An Algebraic Proof of the Continuum Hypothesis

A Proof of the Continuum Hypothesis by Analysis of the Function f(x) = C on the Interval (a, b)

The continuum hypothesis was advanced by Georg Cantor in 1878. It is a hypothesis based on the possibility that infinite sets come in different sizes. Cantor proposed the hypothesis after proving that there is no one to one correspondence between the set of real numbers R and the set of natural numbers N. His conclusion that the real numbers is a larger infinity than the natural numbers was based in part on a constructive proof called the diagonal method.

The diagonal method shows that when trying to list the natural numbers and the real numbers side by side, one for one, that it is possible to construct a real number that won’t appear in the list; for which there is no corresponding natural number. He reasoned that the set of real numbers must be larger than the set of natural numbers even though both sets are infinite. Another way of looking at his results is to assert that it is not possible to list the real numbers using brute force.

Problems arise with the diagonal method due to Cantor’s reliance on:

1.                 The employment of only infinitely long real numbers,
2.                 The use of decimal approximation of real numbers of finite length, and
3.                 The requirement that the real number list be constructed in a very specific manner.

Cantor’s constructive proof is just too contrived. It’s like a trading system that’s been optimized in order to perform well when back tested but can’t generate the same returns in real-time trading.

Kurt Gödel (1940) showed that the CH cannot be disproved from ZF (Zermelo-Fraenkel set theory), even if the axiom of choice (AC) is added. Paul Cohen (1963, 1964) showed that the CH cannot be proven from the ZFC axioms. Since the HC cannot be proved or disproved within ZFC, it is independent with regard to ZFC.

If the HC is independent of the axioms of set theory is there another method available that can be employed to settle the question of its truth or falsity? Using a simple algebraic function, we can show that it is possible to map the natural numbers onto an infinite subset of the real numbers in a very straight forward manner; without resorting to the tricks and restrictions of the diagonal method.

Given – 

1. The set of natural numbers N. Let n stand for an element of N so that

N, {n ∊ N | 1 ≤ n}

2. The set of real numbers R. Let r stand for an element of R so that

R, {r ϵ R | a < r < b}

where a and b are real numbers.

3. The continuous function

f(x) = C where C is any constant and a < x < b.

4. The Evaluate Function Operator – Let E f(x) d_i be taken to mean “evaluate the function f(x) over the domain d where i is an index of the number of iterations of E and i = (1, 2 …).

Equivalencies:

1. The domain d of f(x) is the set R, {r ϵ R | a < r < b} that is d = R.

2. The index i is the set N, {n ∊ N | 1 ≤ n} that is i = N

Since the function f(x) = C is continuous over the interval (a, b) the domain d of f(x) = C must, by definition, contain all real numbers between a and b. This fact is central to the proof of the following theorem.

Theorem 1: The domain d of the continuous function f(x) = C is countable over the interval (a, b).

Proof: Invoking the evaluate function operator on f(x) = C we have

E f(x) = C d_i over (a, b) where i = (1, 2 …)

We can plot the data points as an array showing the values of i, x and f(x), depicted below.

i	1	2	3	…	n	…
	↓	↓	↓	…	↓	…
x	r1	r2	r3	…	rn	…
f(x)	C	C	C	…	C	…

The process of plotting data points can go on indefinitely. We will never run out of elements of the domain to plot nor will there be any missing elements, since the domain by virtue of the fact that f(x) = C is continuous over the interval (a, b), contains all real numbers between a and b. Conversely no matter how long the process goes on, we will never run out of values of i. These facts lead to the conclusion that there is exactly one i for every r and exactly one r for every i in the array. We have thus constructed a one to one correspondence between i and d and we can then write the correspondence as a bijective function,

f : i → d

Having shown that  f : i → d exists we can assert that the cardinal numbers of i and d are equal and the sets are the same size.

Theorem 1 can now be used to prove Theorem 2.

Theorem 2: The set of natural numbers

N, {n ∊ N | 1 ≤ n}

and the set of real numbers

R, {r ϵ R | a < r < b}

have the same cardinality and, as a result. are the same size.

Proof: Theorem 1 proves that  f : i → d exits and equivalencies 1 and 2 establish that i = N and d = R. Substituting N for i and R for d in the bijective function

f : i → d

we have

f : N → R

which completes the proof.

Theorem 3: The Continuum Hypothesis: there can be no infinite set with a cardinality strictly between that of the set of natural numbers and the set of real numbers.

Proof: The proof is a straight forward extension of Theorem 2. It has been shown that the cardinal numbers of N and R are the same. It is self-evident that it is impossible for any infinite set to have a cardinal number in between two infinite sets of the same cardinality. We have shown that N and R are of the same cardinality and therefore it is not possible for any infinite set to have a cardinal number between them. This completes the proof and confirms the truth of the hypothesis.

Ron Ragusa
May 27, 2018

Demonstrating the Naturals are not Countable

It's all in how you present the solution. The listing of naturals isn't limited to any particular format. It turns out that if the naturals are listed so that one digit of each number lies on the diagonal then Cantor's diagonal method works just fine. There's no need to reformat the numbers themselves.

Consider the following two demonstrations:

Pattern 1 – Highest digit of each natural number on the diagonal (absolute). The number Y differs from every number in the list by at least one digit, just as in Cantor's diagonal method on the (0, 1) real number interval.

1   7   3                                                                   
    2   1   1   1   1   1   1                                               
        3   2                                                               
            4   6   5                                                       
                5   0                                                       
                    6   7   3   2                                           
                        7   8   8   8                                       
                            3   1   4   8   7   1   5                       
…

Y =
3     1       4      8     7      1     5       6 …

Pattern 2 – Any digit of the natural number on the diagonal (clustering). In this case, 1, 2, 3, 4, 5, 6, 7 and 3 all lie along the diagonal. Again, Y differs from every number in the list by at least one digit, just as in Cantor's diagonal method on the (0, 1) real number interval.

1   7   3                                                                   
    2   1   1   1   1   1   1                                               
    2   3                                                                   
    5   6   4   1   3   7                                                   
                5                                                           
            1   7   6   7   3   2                                           
                3   2   7   8   8   8                                       
                            3   1   4   8   7   1   5                       
…

Y =
3     1       4      8     7      1     5       6 …

As with Cantor’s treatment of the reals on (0, 1), Y differs from every entry in the list by at least one digit. Both patterns result in a natural number that will never appear in a list of natural numbers no matter how long the list grows.

I have purposely chosen to present the list of naturals in random order, but the method works if the list is presented well ordered as well.

The above demonstration has some interesting implications regarding the Continuum Hypothesis. I have demonstrated the Natural Numbers are not countable and the fact that the Real Numbers on the (0, 1) interval are equally not countable means that both sets possess the same cardinal number. As a consequence of this fact that there can be no infinite set with a cardinal number between them. The Continuum Hypothesis is true because the naturals and reals have the same cardinal number.

A Different Approach

I am going to approach this from a non-numeric perspective. In 1969 G. Spencer Brown authored a book titled Laws of Form. In the book Spencer Brown lays out ideas behind the calculus of indications, a primary non-numerical arithmetic and algebra. I’m not going to go into a detailed explanation of the calculus he discovered (invented?). You may already be aware of his work but in the event you’re not, his book is still in publication and there’s lots of commentary available on the internet.

I have noted with footnotes where I am quoting LOF verbatim. Unquoted definitions, theorems, instructions etc. are my own extensions to the calculus, introduced in order to facilitate the discussion.

“Definition

Distinction is the perfect continence.

That is to say, a distinction is drawn by arranging a boundary with separate sides so that a point on one side cannot reach the other side without crossing the boundary.”[1]

“Construction

Draw a distinction.”[2]

“Knowledge

Let a state distinguished by the distinction be marked with a mark

┐

of distinction.”[3]

Assignment

Call the mark a cross.

“Arrangement

Call the form of a number of crosses [tokens] considered with regard to one another (that is to say, considered in the same form) an arrangement.”[4]

“Expression

Call any arrangement intended as an indicator an expression.”[5]

“Axiom 1: The Law of Calling

The value of a call made again is the value of the call.”[6]

Length

Length is a state of an expression and is given by the number of crosses that comprise the expression, denoted by L(e).

“Value

Call a state indicated by an expression the value of the expression.”[7]

Expressions of unequal lengths are unequal in value. That is to say,

If L(e_i) ≠ L(e_j) then e_i ≠ e_j

 “Equivalence

Call expressions of the same value equivalent.
Let a sign

=

of equivalence be written between equivalent expressions.”[8]

Initial 1: Number

┐= ┐┐(expansion)

and

┐┐= ┐(contraction)

Repeated applications of Initial 1 (expansion) will create forms of increasing length indefinitely.

┐	=	┐┐
┐┐	=	┐┐┐
┐┐┐	=	┐┐┐┐
	…

                         I1

                         I1

                         I1

Call the set  created by indefinitely invoking Initial 1 on each form in the list the set of Natural Expressions [N_e].

Question: Is it possible to list the elements of [N_e] such that no element will be missing from the list?

Given: The first distinction           ┐

“Theorem 1: Form

The form of any finite cardinal number of crosses can be taken as the form of an expression.

That is to say, any conceivable arrangement of any integral number of crosses can be constructed from a simple expression by the initial steps of the calculus.”[9]

The proof of Theorem 1 is given in the text of LOF so I won’t repeat it here. What’s important to note is that using Initial 1 in conjunction with Theorem 1 it is possible to construct an expression the form of which is a member of [N_e] but will never appear in the listing of [N_e].

Theorem 2: Incompleteness

An attempt to list the elements of [N_e] must be incomplete.

That is to say, it is possible to construct an element ω of [N_e] such that ω will never appear in the list of [N_e] no matter how long the list grows.

Proof: The element ω of [N_e] is constructed as follows:

List the elements of [N_e] one above the other. After each element is listed apply Initial 1 to it to expand its length and append it to ω. Since ω is a single expression the length of which grows over time as the listing of [N_e] proceeds, the length of ω will grow at a rate such that it will always exceed the length of any expression in the listing of [N_e].

e	ω
┐	┐┐
┐┐	┐┐┐┐┐
┐┐┐	┐┐┐┐┐┐┐┐┐
┐┐┐┐	┐┐┐┐┐┐┐┐┐┐┐┐┐┐
…	…

ω = ┐┐┐┐┐┐┐┐┐┐┐┐┐┐after 4 elements of [N_e] have been listed.

It’s clear that no matter how long the listing process goes on that ω, an element of [N_e] will never be an expression that appears in the listing of the elements of [N_e]. Since ω will never appear in the list, the list will always be incomplete which proves Theorem 2.

“Third canon. Convention of substitution

In any expression, let any arrangement be changed for an equivalent arrangement.

Step

Call any such change a step.

Let a sign

   

Stand for the words

is changed to.”[10]

Using the convention of substitution we can now substitute the arrangement of symbols representing the natural numbers for crosses in expressions of the form according the following rule.

Rule of substitution

Let an expression of the form be substituted by a natural number such that the value of the natural number is equal to the length of the expression, so that:

┐		1
┐┐		2
┐┐┐		3
	…

We can now expand the substitution of number for expression to include ω

e	ω
┐	┐┐
┐┐	┐┐┐┐┐
┐┐┐	┐┐┐┐┐┐┐┐┐
┐┐┐┐	┐┐┐┐┐┐┐┐┐┐┐┐┐┐
…	…

n	ω
1	2
2	23
3	234
4	2345
…	…

ω = ┐┐┐┐┐┐┐┐┐┐┐┐┐┐

                                                     

ω = 2345

In both cases as the listing of each set continues ω will never appear in the list. Consequently both the non-numeric and numeric lists will always be incomplete.

---

[1] Laws of Form, G. Spencer Brown page 1

[2] Laws of Form, G. Spencer Brown page 3

[3] Laws of Form, G. Spencer Brown page 4

[4] Laws of Form, G. Spencer Brown page 4

[5] Laws of Form, G. Spencer Brown page 4

[6] Laws of Form, G. Spencer Brown page 1

[7] Laws of Form, G. Spencer Brown page 5

[8] Laws of Form, G. Spencer Brown page 5

[9] Laws of Form, G. Spencer Brown page 12

[10] Laws of Form, G. Spencer Brown page 8