≈ 1.24353.We know B = {−1, 1} and A = {−3, 0, 1}.We wish to find a set number X = n m such that (B n ) We begin by calculating 2.5 2 , to see if we go over 3.125 or not.Multiplying {−1, 1} by itself is equal to the set sum {−2, 0} ⊕ {0, 2} which is {−2, 1, 2}.Since the result is a set number larger than {−3, 0, 1}, we next try Y smaller than {1} and bigger than {0} because 2.5 {0} = 2.5.We try with {0, −1} = 3  2 , so that first we find the third power of 2.5.
Next we find the square root of {−3, −1, , so that 2.5 To summarize, there are two types of set numbers: 1) If −1 ∈ X then we say X is a positive set number 2) If −1 / ∈ X then X is a negative set number.Positive set numbers are of the form , a 2 − (n + 1), a 3 − (n + 1), . . .} is the new representation of the set number X.The set number X can also be identified with another set we will call the negative of X , and it is −X = {−(n + 1), a 1 − (n + 1), a 2 − (n + 1), a 3 − (n + 1), . . .}.To give another example, consider the number X = 7.28125 = {−5, −2, 0, 1, 2}.Its new representation is because the integer part is 7 and therefore n = 8.The negative is −X = {−9, −2 − 9, −5 − 9} = {−9, −11, −14}.It is left as an exercise to prove every subset of −N corresponds to a unique real number, and vice-versa.We can obviously identify every real number with a unique subset of N, now that we can identify it with a unique subset of −N.The main idea behind this construction is that we use the first n natural numbers of a set number to determine the sign and the integer part.
This construction of real numbers has been given mostly for its role in giving graphic representations of the real numbers.In this paragraph we have proven that there is a way of defining an order relation for 2 N , and that this order is isomorphic to the order of the extender real number line R.The operations can be meticulously defined case by case, but we will not unnecessarily extend our discussion.These have a simple interpretation in our graphic representations.The operations of real numbers are developed in our next construction.
Function Space.Our second construction, of negative real numbers, involves inverse functions.Every positive real number x ∈ Z can be associated a unique isomorphism of the form Z → R x where R x ⊂ Z is the collection of positive real numbers that are greater than or equal to x.In other words, an isomorphism Z → {+x}, between the positive real numbers and a collection of functions {+x} is given; the function +x : Z → R x is the bijection that acts by +x(a) = a + x.Let R * = {+x} ∪ {(+x) −1 } the set of functions that consists of the functions +x and their respective inverse function (+x) −1 : R x → Z, plus the identity function.
We will define an operation on the elements of R * .Let x, y ∈ Z, and their functions +x, +y ∈ R * , respectively.We define the operation in R * as the function +x • +y : Z → R x+y .We have two classes of functions in R * , as far as domain and range; the objects we call negative real numbers, as well as the positive real numbers.But, so far we have not defined an operation for all of these objects.Now we must find a suitable definitions for +x • (+y) −1 , and for the sum of two inverse functions in R * .
First, let us find the sum of two negative elements in R * .The sum of two positive elements in R * is the composition, so given two negative real numbers (+x) −1 , (+y) −1 , it is natural to set (+x) −1 • (+y) −1 := (+x • +y) −1 ∈ R * .If y < x, then we can find the number x − y > 0 and a unique function Z → R x−y in R * ; this function is defined to be the result of +y The operation we have defined for R * is isomorphic to addition of real numbers.
Dual Function Space.The elements of R * will be represented with bold letters such as x, y, -x, and y+x = +x • +y.We can build an isomorphism + : R * → R * * , where R * * is a collection of bijections of the form R * → R * .Specifically, the elements of R * * are the bijections +(x) : R * → R * such that y → +(x) y+x.Our real numbers, the elements of R * , are functions on positive numbers.The space R * * is a collection of bijective functions of the form R * → R * , and it is isomorphic to R * under addition.In conclusion R ∼ = R * ∼ = R * * , and R * * gives a complete description of addition for real numbers.It is left as an exercise for the reader to prove s(x) = x + 1 for every 0 ≤ x ≤ 8, and compare the results to the equalities above.We do the first few.Now that we have defined the natural numbers, we can construct the integers.We do this using the method we used in defining negative real numbers.That means, every element x ∈ U 0 is associated with a bijective function +x : U 0 → U x 0 , where U x 0 is the set of elements of U 0 that are greater than x.In particular, ∅ is the identity function, and {∅} is associated to the successor function ⊕{∅}.If x is associated a function +x, then s(x) is associated the function +x • ⊕{∅}.We have inductively defined the functions as compositions.With this we can build a relation order isomorphic to the integers.The objects of this order are the functions just defined (the powers of composition for the successor function) and their respective inverse functions.The order of these objects is defined naturally, and now we consider subsets of this space of ordered functions to be positive real numbers.Of course, we can already have said that the extended real number line is the set of all subsets of U 0 , if we consider the first construction we gave of real numbers.Theorem 1.The set of real numbers is U 1 = 2 U 0 , the power set of U 0 .The suitable functions of addition, product and order exist and are well defined in this universe.The system of real numbers is an extension to infinity, of the natural numbers.
In a sense, we have answered a more general question than Benacerraf's identification problem, since we have provided a canonical set theory for natural numbers, which can be naturally extended to obtain a canonical identification of 2 U 0 with the continuum [0, 1], and later with the extended real number line.Our axioms are stating that every set in U 0 − {∅} is successor of a set in U 0 , and that every set in U 0 has a successor in U 0 − {∅}.Furthermore, there is a general procedure for finding the successor of any set in U 0 ; the successor of x is (x {∅}) ⊕ {y ⊕ {∅}} y∈(x∩{∅}) .Our axioms assure us every set in U 0 is of the form (⊕{∅} We have an identification between 2 −N and the continuum [0, 1] of the unit interval.Therefore we have an identification of [0, 1] with 2 U 0 .Since we can again identify each number of the unit interval (refer back to the first construction of the real number line) with the extended real number line, we conclude there is a natural identification of the extended real number line with the set 2 U 0 .

Graphic Representations
We can carry out our constructions into a series of real-world representations.That is to say, we can give simple rules for manipulating physical systems, and these model numerical systems.
Collections of Arrows.In developing General Theory of Systems, we have to classify a system by its objects and its relations.In category theory, we focus on one special type of relations: binary relations.Of course, binary relations are really collections of arrows.Every number 0 ≤ a ≤ 31 is the collection of arrows {x → a}, for all 0 ≤ x < a.For example, 23 = {0 → 23, 1 → 23, 2 → 23, 4 → 23}.This means we give an isomorphism that sends a ∈ N to the set of arrows {x → a}, for all x element of the set number a.For example, 6 is represented by the collection of arrows {1 → 6, 2 → 6} and 13 is the collection of arrows {0 → 13, 2 → 13, 3 → 13} (Figure 17).
Trees.A tree is a graph of nodes and edges such that 1) We can identify a trunk: a principle edge with a number of branches attached to one of the nodes.2) Each branch on the tree is a tree.3) A single edge is a tree; we call it the 0-tree.4) The successor of a tree is obtained by adding a single edge to the trunk.Adding an edge to the trunk of the 0-tree gives its successor, the 1-tree, which is two edges joined together at one node.Adding an edge to the 1-tree, we find its successor, the 2-tree.If two branches are repeated on the same trunk, we substitute the two repeated branches with a single branch; the successor of these.This is called reduction.If one tree can be reduced to obtain another tree they are in the same equivalence class.An irreducible tree is said to be in canonical form.Reducing the 2-tree, we obtain find the canonical form (Figure 20).Adding a single edge to that, we obtain the canonical form of the 3-tree.If we add an edge to the 3-tree we have to reduce and obtain the canonical form of the 4-tree, etc. Continuing in this manner, we find all trees in sequence.

Rings.
A ring, R, is a circumference passing through denumerable number of rings R i ; the center point of every R i is a point on R; a central circumference passing through the center of a denumerable set of circumferences of degree 1.The central circumference is said to have degree 0. The circumferences R i are rings themselves; each R i is a circumference passing through the center of a denumerable set R j i i of circumferences of degree 2, and so on.A natural number n, with set number N, is represented by an equivalence class of rings.To build the canonical ring corresponding to a natural number, we draw an R i , for each element of the set number N = {a 1 , a 2 , a 3 , . . .}.That is to say, the central ring is a circumference going through #(N) circumferences; each of these a ring R i .Then, R i is a circumference going through the center of #(a i ) circumferences.We apply this recursively, until we bottom out.The equivalence relation is defined analagous to trees.If we have a ring with two identical rings, we substitute these both with a single ring; the successor ring of the repeated ring.The successor of R is found by adding a single 0-ring to the central circumference of R, and reducing.
Consider the ring of the number 27={4,3,1,0}.Then R is a circumference passing through the center of 4 circumferences R 1 , R 2 , R 3 , R 4 ; each one representing a number of the set {4, 3, 1, 0}.Giving a degree of freedom to the rings of degree 0 and 1, we are able to represent the set of real numbers.If the number is negative, we paint the degree 0 ring red; if it is positive the degree 0 ring is blue.The degree 1 rings are allowed to be red or blue in order to represent negative integers; a red degree 1 ring means we have a negative power in the binary representation.A 0 ring of degree 1 is neither red nor blue because 2 0 = 1 is its own reciprocal.

Conclusions
Even though these methods and definitions are elementary, and the presentation may seem trivial and in plain sight, these constructions and proofs are not apparent.The real number system has been revisited on many occasions but has never had simple solution.The initial purpose was to find a system theoretic treatment of numbers, as opposed to the set theoretic foundations of the classical axiomatic systems.It was our aim to propose new models of the relations of numerical systems, and finding their true representation in the universe of sets.This work comes after an initial attempt was made in [8].Our construction of real numbers has provided 1) New algorithms for calculating operations of real numbers 2) Graphic representations of real numbers.We can associate numbers to certain classes of physical models.3) We have provided a canonical set theory for arithmetic of natural numbers, and for analysis; one set being the power set of the other.We have answered Benacerraf's identification problem [9], by giving these canonical set representations of numbers, thus proving there is an intrinsic connection between the universe of sets and arithmetic.An extended version of these results and methods is under work.Several topics are treated in the same spirit.The topics include a calculus defined in terms of the order of N. A number of applications are also being prepared.A computing device that operates using radio frequency signals emitted back and forth between two stations, is tempting.Station A emits two sets of signals to station B, which then emits two signal back to station A. This process continues until the signal stabilizes (we can use this physical process to model addition, using the interpretation of energy levels).Other applications may be explored based on the graphic representations of the last section.

Figure 8 .Figures 9
Figures 9 and 10 are also examples of subtraction.