# A New Set Theory for Analysis

## Abstract

**:**

## 1. Introduction

“…these preliminary investigations are tedious and troublesome, and have actually, it must be confessed, not yet reached any entirely satisfactory conclusion at all.”

## 2. Motivation

## 3. $\mathbb{N}\cong \mathit{Cof}\left(\mathbb{N}\right)$

#### 3.1. Order

#### 3.2. Addition

#### 3.3. Product

#### 3.4. Subtraction

## 4. Continuum

#### 4.1. $[0,1]\cong {2}^{-\mathbb{N}}$

**Supremum and Infimum.**We define a supremum function for ${2}^{-\mathbb{N}}$. The order is defined as before, two sets relate $A<B$ if and only if $\mathrm{max}(A\u25b5B)\in B$. Let $\mathbb{X}\subseteq {2}^{-\mathbb{N}}$, so that $A\subseteq -\mathbb{N}$ for every $A\in \mathbb{X}$. The well ordering principle implies the existence of ${x}_{1}:=\mathrm{max}\bigcup \mathbb{X}$. Define ${\mathbb{Y}}_{1}:=\{A\in \mathbb{X}|{x}_{1}\in A\}$, and ${X}_{1}:=(\bigcup {\mathbb{Y}}_{1})-\left\{{x}_{1}\right\}$, and ${x}_{2}:=\mathrm{max}\left({X}_{1}\right)<{x}_{1}$. Then, define ${\mathbb{Y}}_{2}:=\{A\in {\mathbb{Y}}_{1}|{x}_{2}\in A\}$, and ${X}_{2}:=(\bigcup {\mathbb{Y}}_{2})-\{{x}_{1},{x}_{2}\}$, and ${x}_{3}:=\mathrm{max}\left({X}_{2}\right)$. Continue in this manner, so that ${x}_{n+1}:=\mathrm{max}\left({X}_{n}\right)$, where ${X}_{n}:=\bigcup {\mathbb{Y}}_{n}-{\left\{{x}_{i}\right\}}_{i=1}^{n}$ and ${\mathbb{Y}}_{n}:=\{A\in {\mathbb{Y}}_{n-1}|{x}_{n}\in A\}$. In denumerable steps, we have determined a unique set of integers ${x}_{1}>{x}_{2}>{x}_{3}>\cdots $. Define $\mathrm{sup}\mathbb{X}={\left\{{x}_{i}\right\}}_{i}$, which is, by construction, the smallest set number greater than every set in $\mathbb{X}$ (Figure 14).

**Reciprocals.**If we want to multiply $A\subset \mathbb{N}$ by $B\subseteq -\mathbb{N}$, we proceed as before. The definition is the same. Refer to Figure 7. The unit of product, which should be $\left\{0\right\}$, which is a bounded subset of $\mathbb{Z}$ and therefore has a second representation, which is $-\mathbb{N}$. We use this representation to provide a well-defined algorithm for finding the reciprocal of a natural number. To find the reciprocal of $137=\{0,3,7\}$, we seek out a set number $\frac{1}{N}=\{{x}_{1},{x}_{2},\dots \}\subset -\mathbb{N}$ such that

#### 4.2. ${\mathbb{R}}^{+}\cong {\mathbb{Z}}^{-}$

**Topology of Bounded Subsets of $\mathbb{Z}$.**A set $X\in {\mathbb{Z}}^{-}$ will correspond to a unique number in ${\mathbb{R}}^{+}$. 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 ${\mathbb{Z}}^{-}$. For example, given a binary representation ${110101.00101}_{1}$, the set number is $\{5,4,2,0,-3,-5\}$.

- $\mathbb{N}\cong Cof\left(\mathbb{N}\right)$ where the cofinite topology uses the order and operations of set numbers.
- $[0,1]\cong {2}^{-\mathbb{N}}$, with $\varnothing =0$ and $-\mathbb{N}=1$, and the same definitions for set order and operations.
- The continuum of non-negative real numbers is built as a natural generalization of both $Cof\left(\mathbb{N}\right)$ and ${2}^{-\mathbb{N}}$. We piece together $[0,1],[1,2],[2,3],\dots $, into a single continuum ${\mathbb{R}}^{+}$. This is done by considering the upper bounded sets of $\mathbb{Z}$ and a proper extension of the set number relations.

**Supremum.**In the previous sub section, we provided a well defined algorithm for finding the supremum of a family $\mathbb{X}\subseteq [0,1]$, where the elements of $\mathbb{X}$ are arbitrary subsets of $-\mathbb{N}$. Now, we generalize this process to define the supremum of a bounded set of positive real numbers.

**Division.**We now describe division and rational numbers. Let us find the rational representation of the set number $\{5,0,-2,-4,-5\}=33.34375$. If we apply ${s}^{5}$ to the set number, the result is $\{10,5,3,1,0\}=1067$. The action of adding 5 to the elements of a set number is equivalent to multiplication by ${2}^{5}$. In our example, $\left\{5\right\}\odot \{5,0,-2,-4,-5\}=\{10,5,3,1,0\}$. If $A\in {\mathbb{Z}}^{-}$, we define $\mathbb{A}=\{X\in Cof(\mathbb{N}\left)\right|X\odot A\in Cof\left(\mathbb{N}\right)\}$. If $\mathbb{A}=\varnothing $, we say the set number A is irrational, and if $\mathbb{A}\ne \varnothing $, then A is rational. A fraction representing A is an ordered pair $(M,N)$, where $N\in Cof\left(\mathbb{N}\right)$ is an element of $\mathbb{A}$ and $M=N\odot A$. If A is rational, the well ordering principle implies the existence of $\mathrm{min}\left(\mathbb{A}\right)$, and the corresponding fraction is said to be the irreducible fraction of $\mathbb{A}$. In our example, $\left(\right\{10,5,3,1,0\},\{5\left\}\right)$ is the irreducible fraction of $\{5,0,-2,-4,-5\}$. This can be expressed as $33.34375=\frac{1067}{32}$. Given a finite set number $A\in {\mathbb{Z}}^{-}$, we can give infinite, but equivalent, representations of an irreducible fraction $\frac{m}{n}$.

**Sum and Product of Infinite Sets.**We must define the set number sum of two sets $A,B\in {\mathbb{Z}}^{-}$ each with perhaps infinite elements. A, is the set of integers ${a}_{1}>{a}_{2}>{a}_{3}>\cdots $, and B the set of integers ${b}_{1}>{b}_{2}>{b}_{3}>\cdots $. Define ${A}_{n}={\left\{{a}_{i}\right\}}_{i=1}^{n}$, and in a similar manner ${B}_{m}={\left\{{b}_{m}\right\}}_{j=1}^{m}$. The sum is defined as $A\oplus B:={\mathrm{sup}}_{n,m}({A}_{n}\oplus {B}_{m})$ for all $n,m\in N$. 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 $A,B\subset \mathbb{N}$. 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 ${A}^{\varnothing}=\left\{0\right\}$. We define the power ${A}^{\left\{0\right\}}=A$ because $\left\{0\right\}=1$. With this, we are able to give a recursive formula ${A}^{B}=A\odot {A}^{B\ominus \left\{0\right\}}$. To find the power of ${127}^{4}={\{0,1,2,3,4,5,6\}}^{\left\{2\right\}},$ we first reduce the expression. We know ${A}^{\left\{2\right\}}=A\odot {A}^{\left\{2\right\}\ominus \left\{0\right\}}$. We know, from the subtraction of set numbers, that $\left\{2\right\}\ominus \left\{0\right\}=\{0,1\}$, then ${A}^{\left\{2\right\}}=A\odot {A}^{\{0,1\}}$. Then, we find ${A}^{\{0,1\}}=A\odot {A}^{\{0,1\}\ominus \left\{0\right\}}=A\odot {A}^{\left\{1\right\}}=A\odot (A\odot {A}^{\left\{1\right\}\ominus \left\{0\right\}})=A\odot (A\odot A)$. Finally, ${A}^{\left\{2\right\}}=A\odot (A\odot (A\odot A))$, as we expect since $\left\{2\right\}=4$.

**Roots.**The process of finding a root is finding a reciprocal power, that is to say ${A}^{\frac{1}{B}}$, for some set $B\in Cof\left(\mathbb{N}\right)$. To find $X={A}^{\frac{1}{B}}$, we must find a set $X\in {\mathbb{Z}}^{-}$ such that ${X}^{B}=A$. The set $P(A,B)$ of all set numbers X such that ${X}^{B}<A$ is bounded above. Define ${A}^{\frac{1}{B}}=\mathrm{sup}P(A,B)$. We have three cases; $A>\left\{0\right\}$, $A=\left\{0\right\}$, and $A<\left\{0\right\}$. In the first case, ${A}^{\frac{1}{B}}<A$, while in the last case ${A}^{\frac{1}{B}}>A$. The roots of $\left\{0\right\}$ are ${\left\{0\right\}}^{\frac{1}{B}}=\left\{0\right\}$, for every $B\in Cof\left(\mathbb{N}\right)$. We give an example, ${3}^{1/4}={\{0,1\}}^{1/\left\{2\right\}}$. The fourth power of $\{0,-1\}$ is greater than $\{0,1\}$. We find that ${\{0,-2\}}^{4}<\{0,1\}$, and $\{0,-2,-3\}>\{0,1\}$. Then, we find $\{0,-2,-4\}<\{0,1\}$. We continue in this manner, with trial and error; finding the fourth power of sets such that the fourth power is less than {0,1}. We are finding a number whose fourth power is equal to the natural number $3=\{0,1\}$.

**Logarithms.**In the last section, we extended the definition of powers to include rational numbers and, finally, irrational powers as well. Now, we explore the inverse function. To find $X={\mathrm{log}}_{B}A$, we find a set number X such that ${B}^{X}=A$. It is not difficult to prove the following statements.

**Properties of Operation.**The axiomatic properties of the field of real numbers hold, taking into account that we have not yet described negative real numbers. The identity for addition is ∅, while the identity for product is $\left\{0\right\}$. The commutative property of addition is trivial because of the commutative properties of ▵ and ∩. It is not easy to give a direct proof of the associative property for set number addition. We first have to show $\left\{n\right\}\oplus (A\oplus \{m\left\}\right)=\left(\right\{n\}\oplus A)\oplus \left\{m\right\}$, for any singletons $\left\{n\right\},\left\{m\right\}\subset \mathbb{Z}$. Let $N=\{{x}_{1},{x}_{2},\dots ,{x}_{n}\}$, $M=\{{y}_{,}{1}_{2},\dots ,{y}_{m}\}$ and $A=\{{a}_{1},{a}_{2},\dots ,{a}_{p}\}$ three finite subsets of $\mathbb{Z}$. The sum of these can be written

## 5. Construction of $\mathbb{R}$

**Unit Interval.**The real numbers of ${I}_{1}$ are set numbers of the form $\{-1,{x}_{1},{x}_{2},{x}_{3},\dots \}$, with ${x}_{i}\le -3$. Next, we define ${I}_{2}$ as the collection of set numbers of the form $\{-1,-2,{x}_{1},{x}_{2},{x}_{3},\dots \}$, with ${x}_{i}\le -4$. The interval ${I}_{3}$ is the collection of set numbers $\{-1,-2,-3,{x}_{1},{x}_{2},{x}_{3},\dots \}$, with ${x}_{i}\le -5$. The negative interval $-{I}_{1}=[-1,0)$ is the collection of set numbers of the form $\{-2,{x}_{1},{x}_{2},{x}_{3},\dots \}$ with ${x}_{i}\le -3$. The real number $-1\in \mathbb{R}$ is identified with $\frac{1}{4}=\{-2\}$, and $-\frac{1}{2}\in \mathbb{R}$ is the set number $\{-2,-3\}$. The interval $-{I}_{2}$ is defined as the family of sets $\{-3,{x}_{1},{x}_{2},{x}_{3},\dots \}$, with ${x}_{i}\le -4$

**Function Space.**Our second construction, of negative real numbers, involves inverse functions. Every positive real number $X\in {\mathbb{Z}}^{-}$ can be associated a unique isomorphism $\oplus X:{\mathbb{Z}}^{-}\to {\mathbb{R}}_{X}$, where ${\mathbb{R}}_{X}\subset {\mathbb{Z}}^{-}$ is the collection of positive real numbers that are greater than or equal to X. In other words, an isomorphism ${\mathbb{R}}^{+}\to \{\oplus X\}$ is given and $\oplus X:{\mathbb{Z}}^{-}\to {\mathbb{R}}_{X}$ is the bijection that acts by $\oplus X\left(A\right)=A\oplus X$. Let ${\mathbb{R}}^{\ast}=\{\oplus X\}\cup \left\{{(\oplus X)}^{-1}\right\}\cup \left\{Id\right\}$. We have two classes of functions in ${\mathbb{R}}^{\ast}$, as far as domain and range—the objects we call negative real numbers, and the positive real numbers.

## 6. Universe of Finite Sets

**Definition**

**1.**

- 1.
- $\varnothing \in {\mathbb{U}}_{0}$
- 2.
- ${x}_{1},{x}_{2},{x}_{3},\dots ,{x}_{n}\in {\mathbb{U}}_{0}$, then $\{{x}_{1},{x}_{2},{x}_{3},\dots ,{x}_{n}\}\in {\mathbb{U}}_{0}$
- 3.
- ${\mathbb{U}}_{0}$ is the set of objects that satisfy 1. or 2.

**Definition**

**2.**

**Axiom**

**1.**

**Axiom**

**2.**

**Theorem**

**1.**

## 7. Graphic Representations

**Collections of Arrows.**In developing General Theory of Systems, we have to classify a system by its objects and its relations. Binary relations are collections of arrows. Every number $0\le a\le 31$ is the collection of arrows $\{x\to a\}$, for all $0\le x<a$. For example, $23=\{0\to 23,1\to 23,2\to 23,4\to 23\}$. This means we provide an isomorphism that sends $X\in \mathbb{N}$ to the set of arrows $\{x\to X\}$, for all x element of the set number X. For example, 6 is represented by the collection of arrows $\{1\to 6,2\to 6\}$ and 13 is the collection of arrows $\{0\to 13,2\to 13,3\to 13\}$ (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 finite number of branches attached to one of the nodes. All branches are attached to the same node of the trunk. (2) Each branch on the tree is a tree. (3) A single edge is a tree; we call it the 0-tree. 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 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. Every natural number is associated an equivalence class of trees. Every branch on the canonical tree of a set number X corresponds to a natural number $k\in X$. Every tree is made up of smaller trees, and we give a well defined method of building trees. The canonical tree corresponding to the set number X has $\#\left(X\right)$ branches. Each branch is defined in the same way.

**Rings.**A ring, R, is a circumference passing through the center of a denumerable number of rings ${R}_{i}$. The central circumference R is said to have degree 0. The circumferences ${R}_{i}$ are rings themselves; we say they have degree 1. Each ${R}_{i}$ is a circumference passing through the center of a denumerable set ${R}_{i}^{{j}_{i}}$ 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 $N=\{{a}_{1},{a}_{2},{a}_{3},\dots \}$. That is to say, the central ring is a circumference going through $\#\left(N\right)$ circumferences; each of these a ring ${R}_{i}$. Then, ${R}_{i}$ is a circumference going through the center of $\#\left({a}_{i}\right)$ circumferences. We apply this recursively, until we bottom out.

## 8. Conclusions

## Funding

## Conflicts of Interest

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**Figure 4.**Graphic representation of $4=2+2=2+(1+1)=2+1+(0.5+0.5)=2+1+0.5+(0.25+0.25)=2+1+0.5+0.25+(0.125+0.125)\dots $

**Figure 6.**We illustrate $7\xb79$. The first and second columns are the pivot and base, respectively. The next three columns correspond to the displacements of our base. The last column is the sum of the displacements. The result is $63=\{0,1,2,3,4,5\}$.

**Figure 14.**We illustrate the process for finding the supremum of a family $\mathbb{X}$. It is left to the reader to the complete the procedure and find $\mathrm{sup}\left(\mathbb{X}\right)=B$.

**Figure 15.**We find the reciprocal of $N=137=\{0,3,7\}$. Column ${A}_{1}$ is the sum of three negative displacements of N, and it is a set number less than $1=-\mathbb{N}$. We try adding another displacement to ${A}_{1}$, without going over $-\mathbb{N}$. We find ${A}_{3}$ works, and continue in this manner. The number $\frac{1}{N}$ is the set of negative integers that indicate the valid displacements.

**Figure 16.**The power set ${2}^{-\mathbb{N}}$ will be given the structure of the real number system through our bijection $\mathbb{R}\to [0,1]$.

**Figure 17.**We represent the natural numbers from 0 to 15. Every number is connected to its elements.

**Figure 18.**We can represent several numbers in a single diagram. We label 32 points on the circumference, and start adding arrows. Here we see the representation of numbers $0-31$.

**Figure 19.**Graphic representation of three real numbers. Graph (

**a**) is an approximation of $\pi $. The second graph (

**b**) is an approximation of $-e$, and (

**c**) represents $-4.375$.

**Figure 20.**Canonical trees can be built easily, given a set number. The canonical tree for $7=\{0,1,2\}$ has three branches. One branch is the 0-tree, the second branch is the 1-tree and the third branch is the 2-tree. The canonical tree of $8=\left\{3\right\}$ is a trunk with one branch, which is the 3-tree.

**Figure 21.**The construction of rings is carried out similarly to the recursive construction of trees. The canonical form of $8=\left\{3\right\}$ is a ring R with one object ${a}_{1}$. The ring ${a}_{1}$ is the ring for 3. The ring for 3 is itself a ring with two subrings ${a}_{1,1}$ and ${a}_{1,2}$ which are the rings for 0 and 1, respectively.

**Figure 22.**Here we present the canonical ring of 27. Each of the rings of degree 1, correspond to an elment of $27=\{4,3,1,0\}$.

**Figure 23.**Rings of degree greater than or equal to 2 are colorless. All 0 rings are colorless. Red indicates negative and blue, positive. If the degree 0 ring is blue, x is positive real number. If the degree 0 ring is red, then x is a negative number. Each of the degree 1 rings (except for the 0 ring) are colored red or blue because the powers of 2 that represent our set number are positive and negative integers. No more coloring is needed to represent x.

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**MDPI and ACS Style**

Ramírez, J.P.
A New Set Theory for Analysis. *Axioms* **2019**, *8*, 31.
https://doi.org/10.3390/axioms8010031

**AMA Style**

Ramírez JP.
A New Set Theory for Analysis. *Axioms*. 2019; 8(1):31.
https://doi.org/10.3390/axioms8010031

**Chicago/Turabian Style**

Ramírez, Juan Pablo.
2019. "A New Set Theory for Analysis" *Axioms* 8, no. 1: 31.
https://doi.org/10.3390/axioms8010031