B. Vlahović
Professor of physics, research interest: Nuclear physics, material science, cosmology, and number theory.
less
Related Authors
Cesar Silva
Williams College
Wolfgang Mückenheim
Augsburg University of Applied Sciences
Jaykov Foukzon
Technion - Israel Institute of Technology
David Sauzin
Centre National de la Recherche Scientifique / French National Centre for Scientific Research
InterestsView All (6)
Uploads
Papers by B. Vlahović
Through rigorous proofs, we demonstrate that \( S \) encompasses all real numbers by recognizing each within a nested sequence of intervals, corresponding to an element \( x_\alpha \) in \( S \). Furthermore, we establish that any Dedekind cut in \( S \) does not produce gaps, thereby confirming \( S \) as a complete and countable representation of the real line.
In this article, we provide a surprising result that the denumerability of real numbers is possible using the Axiom of Choice. The proof is sketched here and elaborated further in the text. First, we consider an interval $[a,b]$, where $a<b$, that contains only algebraic numbers and define a set $S$, which by definition contains all possible algebraic intervals $I_n=[a_n, b_n]$, and $a_n \leqslant b_n$, where $a_n$ and $b_n$, are algebraic numbers from $[a,b]$. Then defined is a function $f$ that for each interval $I_n$ assigns a transcendental number $t_n$, so that $t_n$ is within the interval on which function $f$ is acting, i.e.
$f(I_n)=t_n$, and $a_n \leqslant t_n \leqslant b_n$. It is also required that $t _n$ is in all intervals containing that interval, i.e. in all superintervals. The numbers $t_n$ are then added to the set $S$, and to all algebraic intervals $I_n$ of the set defining the set $S’$, which is countable and complete. The proof is simple and based on contra example. Assume that there is a missing transcendental number $T_x$ that is different from all $t_n$ numbers generated by the function $f$ and therefore from all numbers in the set $S’$, and define that missing number $T_x$ with a sequence of nested closed intervals with algebraic endpoints $\alpha_n$ and $\beta_n$, so
$T_x=[\alpha_1, \beta_1]\supset [\alpha_2, \beta_2] \supset [\alpha_3, \beta_3]… \supset [\alpha_n, \beta_n]…$.
Using the Axiom of Choice, one can recognize in the set $S’$ closed intervals $I_n=[a_n,b_n]$ that have the same endpoints as the closed intervals $[\alpha_n, \beta_n]$ in the sequence that defines the missing number $T_x$. So, one can find in the set $S’$ all closed intervals $I_n$ with the same endpoints $a_n=\alpha_n$ and $b_n=\beta_n$,
which can be placed together to create a sequence
$T_x=[a_1,b_1]\supset [a_2,b_2] \supset [a_3,b_3]… \supset [a_n,b_n]…$
So, using the AC, one can recognize in the set $S’$ sequence of intervals $[a_n,b_n]$ that is exactly the same as the sequence of intervals $[\alpha_n, \beta_n]$ that represents the missing number, i.e. it has closed intervals with the same endpoints. To show that set $S'$ is complete, that there is no missing number $T_x$, that $T_x$ is in the set $S’$, it is enough to show that the intersection of the sequence of closed intervals $[a_n,b_n]$ is not empty, that it contains the number $T_x$. It is shown using the Intersection Property Theorem and applying it to the sequence of intervals $[\alpha_n, \beta_n]$ that defines the $T_x$ number. To satisfy the requirement that the function $f$ acted on all $I_n$ intervals in the set $S$, the same $T_x$ number must be in the sequence with the $[a_n,b_n]$ intervals that are from the set $S'$.
The power of AC is that it does not require to define sequence of any real number. Instead, it allows to recognize in the set $S'$ any sequence representing any real number. The sequence of any real number is implicitly in the set $S'$ because all possible intervals with algebraic endpoints are in the set $S'$. Since by AC in the set $S'$ can be recognized any arbitrary sequence representing any number, and since these sequences have nonempty intersections, the set $S'$ is complete.
Assuming the validity of AC and its consistency with ZF axioms, the discrepancy with the diagonal argument and other proofs of nondenumerability must originate in logical or technical problems common for these proofs, which we discuss in the second part of the article. In each of them, there is either a logical problem that the inability to assign in a sequence a specific n-th place for a number implies that that number is not in the sequence or a technical problem related to ignoring the Intersection Property Theorem in some elements of the proofs, or both of these problems.
The proof of denumerability of real numbers provided in the first part of the article does not specify the sequence of real numbers since numbers $t_n$ are chosen arbitrarily. In the third part of the article, we provide an explicit sequence of all real numbers. It is done by modifying the G\"{o}del constructible model universe and generating a set of real numbers, starting with a set of algebraic numbers and expanding it iteratively by adding in each step transcendental numbers and creating bigger and bigger sets of real numbers. That iterative process that by diagonal procedure creates in each step a new layer of transcendental numbers will cumulatively, similarly to Wang's $\Sigma$ constructive model, which is different from ZF, result in a set of all real numbers. For the first time, provided is an explicit example that satisfies Wang's model.
Through rigorous proofs, we demonstrate that \( S \) encompasses all real numbers by recognizing each within a nested sequence of intervals, corresponding to an element \( x_\alpha \) in \( S \). Furthermore, we establish that any Dedekind cut in \( S \) does not produce gaps, thereby confirming \( S \) as a complete and countable representation of the real line.
In this article, we provide a surprising result that the denumerability of real numbers is possible using the Axiom of Choice. The proof is sketched here and elaborated further in the text. First, we consider an interval $[a,b]$, where $a<b$, that contains only algebraic numbers and define a set $S$, which by definition contains all possible algebraic intervals $I_n=[a_n, b_n]$, and $a_n \leqslant b_n$, where $a_n$ and $b_n$, are algebraic numbers from $[a,b]$. Then defined is a function $f$ that for each interval $I_n$ assigns a transcendental number $t_n$, so that $t_n$ is within the interval on which function $f$ is acting, i.e.
$f(I_n)=t_n$, and $a_n \leqslant t_n \leqslant b_n$. It is also required that $t _n$ is in all intervals containing that interval, i.e. in all superintervals. The numbers $t_n$ are then added to the set $S$, and to all algebraic intervals $I_n$ of the set defining the set $S’$, which is countable and complete. The proof is simple and based on contra example. Assume that there is a missing transcendental number $T_x$ that is different from all $t_n$ numbers generated by the function $f$ and therefore from all numbers in the set $S’$, and define that missing number $T_x$ with a sequence of nested closed intervals with algebraic endpoints $\alpha_n$ and $\beta_n$, so
$T_x=[\alpha_1, \beta_1]\supset [\alpha_2, \beta_2] \supset [\alpha_3, \beta_3]… \supset [\alpha_n, \beta_n]…$.
Using the Axiom of Choice, one can recognize in the set $S’$ closed intervals $I_n=[a_n,b_n]$ that have the same endpoints as the closed intervals $[\alpha_n, \beta_n]$ in the sequence that defines the missing number $T_x$. So, one can find in the set $S’$ all closed intervals $I_n$ with the same endpoints $a_n=\alpha_n$ and $b_n=\beta_n$,
which can be placed together to create a sequence
$T_x=[a_1,b_1]\supset [a_2,b_2] \supset [a_3,b_3]… \supset [a_n,b_n]…$
So, using the AC, one can recognize in the set $S’$ sequence of intervals $[a_n,b_n]$ that is exactly the same as the sequence of intervals $[\alpha_n, \beta_n]$ that represents the missing number, i.e. it has closed intervals with the same endpoints. To show that set $S'$ is complete, that there is no missing number $T_x$, that $T_x$ is in the set $S’$, it is enough to show that the intersection of the sequence of closed intervals $[a_n,b_n]$ is not empty, that it contains the number $T_x$. It is shown using the Intersection Property Theorem and applying it to the sequence of intervals $[\alpha_n, \beta_n]$ that defines the $T_x$ number. To satisfy the requirement that the function $f$ acted on all $I_n$ intervals in the set $S$, the same $T_x$ number must be in the sequence with the $[a_n,b_n]$ intervals that are from the set $S'$.
The power of AC is that it does not require to define sequence of any real number. Instead, it allows to recognize in the set $S'$ any sequence representing any real number. The sequence of any real number is implicitly in the set $S'$ because all possible intervals with algebraic endpoints are in the set $S'$. Since by AC in the set $S'$ can be recognized any arbitrary sequence representing any number, and since these sequences have nonempty intersections, the set $S'$ is complete.
Assuming the validity of AC and its consistency with ZF axioms, the discrepancy with the diagonal argument and other proofs of nondenumerability must originate in logical or technical problems common for these proofs, which we discuss in the second part of the article. In each of them, there is either a logical problem that the inability to assign in a sequence a specific n-th place for a number implies that that number is not in the sequence or a technical problem related to ignoring the Intersection Property Theorem in some elements of the proofs, or both of these problems.
The proof of denumerability of real numbers provided in the first part of the article does not specify the sequence of real numbers since numbers $t_n$ are chosen arbitrarily. In the third part of the article, we provide an explicit sequence of all real numbers. It is done by modifying the G\"{o}del constructible model universe and generating a set of real numbers, starting with a set of algebraic numbers and expanding it iteratively by adding in each step transcendental numbers and creating bigger and bigger sets of real numbers. That iterative process that by diagonal procedure creates in each step a new layer of transcendental numbers will cumulatively, similarly to Wang's $\Sigma$ constructive model, which is different from ZF, result in a set of all real numbers. For the first time, provided is an explicit example that satisfies Wang's model.
The second set, derived by inserting specifically defined transcendental numbers between algebraic numbers, begins with the sequence of algebraic numbers as the 0-th layer in Wang's $\Sigma$ model. A diagonal procedure applied to that layer creates a set of transcendental numbers, which combined with the 0-th layer form the $\Sigma_1$ layer. Continuing this recursive procedure results in a set equivalent to set $S'$. This suggests that all constructive real numbers are countable, challenging the traditional diagonal argument. The application of the AC is crucial for selecting transcendental numbers and identifying intervals and sequences, making the set $S'$ complete and countable. This might resonate with constructivists who oppose the AC. However, further exploration revealed conflicts in non-denumerability proofs with the Nested Interval Property, indicating a need for reevaluation of the countability of real numbers.
The second set, derived by inserting specifically defined transcendental numbers between algebraic numbers, begins with the sequence of algebraic numbers as the 0-th layer in Wang's $\Sigma$ model. A diagonal procedure applied to that layer creates a set of transcendental numbers, which combined with the 0-th layer form the $\Sigma_1$ layer. Continuing this recursive procedure results in a set equivalent to set $S'$. This suggests that all constructive real numbers are countable, challenging the traditional diagonal argument. The application of the AC is crucial for selecting transcendental numbers and identifying intervals and sequences, making the set $S'$ complete and countable. This might resonate with constructivists who oppose the AC. However, further exploration revealed conflicts in non-denumerability proofs with the Nested Interval Property, indicating a need for reevaluation of the countability of real numbers.