An Objection to Cantor's Diagonal Argument

    Cantor's diagonal argument is a proof devised by Georg Cantor to demonstrate that the real numbers are not countably infinite. Suppose that all real numbers in the section from 0 to 1 are shown by infinite binary fractions, and they are put into one-to-one correspondence with all natural numbers. We could then enumerate all real numbers in this interval as a sequence, R₁,R₂,R₃, ... :

    In this sequence, a_nm is the m-th digit of the R_n and diagonal digits are enclosed in square brackets. Consider an infinite binary fraction  X =0.b₁b₂...b_n... . We can compose the new number X as follows: if a₁₁ of R₁ is 0 then b₁ is 1, if a₂₂ of R₂ is 1 then b₂ is 0, and so on. Finally, we can get the new number X: X differs in the n-th binary place from R_n. So this new number X is not in our sequence. In conclusion, the assumption, that all real numbers in the interval [0,1] are put into one-to-one correspondence with all natural numbers, must be false. Then, all real numbers between 0 and 1 is strictly larger than all natural numbers. Q.E.D.

    Cantor’s diagonal argument uses the completion of infinite processes. However, the mathematicians before Cantor did not accept the completion of infinite processes. This is the controversial point of Cantor’s diagonal argument. To avoid this point, we shall discuss within the range of finite binary fractions. In addition, we shall write numbers in binary notation in equations. First, we enumerate all 11-digit binary fractions in the interval [0,1] in ascending order:

    In this sequence, F₁₁(m) is the m-th 11-digit binary fraction and diagonal digits are enclosed in square brackets. Using Cantor’s diagonal procedure, we can compose the number X = 0.111, which differs from the diagonal number in each digit. However, X is different from only the first three numbers in this sequence and then X equals to F₁₁(1000).

    Next, we enumerate all n-bit binary fractions in the interval [0,1] in ascending order:

    In this sequence, F_n(m) is the m-th n-bit binary fraction and diagonal digits are enclosed in square brackets. We compose the number X in the above mentioned manner. However X differs from only first n numbers in this sequence. This sequence contains X because it includes all n-bit binary fractions in the interval [0,1].
    Next, the diagonal number is only n-bits after the binary point. On the contrary, this sequence contains 10ⁿ+1 binary fractions. The ratio n/(10ⁿ+1) is monotonically decreasing. As n increases, n/(10ⁿ+ 1) approaches 0:

    Therefore, Cantor's diagonal argument has no application to all n-bit binary fractions in the interval [0,1].

    Consider a real number α in the interval [0,1]. If α is not a binary fraction, then α lies between two n-bit binary fractions and there exists a natural number m: 

We can approximate the real number α by F_n(m) and then the error is less than 1/10ⁿ:

As n increases, then F_n(m) approaches α

    At the time, the number of all n-bit binary fractions in the interval [0,1] is consistently 10ⁿ+1. No matter how large n is, 10ⁿ+1 is a natural number.  In conclusion, all n-bit binary fractions in the interval [0,1] are constantly countable.

    We shall use decimal notation in last two sections. Consider the sequence of all 150-digit binary fractions in the interval [0,1]. Our sequence consists of approximately 1.4X10⁴⁵ binary fractions. Let us compare this number with the diameter of the observable universe. We choose the diameter of a proton as the reference length for measuring the diameter of the observable universe. The ratio of the diameter of the observable universe to the diameter of a proton is roughly 2.6X10⁴¹. Even if we could use proton sized digits, we could not write our sequence in one column in our observable universe. At the time, the diagonal number differs from only 150 binary fractions of our sequence. This means that Cantor's diagonal argument cannot apply in our observable universe. Where is Cantor's paradise?

    Next, we shall compare the number of irrational numbers with that of binary fractions in the interval [0,1]. Let us regard an irrational number as a pigeon and an interval of adjacent two n-bit binary fractions as a pigeonhole. According to the pigeonhole principle, if the number of irrational numbers were larger than that of binary fractions, there would be at least two irrational numbers, between which there is no binary fraction. However, there is not such a pair of irrational numbers. The proof is as follows. Consider arbitrary two irrational numbers α and β in the interval [0,1]. If the difference of them is zero, they are the same number. If they are different numbers, the difference isn't zero. Let us assume that the difference is ε:

There exists a natural number n such that:

For enough large n, there are two natural numbers l and m such that:

So, we can sandwich any two irrational numbers by different pairs of n-bit binary fractions. According to the pigeonhole principle, the number of irrational numbers is not larger than that of binary fractions in the interval [0,1]. Evidently, they are countable. Therefore, irrational numbers in the interval [0,1] are countable.