Contra Cantor -- 15-Aug-19
My theory stands as firm as a rock; every arrow directed against it will return
quickly to its archer. How do I know this? Because I have studied it from all
sides for many years; because I have examined all objections which have ever
been made against the infinite numbers; and above all because I have followed
its roots, so to speak, to the first infallible cause of all created things.
Georg Cantor
No one will drive us from the paradise which Cantor created for us.
David Hilbert
There is nothing wrong with Cantor’s argument.
Wilfrid Hodges
CONTRA CANTOR: HOW TO COUNT THE “UNCOUNTABLY INFINITE”
Erdinç Sayan
Short Description. Demonstration that Cantor’s diagonal argument is flawed and that
real numbers, power set of natural numbers and power set of real numbers have the same
cardinality as natural numbers.
Abstract. Cantor’s diagonal argument purports to prove that the set of real numbers is
nondenumerably infinite. I contend that the set of real numbers is, just like natural numbers,
denumerably infinite. In the first part of my paper, I show that Cantor’s diagonal argument is
flawed, and point to the exact source and extent of its failure. Then I use a similar criticism
against another version of Cantor’s diagonalization maneuver, which he uses to prove that the
power set of natural numbers is nondenumerably infinite.
In the second part of the paper, I propose an indirect method of establishing the
denumerable infinity of real numbers (rather than directly finding a bijection between naturals
and reals). My method utilizes tree diagrams. For convenience and simplicity, I employ a
binary tree wherein the branches represent the real numbers in the interval [0, 1) written in
binary symbolism. Cantor’s diagonal argument concludes that the real numbers in the interval
[0, 1) are nondenumerably infinite, and this suffices to establish that the entire set of real
numbers are nondenumerably infinite. I first take up the task of establishing that the real
numbers in the interval [0, 1) are denumerably infinite, by way of showing that the segments
of the binary tree (a segment is the line between any two consecutive nodes of the tree) can be
paired off with natural numbers. This way of “counting” the segments establishes that the set
of segments is denumerably infinite. I then prove that the set of branches of the tree, and
thereby the reals in [0,1) they represent, cannot have a higher cardinality than the set of
1
Contra Cantor -- 15-Aug-19
segments of the tree. Hence the branches, and therefore the reals in [0, 1) represented by the
braches, can only be denumerably infinite. I then move on to show that not only the set of
reals in [0, 1) but the entire set of reals is denumerably infinite.
Next I argue that two other contentions of Cantor’s are erroneous. Against his claim
that the power set of real numbers is nondenumerably infinite, with cardinality ‫א‬2, I show that
that set too is denumerably infinite, just as is the power set of natural numbers. I end with
some concluding remarks.
Keywords. Cantor’s diagonal argument, binary tree, cardinality of natural numbers,
cardinality of real numbers, cardinality of the power set of natural numbers, cardinality of the
power set of real numbers, Continuum Hypothesis
Georg Cantor’s celebrated diagonalization argument purports to demonstrate that the
set of real numbers has a higher cardinality than the set of natural numbers: real numbers are
“uncountably (or nondenumerably) infinite” whereas natural numbers are “countably (or
denumerably) infinite.” I want to establish that the set of real numbers is, like the set of
natural numbers (or the set of counting numbers), denumerably infinite, not nondenumerably
infinite. But first, in sections Ib and Ic, I argue that Cantor’s diagonal argument is fallacious,
and I point to the exact source and extent of its failure. In section II, I criticize another
version of the Cantorian diagonalization maneuver to prove that the power set (i.e. the set of
all subsets) of natural numbers is nondenumerably infinite.
In the more positive subsequent part of my paper, I propose, in section IIIa.2, an idea
about an indirect way of establishing the denumerable infinity of real numbers between 0 and
0.999… (rather than attempting the difficult task of directly finding a bijective function
between natural numbers and real numbers in that interval). My method is to use tree
diagrams. For convenience and simplicity of the diagrams, I employ the binary number
system to represent the set of real numbers with my trees rather than the ordinary decimal
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Contra Cantor -- 15-Aug-19
number system, but the procedures I describe can also be carried out—though in a much
messier way—in the decimal system. In section IIIb, I go on to show that not only the real
numbers between 0 and 0.999…, but the entire set of real numbers1 is denumerably infinite.
In section IV, I argue that two other contentions of Cantor’s are mistaken. Against his claim
that the power set of real numbers is nondenumerably infinite, with cardinality ‫א‬2, I show that
that set too is only denumerably infinite, and so is the power set of natural numbers. I end
with some concluding remarks in section V.
Ia. Cantor’s diagonal argument
Cantor gave two purported proofs for the claim that the cardinality of the set of real
numbers is greater than that of the set of natural numbers.2 According to a popular
reconstruction of the more widely known of these proofs, his diagonal argument, we
randomly tabulate the real numbers in the interval [0, 1) in an array. I will use a binary
version of the table for ease of exposition and give Table 1 as an example:
In this paper when I talk about “real numbers” I will mean positive real numbers, but what I have to say can be
extended to the union of positive and negative numbers.
1
2
Cantor’s first proof is found in his [1]. His second, diagonal, proof appeared in his [2]. An English translation
of the latter can be found in [3], pp.99-102.
3
Contra Cantor -- 15-Aug-19
1
2
3
4
5
6
7
8
9
10
11
0.10111011000...
0.11010111000...
0.11100100110... = d
0.01101110001...
0.10000100011...
0.11000110100...
0.10010001110...
0.00101110001...
0.01111011100...
0.00111001011...
0.01010001010...
.
.
.
Table 1
Cantor would ask us to assume for purposes of reductio ad absurdum that this table is
“complete” in the sense that it lists every real number in the interval [0, 1). The items in this
list are enumerated by the natural numbers shown in the column on the left-hand side of the
table. The digits on the diagonal of the list (boldfaced) give us the infinitely long number d:
d = 0.11100100110...
Since d is a real number in the interval [0, 1) it must be contained in the list. Let us suppose it
is on line 3. We then construct a new number i by substituting every 0 after the binary point
in d by 1, and every 1 by 0:
i = 0.00011011001...
Cantor asserts that i is a real number in [0, 1) but it cannot be found in the putatively complete
list, because i is guaranteed to differ from every number in the list, since every number’s digit
that falls on the diagonal is changed in the corresponding digit of i. For example, i differs
from the number on the 7th line in at least in its 7th digit: the 7th digit of that number, viz. 0,
which is also the 7th digit of the diagonal, is turned into 1 in the 7th digit of i. Similarly for
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Contra Cantor -- 15-Aug-19
each and every one of the other numbers in the list. So i cannot be a member of the list, even
though we had assumed initially that the list was complete.
No list of real numbers in [0, 1) to be set up in a similar table (but with a different
ordering of reals) will manage to include all of the real numbers in [0, 1), for we can always
construct some number i for any given list, which will be left out in the list. It follows,
according to Cantor, that no such list will have the items in it being enumerable by natural
numbers. Hence, there can be no one-to-one correspondence between the set of natural
numbers and the set of real numbers in [0, 1). His conclusion is that the size, or cardinality, of
the set of reals even only in the interval [0, 1) is greater than the cardinality of the whole set of
naturals. This suffices to establish that the entire set of real numbers are also nondenumerably
infinite.3
Ib. What is wrong with Cantor’s diagonal argument?
There is a problem with Cantor’s argument above. First, let me call two binary
numbers in [0, 1) “inverses” of each other if one number has 1 in a certain location after the
binary point, the other number has 0 in the corresponding location; and if it has 0 in that
location, the other number has 1. Thus the numbers on lines 4 and 7 in Table 1, for example,
are inverses of each other. So are the numbers on lines 6 and 10. Crucially, d and i are also
inverses of each other. If we are to begin by assuming this list is complete, it is natural to
require that every number in the list must have its inverse also included in the list—otherwise
3
Cantor’s proof in [2] contained some differences from the modern popular exposition of that proof that we used
here. He considered sequences of infinite-length strings, where the strings are made of possible combinations of
0’s and 1’s, in his proof, rather than real numbers in [0, 1) written in binary system, which is what we did in the
table above. The prevalent assumption is that the popular expositions of the proof in modern texts are equivalent
in essence to Cantor’s original proof.
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Contra Cantor -- 15-Aug-19
we would have a quick objection to the completeness of the list. Cantor says i is not included
in the list. But how come? The number i’s inverse, namely d, is in the list in Table 1—we
assumed it is on line 3. Since its inverse is in the list, i must also be somewhere in the list,
like every other pair of inverses. But if i cannot be in the list, as Cantor insists, then this
should be indicative of something suspicious with his entire procedure.
We said that not only d but also its inverse i has to be in the list, in order for the initial
assumption (for reductio) of the completeness of the list to be unobjectionable. What happens
if i is in the list as the completeness assumption requires? Suppose the inverse i of the
diagonal is on line 8 in Table 2:
1
2
3
4
5
6
7
8
9
10
11
0.10111011000...
0.11010111000...
0.1110010!110... = d
0.01101110001...
0.10000100011...
0.11000110100...
0.10010001110...
0.0001101!001... = i
0.01111011100...
0.00111001011...
0.01010001010...
.
.
.
Table 2
Now, because d and i are inverses of each other, if the diagonal’s 8th digit (highlighted red) is
1, i’s 8th digit has to be 0, and if the diagonal’s 8th digit is 0, i’s 8th digit has to be 1. Since
the diagonal’s and i’s 8th digits overlap, this 8th digit can be neither 0 nor 1, which means
that the diagonal’s and i’s common 8th digit is paradoxical. Hence inclusion of i in Cantor’s
table leads to paradox in the 8th digit of the 8th line. This entails that the 8th digit of d
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Contra Cantor -- 15-Aug-19
(highlighted red) also harbors paradox. So when i is included in the table, d, and crucially i,
cease to be numbers—because they both have one paradoxical digit—and this circumstance
ruins Cantor’s ploy of inverting the diagonal.
Let me recast the flaw in the diagonal argument as follows:
(1) Assume that we have a complete table of real numbers in the interval [0, 1).
(2) Every real number in the said interval and its inverse have to be included in the table.
(3) d is a real number in the said interval.
(4) Therefore d has to be included in the table.4
(5) Since d is in the table, its inverse i also has to be in the table.
(6) But when both d and i are included in the table, they are no longer numbers, as each
now contains a paradoxical digit. (All of this, of course, derails Cantor’s strategy of
inverting the diagonal.)
(7) Therefore, a putatively complete list of reals in [0, 1) is impossible—even as an initial
assumption for a Cantorian reductio argument.
A Cantorian might think that I have not refuted what Cantor proved but simply found
an alternative way of proving his result that i cannot be contained in the list. I tend to think
instead that Cantor came up with the correct result with a faulty reasoning or invalid
reductio—and it is not uncommon for someone to arrive at a true conclusion with an invalid
reasoning. His faulty reasoning was to think that he could assume initially that the table was
complete, apparently being oblivious to the complication stemming from such an
assumption’s requirement that d and its inverse i must be included in the table. This
4
If d is not included in the table, even though we know that it satisfies the condition of being a real number in [0,
1), then Cantor wouldn’t need to bother to find a number i which is not in the table—d would fill the bill. And
constructing i would be superfluous.
7
Contra Cantor -- 15-Aug-19
complication is, to repeat, the clash of i and the diagonal at one digit (as shown in Table 2),
which makes that digit of the diagonal paradoxical. Taking inverse of a diagonal that
embodies paradox, of course, doesn’t make sense. Cantor’s constructing i would not have
looked so straightforward and unproblematic, if he had taken cognizance of this complication
brought about by the completeness assumption. My and Cantor’s reasons for why i cannot be
contained in the table of reals in [0, 1) are not the same. The difference between my position
and Cantor’s may sound subtle but is nevertheless real. After all, my reasoning (1)-(7) for the
conclusion that i cannot be contained in the table is totally different from Cantor’s.5 Cantor
thinks, (a) there is no problem with assuming that we could take the inverse of the diagonal of
the table, and (b) i is a real number (although one that is not in the table). I claim that,
because of the initial assumption of completeness, (a) the diagonal of the table becomes
unsuitable for Cantorian inversion (recall the ‘!’ sign in Table 2 on the path of the diagonal),
and (b) i and d become nonnumbers.6
Ic. Why Cantor’s “complete” table generates infinitely many nonnumbers
Actually not only the diagonal, but infinitely many other lines in a Cantorian table also
give rise to the problem of paradoxical digit. In the example in Table 3 below one of the lines
that run parallel to the diagonal, call it a “pseudo-diagonal line,” is indicated with boldface.
Let the number d* (highlighted yellow) consist of the first two digits of the first line appended
to the digits of the pseudo-diagonal line. If the table is to be assumed complete, number d*
5
Notice that (3) and (6) contradict, hence my (1)-(7) has the structure of a reductio argument. Although my
initial assumption (1) is the same as Cantor’s initial assumption, my reductio is clearly different from Cantor’s.
Indeed, my (6) is in conflict with Cantor’s diagonalization strategy.
6
See APPENDIX for another look at why Cantor’s reductio is invalid.
8
Contra Cantor -- 15-Aug-19
will have to be a member of the table. Suppose it is on line 4. Again, as a requirement of the
completeness assumption, the inverse i* of number d* also has to be in the table; so, let us
suppose it is on line 7.
1
2
3
4
5
6
7
8
9
10
0.101110110001...
0.110101110000...
0.110100011101...
0.10110110!010... = d*
0.011011100010...
0.100001000111...
0.01001001!011... = i*
0.001011100010...
0.111011001011...
0.010111011100...
.
.
.
Table 3
Thus, Cantor’s table contains a paradoxical digit not only on the path of its proper diagonal.
There are other elements of the table, like the pseudo-diagonal, number i* and the pseudodiagonal number d*, which are inflicted with the same problem of paradoxicality. And since
there are infinitely many pseudo-diagonal lines in the table, the Cantorian table contains
infinitely many rows, each with a paradoxical digit. Those infinitely many rows therefore
embody nonnumbers instead of numbers. So, the Cantorian table actually turns out to be a
display of infinitely many nonnumbers because of the initial completeness assumption. This
is another blow to the diagonalization argument, because we see that we don’t really have in
our hands a flawlessly complete table of reals in [0, 1) on which to perform the
diagonalization procedure.7
7
We can restate the problem as the following paradox. If d and d* (and the infinitely many others like d*) are
numbers, then they need to be included in a table that claims to be complete. But when they are included in the
table, they cease to be numbers, and hence should be excluded from the table.
9
Contra Cantor -- 15-Aug-19
But why insist on listing the reals in [0, 1) in a table? Cantor does so and this is where
the diagonal argument’s big fallacy lies. The problem is not that reals are nondenumerable as
Cantor thought; the problem is with listing reals in the form of a table. Such a table appears to
be a natural and innocuous way of displaying the totality of real numbers, but this appearance
is deceptive. In a supposedly exhaustive table of reals there would have to be elements—the
diagonal and the infinitely many pseudo-diagonals—that must cross each of the other
numbers in the table, including d and infinitely many numbers like d*, and their inverses, at
some digit, and this gives rise to the problem of paradoxical digits we have pointed out. As
we will see in sections IIIa and IIIb, there is an alternative way of displaying the totality of
real numbers—via binary trees—which raises no problems or paradoxes. (More on this point
in section IIIa.3.) Moreover, as we will show later, real numbers can be enumerated by
natural numbers by using this alternative method.
II. The diagonal argument regarding the cardinality of the
power set of natural numbers
Cantor uses a version of his diagonal argument to prove that the cardinality of the
power set of natural numbers is greater than the cardinality of natural numbers themselves. I
will show that our criticisms of Cantor’s diagonalization strategy can be applied here too.
Once again, I will use a version of his proof which is easier to follow than his original proof in
[2].
In Table 4 we assume that the very top row contains the complete set of members of
the power set of natural numbers in random order. On every other row of the table, we see the
symbols ‘∈’ and ‘∉’. These symbols indicate whether the natural number assigned to the row
10
Contra Cantor -- 15-Aug-19
which the symbol belongs to is a member or nonmember, respectively, of the subset found on
the corresponding column. For example, on the row to which the natural number 3 is
assigned, we see ∈ under the subset {2,3,7,9}, because that subset contains 3. On the same
row we see ∉ under the subset {2,5,68}, because it does not contain 3. And similarly for all of
the rows in the table. Cantor would then focus on the diagonal of the table (highlighted with
blue) and form a new set Si as follows. For each subset on top of the table we look at the
diagonal element under that subset. Si includes the natural number on that diagonal element’s
row, if the diagonal element is ∉, and excludes that natural number if the diagonal element is
∈. For example, the diagonal element under {4,6} is ∉, and it occurs on number 2’s row; so Si
contains 2. The diagonal element under {2,5,68} is ∈, and it is on number 5’s row; so Si does
not contain 5. And so on. The set Si obtained in this way, which in our example is
{1,2,4,6,7,… }, contains infinitely many members and is obviously a subset of the set of
natural numbers.
{2,3,7,9} {4,6}
1
2
3
4
5
6
7
8
∉
∈
∈
∉
∉
∉
∈
∉
∉
∉
∉
∈
∉
∈
∉
∉
{1,3,15,72,603}
Ø
{2,5,68}
{7}
{15,360,455}
{4,8}
…
∈
∉
∈
∉
∉
∉
∉
∉
∉
∉
∉
∉
∉
∉
∉
∉
∉
∈
∉
∉
∈
∉
∉
∉
∉
∉
∉
∉
∉
∉
∈
∉
∉
∉
∉
∉
∉
∉
∉
∉
∉
∉
∉
∈
∉
∉
∉
∈
…
…
…
.
.
.
Sd = {3,5,8,…}
Si = {1,2,4,6,7,… }
Table 4
11
…
…
…
…
…
Contra Cantor -- 15-Aug-19
Cantor claims that Si cannot be found among the original sequence of subsets on top of the
table, because it is guaranteed to differ from each subset there in at least one member. For
example, Si differs from {4,6} since 2 is a member of Si but is not a member of {4,6}; and Si
differs from {2,5,68} since 5 is a not member of Si but is a member of {2,5,68}. Because the
sequence of subsets on top of the table are enumerable by natural numbers, Cantor concludes
that there are more subsets in the power set of naturals than there are naturals. Therefore, the
power set of naturals is nondenumerably infinite.
To show that Cantor’s conclusion here too is faulty, I want to first call two subsets of
natural numbers “inverses” of each other if one has all the naturals except the naturals the
other has. (In more familiar terminology, two sets are inverses if they are complements of one
another with respect to the universal set of natural numbers.) Thus the inverse of {1} contains
all the naturals except 1. The inverse of {2,3,7,9} is the infinite set {1,4,5,6,8,10,11,12,…}.
The inverse of the empty set Ø is the entire set of naturals.
It is obvious that for every member of the power set of natural numbers, the inverse (or
complement) of it is also a member of the power set, and both the member and its inverse
must be found in the sequence of subsets on top of Table 4, which is assumed to be complete
for reductio. Moreover, the set Sd = {3,5,8,…}, which corresponds to the diagonal, must be a
member of that sequence. But then Sd’ inverse Si = {1,2,4,6,7,… } must also be included in
that supposedly complete sequence. Cantor’s diagonalization method here leads him to the
conclusion that Si is not among the subsets of natural numbers enumerated on top of Table 4.
But why should Si not be there while its inverse Sd is? Once again, Cantor’s conclusion is a
symptom of the defectiveness of his diagonalization strategy. (At the end of section IV we
show that the power set of natural numbers is in fact denumerably infinite.)
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Contra Cantor -- 15-Aug-19
A version of the problem of “paradoxical digits” shows up also in this version of
Cantor’s diagonal procedure. Suppose that the set Sd, which corresponds to the diagonal,
happens to be in the 5th column as shown in Table 5 (highlighted yellow), and its inverse
(complement) Si is on the 9th column of the table (highlighted green). If there is the symbol ∈
in the intersection of the 9th row and the 9th column, this means Sd contains the number 9.
But then Si does not contain the number 9, which implies that there must be the symbol ∉ in
the intersection of the 9th row and the 9th column. Thus the intersection of the 9th row and
the 9th column can be neither ∈ nor ∉. Once again, this is an unavoidable consequence of the
assumption that the set of subsets of natural numbers be complete. And as before, Cantor’s
supposition that he can unproblematically take the inverse of the diagonal of the table is false.
{2,3,7,9} {4,6} {1,3,15,72,603} Ø Sd ={3,5,7,!,…} {2,5,68} {7} {15,360,455} Si ={1,2,4,6,8,!,10,… } {4,8} …
1
2
3
4
5
6
7
8
9
10
∉
∈
∈
∉
∉
∉
∈
∉
∈
∉
∉
∉
∉
∈
∉
∈
∉
∉
∉
∉
∈
∉
∈
∉
∉
∉
∉
∉
∉
∉
∉
∉
∉
∉
∉
∉
∉
∉
∉
∉
∉
∉
∈
∉
∈
∉
∉
∈
∈
∉
∉
∈
∉
∉
∈
∉
∉
∉
∉
∉
.
.
.
∉
∉
∉
∉
∉
∉
∈
∉
∉
∉
Sd = {3,5,7,!,…}
Si = {1,2,4,6,8,!,10,… }
Table 5
13
∉
∉
∉
∉
∉
∉
∉
∉
∉
∉
∈
∈
∉
∈
∉
∈
∉
∈
!
∈
∉
∉
∉
∈
∉
∉
∉
∈
∉
∉
…
…
…
…
…
…
…
…
…
…
Contra Cantor -- 15-Aug-19
IIIa. Counting the real numbers between 0 and 0.999…
IIIa.1. The tree diagram in Figure 1 represents the numbers 0.00 in the uppermost branch,
0.01 in the first lower branch, 0.10 in the second lower branch, and 0.11 in the lowermost
branch.
0
0
1
0.
0
1
1
Figure 1
A similar tree diagram in Figure 2 below displays each and every real number in the
interval [0, 1) written in binary symbolism. It will be observed that no real number in that
interval, whether rational or irrational, is missing in that tree, since the tree a systematically
constructed display of all possible reals. Thus this infinite tree, whose only a small initial
portion is shown in Figure 2, is a model of the complete stock of real numbers in [0, 1).
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Contra Cantor -- 15-Aug-19
0
0
1
0
0
1
1
0.
0
1
1
0
0
…
1
…
0
…
1
…
0
…
1
0
…
…
1
…
0
…
1
1
0
…
…
0
1
0
…
…
1
…
0
…
1
…
1
Figure 2
IIIa.2. In Figure 3, we use the dashed red line to trace the branch segments of the binary tree
(a segment is the line between any two consecutive nodes of the tree) to number, hence count,
those segments. A segment crossed by the red line is labeled with ‘si,’ where ‘i’ is the countnumber of that segment. We can thus correspond each segment in this infinite tree to a unique
natural number. Note that we are counting every segment, whether the segment belongs to a
branch that terminates in infinitely many 0’s or in whatever combination of 0’s and 1’s. Also
note that our procedure of counting relies on the idea that natural numbers are not any more
exhaustible than the segments of the tree.
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Contra Cantor -- 15-Aug-19
…
0
…
1
…
s28
0
…
s27
1
…
s26
0
…
s25
1
…
0
…
1
…
s22
0
…
s21
1
…
0
…
1
…
0
…
1
…
0
…
1
…
s30
s7
0
0
1
s8
s6
s9
0
1
0
s5
s10
s1
s29
1
s24
s23
0.
s11
s4
s2
0
0
1
s12
1
s20
s19
s3
s18
s13
1
0
s17
s14
1
s16
s15
Figure 3
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Contra Cantor -- 15-Aug-19
Observe the family resemblance between the above strategy of numbering the
segments and Cantor’s way of proving that rational numbers have the same cardinality as
natural numbers, as illustrated in the following famous table:
1/1
1/2
1/3
1/4
1/5
1/6 . . .
2/1
2/2
2/3
2/4
2/5
2/6 . . .
3/1
3/2
3/3
3/4
3/5
3/6 . . .
4/1
4/2
4/3
4/4
4/5
4/6 . . .
5/1
.
.
.
5/2
.
.
.
5/3
.
.
.
5/4
.
.
.
5/5
.
.
.
5/6 . . .
.
.
.
The dashed red line counts the rational numbers as it traces them.8
From the fact that the segments of the tree in Figure 3 are enumerable by natural
numbers, and therefore the set of segments of it has the same cardinality as the set of natural
numbers9, can we derive the conclusion that real numbers in [0, 1) have the same cardinality
as natural numbers? Let us show that the answer is “Yes.” Note that the number of branches
of a binary tree is given by 2n, where n is the node level in the tree, whereas the number of
segments of it is given by ∑𝑛𝑘=1 2𝑘 . The general expression for the difference between the
Different numerals representing the same numbers, such as 1/2, 2/4, 3/6, …, need be counted only once. But
even if the infinitely many different numeral representations for each rational number are included in the
counting, the cardinality of the set of numerals in the table is still the same as the cardinality of natural numbers.
8
9
The countability of the nodes of the tree in Figure 3 follows from a theorem of set theory that states, “A
countable union of countable sets is countable.” For, the nodes at every node level are countable (because there
are finite number of them at every level) and there are countably infinitely many node levels in the tree. The
same theorem applies to the segments too.
17
Contra Cantor -- 15-Aug-19
number of segments and the number of branches is (∑𝑛𝑘=1 2𝑘 ) ̶ 2n = (2+4+8+…+2n-1+2n) ̶ 2n
= 2+4+8+…+2n-1. So, the difference goes to infinity as n goes to infinity, i.e.,
Lim n → ∞ [(∑𝑛𝑘=1 2𝑘 ) ̶ 2n] = Lim n → ∞ (2+4+8+…+2n-1)
goes to infinity. (Notice that since there are infinitely many node levels in our tree, it is an
infinite tree.) Hence there will be infinitely many more segments than branches as n goes to
infinity.10 Therefore the branches of the tree in Figure 3 are not more numerous than its
segments.11
Thus our method of counting the segments and finding them to be denumerably
infinite allows us to establish that the branches cannot have higher than denumerably infinite
cardinality. Hence the reals represented by those branches cannot have higher than
denumerably infinite cardinality. It will be noticed that we are not trying to directly show
existence of a bijective function between reals in [0, 1) and naturals. Instead, our tactic is an
indirect one: we show that the set of segments is denumerably infinite by counting them, and
prove (with the limit argument in the previous paragraph) that the cardinality of the set of
branches, and hence the cardinality of the reals represented by those branches, cannot be
greater than denumerable infinity. Now, to say that the cardinality of the set of branches
cannot be greater than denumerable infinity is to say that either, (i) the cardinality of the
branches is a finite cardinality, or (ii) it is denumerably infinite cardinality. Since the set of
10
If Cantor were right in his claim that the set of reals in [0, 1) is nondenumerable, then the set of branches
would be nondenumerable too (while the set of segments is denumerable), because the branches stand for the
reals in [0, 1). But one wouldn’t be warranted to presume that the set of branches are nondenumerable on the
grounds of Cantor’s diagonal argument’s conclusion that the set of reals is nondenumerable. For we contested
earlier, in sections Ib and Ic, the validity of Cantor’s diagonal argument.
11
It can also be proved by mathematical induction that, for every one of the denumerably infinite node levels in
the tree, the number of branches never exceeds the number of segments.
18
Contra Cantor -- 15-Aug-19
branches is an infinite set, (i) cannot be the case. Therefore, the cardinality of the set of
branches, and hence the cardinality of the set of reals in [0, 1), is denumerable infinity.
At this point one might want to argue that, if the set of reals and the set of naturals are
both denumerably infinite, then it should be possible, in principle, to list all the reals in a
table, because being in a list is the essence of being denumerable. And when we tabulate the
reals, we are back to the problem of clash between the diagonal and i that we discussed in
section Ib. My response would be to reject the claim that being denumerable entails being
tabulable. To say that reals are denumerable is nothing more to say than that there is a
bijective correspondence function between natural numbers and real numbers. A table, on the
other hand, involves more than a simple functional relation: a table, being a certain kind of
arrangement, contains a diagonal element, whereas there is no diagonal (and pseudodiagonals) in a simple, pure bijective function. Hence to assert that if reals and naturals are
both denumerable then we can set up a table pairing them off is unwarranted.12 The
appropriate way to arrange the whole set of reals is trees, not tables. Cantor’s starting point
that listing all the reals in a table would have been a natural way to enumerate the reals was
therefore mistaken.
IIIa.3. We saw in sections Ib and Ic that the Cantorian table is beset with problems. In
contrast to Cantor’s table, our tree diagram in Figure 3 does not involve any such problems,
because it is absolutely complete: no real number is left out in the tree, as the tree is a
systematical construction of all possible reals, and every real has its inverse included in the
12
The whole set of reals contains all the real numbers and their inverses, which means that there are infinitely
many potential diagonals and their inverses (and infinitely many d*’s and i*’s) for infinitely many possible
attempts to tabulate those reals. But every tabulation leads to a list where we will have the clash problem and the
diagonal and the inverse of the diagonal turning into nonnumbers. Therefore, the set of reals cannot be
displayed in any table. It is an enumerable set but this set cannot be put into the format of a Cantorian table.
19
Contra Cantor -- 15-Aug-19
tree—as it should. For this reason, our tree is not susceptible to taking “diagonals” on it. No
ploy reminiscent of Cantorian diagonalization on our binary tree will work, because if any
number were to be somehow designated as the “diagonal” of a tree, d’s inverse i will always
be in that tree. Nor can our tree be transformed into a Cantorian table, because a putatively
exhaustive table of reals and a tree representation of all reals are radically different structures:
in a table of reals there would have to be an element—the diagonal—that must cross each of
the numbers in the table, including the diagonal number d and its inverse i, at some digit
(which leads to the problems we have pointed out in section Ib), whereas in a tree there is no
branch that crosses all the other branches. Therefore, a tree and a table are crucially
asymmetrical arrangements.13
IIIb. Counting the entire set of real numbers
We will now use our method of counting the segments of a tree to do more than show
that the real numbers in [0, 1) have the same cardinality as natural numbers. This method will
enable us to demonstrate that the entire set of real numbers also has the same cardinality as
natural numbers. To do this, we will set up a one-to-one correspondence between naturals and
all the segments contained in an infinite series of binary trees, which together represent the
totality of real numbers. The infinite series of tree diagrams, of which only a small initial
13
A Cantorian might argue that our tree can (in principle) be converted into a table, by displaying the reals
represented by the branches in a list, and Cantor’s diagonization procedure will take it from there. Hence—the
contention might go—our tree is no help in refuting Cantor’s conclusions about the cardinality of the set of real
numbers in [0, 1). My reply would be that the supposition that our binary tree can be converted into a Cantorian
table is simply false because, as I pointed out above, there is an important asymmetry between a table and a tree.
Here is an a priori reason why Cantor cannot be rescued by this kind of conversion argument. Suppose we
converted our tree into a table that contains all the reals in our tree, i.e. suppose the conversion was total and
without loss of any real numbers. Assume that we then constructed the number i from the diagonal of this table.
The number i obtained after the supposed conversion will be in the original tree, since the tree is an absolutely
exhaustive display of reals in [0, 1). And since the conversion is supposed to be without loss, i has to be in the
table too. This contradicts Cantor’s claim that i cannot be in the table.
20
Contra Cantor -- 15-Aug-19
portion is shown in Figure 4, contains all of the real numbers by displaying each one of the
consecutive real number intervals [0, 1), [1, 10), [10, 11), [11, 100), etc. by a separate tree.
The tree for a given real number interval is shifted to the right with respect to the tree for the
real number interval above it. The rest of the numbering procedure should be selfexplanatory—just follow the dashed red arrows. In this way of numbering the segments there
is of course no risk of running out of natural numbers to go around…
21
Contra Cantor -- 15-Aug-19
Figure 4
22
Contra Cantor -- 15-Aug-19
Let us summarize:
(1) Each branch of every tree in Figure 4 represents a real number and every branch is
made up of segments.
(2) We can count each segment in Figure 4 by corresponding it to a distinct natural
number. Hence there are denumerably infinitely many segments in the entire series
of trees in Figure 4.14
(3) We can assert for each tree in the infinite series of trees in Figure 4, that there can be
no more branches of it than its segments (for reasons we give in section IIIa.2).
(4) Hence the cardinality of the whole set of branches in Figure 4 cannot be greater than
the cardinality of the whole set of segments, the latter being denumerable infinity (as
we said in (2) above).
(5) Since the branches represent reals, the cardinality of the entire set of reals cannot be
greater than denumerable infinity.
Now, (5) means either, (i) the cardinality of the entire set of reals is a finite cardinality, or (ii)
it is denumerably infinite cardinality. Since (i) cannot be the case, the cardinality of the entire
set of real numbers must be denumerable infinity.
Our result that there are no more real numbers than natural numbers might cause some
puzzlement. Natural numbers are a proper subset of reals and there are infinitely many
numbers other than naturals among reals. So how can we claim that real numbers are not
more numerous than natural numbers? But this is like asking, “How can natural numbers not
14
That there are countably infinitely many segments in the entire series of trees in Figure 4 also follows from the
theorem we mentioned in footnote 9, which states that a countable union of countable sets is countable. For,
each tree in the series contains countably infinitely many segments (as we saw in footnote 9) and there are
countably many trees in the whole series.
23
Contra Cantor -- 15-Aug-19
be more numerous than odd natural numbers?” Naturals are equinumerous with odd naturals,
even though the latter is a proper subset of the former. This is an example of what is known
as Galileo’s “paradox.” This phenomenon is not regarded as a paradox or puzzle anymore;
we have learned how to live with it, so to speak. Neither do we need to get puzzled about the
real numbers not being more numerous than natural numbers.
IV. Cardinality of the power sets of real numbers and natural numbers
Cantor is famously taken to have proven that the power set ℘(S) of any set S has more
members than S itself. In the more interesting case of infinite sets like the set of natural
numbers or real numbers, the cardinality of their power set is higher, according to him, than
the cardinality of the sets themselves.
I want now to argue, against Cantor, that the power set of real numbers in the interval
[0, 1) cannot have a higher cardinality than natural numbers. I will do this by first showing
that the members of the power set ℘(𝒮) of the set 𝒮 of segments of the binary tree in Figure 3
can be correlated one-to-one with the real numbers in [0, 1). Below are some examples of
how the members of ℘(𝒮) are to be correlated with the real numbers in [0, 1). In these
examples, a boldfaced 1 in the i-th digit of the real number on the right column signifies that
segment si is a member of the subset of 𝒮 corresponding to that real number, and a 0 in the
j-th digit of the same real number signifies that sj is not a member of that subset. Thus,
{s1} corresponds to the real number
0.100000000000000…
{s2} corresponds to the real number
0.010000000000000…
{s3, s7, s11} corresponds to the real number
0.001000100010000…
24
Contra Cantor -- 15-Aug-19
{s1, s5, s6, s11, s14} corresponds to the real number
0.100011000010010…
Infinite subset {s1, s3, s5, s7, s9, s11, s13, …} of 𝒮 corresponds to the real number
0.101010101010101…
Infinite subset {s2, s4, s6, s8, s10, s12, s14, …} of 𝒮 corresponds to the real number
0.010101010101010…
Infinite subset {s1, s3, s4, s6, s7, s8, s10, s11, s12, s13, …} of 𝒮 corresponds to the real number
0.10110111011110111110…
0.000000000000000…
Ø corresponds to the real number
Entire set 𝒮 (which is a subset of itself) {s1, s2, s3, s4, s5, s6, s7, s8, s9, s10, s11, s12, s13, …}
corresponds to the real number
0.1111111111111…
Hence we can represent, by the procedure above, every member of ℘(𝒮) by a unique
real number in the interval [0, 1). The converse is also true: for every real number in [0, 1]
there corresponds a unique member of ℘(𝒮). Our argument for the conclusion that the
cardinality of the power set of reals in [0, 1) is denumerable infinity is as follows:
(1) The set of segments 𝒮 of the tree in Figure 3 is denumerably infinite.
(2) The set of branches of the tree in Figure 3 is denumerably infinite. (As we argued
in IIIa.2.)
(3) Since the branches in Figure 3 represent the reals in [0, 1), the set of reals in [0, 1)
is denumerably infinite. (From 2)
(4) The members of the power set ℘(𝒮) can be correlated one-to-one with the reals in
[0, 1). (As we argued in the previous paragraph.)
(5) ℘(𝒮) and the reals in [0, 1) have the same cardinality. (From (4))
(6) ℘(𝒮) is denumerably infinite. (From (3) and (5))
(7) If two sets have the same cardinality, their power sets have the same cardinality.
(Theorem of set theory)
(8) 𝒮 and the set of reals in [0, 1) have the same cardinality. (From (1) and (3))
(9) The power set of reals in [0, 1) is denumerably infinite. (From (6), (7) and (8))
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Contra Cantor -- 15-Aug-19
How about the power set of all reals, and not just the ones in [0, 1)? We know from
section IIIb that the cardinality of the set of all reals is denumerable infinity, which is the
same as the cardinality of the set of segments 𝒮 of the tree in Figure 4. Since the power sets
of two sets have the same cardinality if the sets themselves have the same cardinality, and
since we know that the cardinality of ℘(𝒮) is denumerable infinity from line (6) above, we
can deduce that the power set of all real numbers is also denumerably infinite—contrary to
what Cantor claims.
By the way, since both the set of natural numbers and the set of segments 𝒮 of the tree
in Figure 3 have the same cardinality, both being denumerably infinite, from the theorem we
used on line (7) above, it follows that the power set of natural numbers and ℘(𝒮) have the
same cardinality. Moreover, since we know from line (6) that ℘(𝒮) is denumerably infinite,
we can conclude that the power set of natural numbers is also denumerably infinite.
V. Concluding remarks
When he formulated his theory of “transfinite numbers” based on his diagonal
arguments, Cantor proposed his “continuum hypothesis.” This hypothesis asserts that there is
no intermediate cardinality between the cardinality of natural numbers (symbolized by ‫א‬0, the
first transfinite number) and the cardinality of real numbers (symbolized by ‫א‬1, the
succeeding transfinite number). He thought the continuum hypothesis is true, but neither he
nor any other mathematician has been able to prove it. If my arguments above are correct we
know why they all failed. No wonder the continuum hypothesis hasn’t ever been proved or
26
Contra Cantor -- 15-Aug-19
disproved, because there is nothing to prove or disprove!15 Natural numbers and real
numbers, and even the power sets of the two, have, as we have shown, the same cardinality
‫א‬0.16
APPENDIX
Here is another look at why Cantor’s reductio is an invalid one. Let me reconstruct
Cantor’s argument in [2], this time in terms of binary strings:
(1) A complete sequence (complete list) L of the set M of all binary strings exists.
(Assumption for reductio)
(2) We can invert the diagonal of any infinite sequence (list) of binary strings to get a
string which is not in the sequence (list).
(Theorem Cantor thinks he has proven in
the first part of his paper.)
(3) We can invert the diagonal of L to obtain a string i which is not in L.
(From 1 and 2,
since L is a particular instance of “any infinite list of binary strings” mentioned in 2)
(4) i is not in L.
(5) i is in L.
(From 3)
(From 1, since L is assumed to be a complete list of all binary strings)
(6) Contradiction!
(From 4 and 5)
(7) No complete sequence (complete list) L of the set M of all binary strings exists.
(From 1-6, rejection of 1 by reductio)
15
In 1940, Gödel demonstrated that the continuum hypothesis cannot be disproved using ZFC (ZermeloFraenkel set theory together with the axiom of choice) and in 1963 Paul Cohen showed that it cannot be proved
using ZFC.
16
I want to express my thanks to İlhan İnan, Hasan Keler, Hilmi Demir, Ahmet Çevik, Hasan Çağatay, Özgür
Aydoğmuş for their comments.
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Contra Cantor -- 15-Aug-19
My reconstruction of Cantor’s argument above is based on his following words:
From this proposition [i.e. proposition in step (2) above] it follows immediately that the
totality of elements of M cannot be brought into [a] sequential form [i.e. list] … ;
otherwise, we would have the contradiction that a thing [i] … would be an element of M
as well as not an element of M.” ([3], p.921; my emphases)
The problem with the above argument is the following. In a complete list L of M we
have to have a string that corresponds to the inverse i of the diagonal. (Note that in (5) Cantor
assents to i being in L.) But, as we have seen, i and the diagonal coincide at some digit (at
their 8th digit in our example of Table 2) and this results in their common digit becoming
paradoxical. Because of this, Cantor’s diagonalization procedure becomes unusable on L. It
follows that (3) is false and so is the more general statement (2). Consequently, (1)-(7) is an
invalid reductio argument.
Let us put it all in outline. According to Cantor, the contradictory pair of propositions
(4) and (5) follow from the diagonalization procedure together with the assumption that we
can put all binary strings in a list; so, by reductio we can deny our assumption that the totality
of binary strings is listable. Our objection to this argument is that we cannot derive both (4)
and (5), because derivation of (4) (which presupposes diagonalization) is blocked by (5)
which renders diagonalization on the supposed list void by making the diagonal contain a
paradoxical digit. Cantor’s reductio is thereby invalidated, as it fails to derive one member of
the contradictory pair, namely, (4).
In modern popular expositions we have the following portion of Cantor’s argument
(except those expositions usually cast it in terms of binary or decimal real numbers between 0
and 1 instead of binary strings):
28
Contra Cantor -- 15-Aug-19
(1) A complete list L of all binary strings exists.
(Assumption for reductio)
(2) We can invert the diagonal of L to obtain a string i.
(By diagonalization method)
(3) i is not in L.
(4) Contradiction!
(Since L was assumed to be a complete list of all binary strings)
(5) Therefore, no complete list of all binary strings exists.
I focus on criticizing this portion (or rather, the binary-number version of it) in my paper, but
my criticisms also apply to the full original argument, as I explain above.
REFERENCES
[1] Cantor, G. (1874). Ueber eine Eigenschaft des Inbegriffes aller reellen algebraischen
Zahlen (On a Property of the Set of All Real Algebraic Numbers). Journal für die reine und
angewandte Mathematik. 77: 258-262.
[2] Cantor, G. (1890-1891). Ueber eine elementare Frage der Mannigfaltigkeitslehre (On an
Elementary Question of Set Theory). Jahresbericht der Deutschen MathematikerVereinigung. 1: 75-78.
[3] Ewald, William. (1996). From Kant to Hilbert: A Source Book in the Foundations of
Mathematics, Vol. 2. New York: Oxford U. P.
29