A Solution to the Continuum Hypothesis

ABSTRACT:

Re: the Continuum Hypothesis: Is there any set which has more members than the set of natural numbers (N), but fewer members than the set of real numbers (R)? The short answer is no. The long answer, that includes mathematical proof, shall be set forth in this paper.

Imagine a finite set of natural numbers (N') with n' members. Imagine another finite set of natural numbers (N) with 2^n' members. Also imagine a set of real numbers (R) with c members:

Even though N and N' are both sets of natural numbers, they do not bijectively map to each other. However, there is a bijective mapping between N and the power set of N' (P(N')), since P(N') also contains 2^n' members:

Now let's take n' to its infinite limit. R has c members, where c = 2^n'. N and R have the same cardinality:

N and R have the same cardinality? This is not consistent with Cantor's theorem which states: ""Let f be a map from set N' to its power set P(N'). Then f: N'-->P(N') is not surjective. As a consequence, the cardinality of N' is less than the cardinality of P(N') holds for any set N'."

Clearly P(N') has more members than N'. We can prove this by counting the members of P(N'). We start with 1 and count all the way to 2^n'... but this implies the existence of a bigger set of natural numbers (N) with 2^n' members. Albeit, the power set of N (P(N)) has more members than N, but these "more members" imply the existence of an even larger set of natural numbers and so on and so on to infinity. Speaking of infinity, the cardinality of R (c) equals aleph-1 which is believed to be greater than aleph-0:

If aleph-1 is greater than aleph-0, then the reciprocal of aleph-1 is less than the reciprocal of aleph-0 (see 5 below). However, if the absolute value of the reciprocal of aleph-0 equals zero, then the absolute value of aleph-1's reciprocal is less than zero (see 6). But no number has an absolute value less than zero, so aleph-1 can't be greater than aleph-0 (see 7, 8, 9).

When comparing infinite sets, it appears that N can have the same cardinality as P(N) and R. Cantor, of course, set forth a brilliant counter-argument in support of his theorem. His argument (or a contemporary version of it) begins with a complete list of members of the power set:

Each number is classified as either "selfish" or "non-selfish." A number is selfish if it is a member of a subset and that same number is the natural number paired with the subset; otherwise, it is non-selfish. In the list above, rows 1 and 3 are examples where 1 and 3 are selfish. They are each members of their respective sets as well as natural numbers paired with those sets. Rows 2 and 4 are examples of 2 and 4 being non-selfish numbers. Now, by definition of the power set, there allegedly should be a subset S that contains all non-selfish numbers. Here's where Cantor creates the template for Russel's paradox: the natural number s can't be a member of S or S would not be the set of all non-selfish numbers. Also, the natural number s can't be a non-selfish number either, since it would be a member of S. A third option is a selfish number that is not a member of S. Unfortunately, such a number is a natural number paired with a different subset. Thus, Cantor argues that there is no possible natural number s that can pair with S. Therefore, there is no bijective mapping between N and P(N).

Yes, a brilliant argument! But Cantor had to move the goalpost to make it. If we move the goalpost back to its original position we see that a bijective mapping between two sets only requires that the sets have the same cardinality, i.e., the same number of members. It does not require the numbers and subsets to have certain attributes like "selfishness" or "non-selfishness." As an illustration, suppose we have a subset S={1, 2, 3, ..., n} that is the only subset that hasn't been paired with a natural number. Subset S is defined as the set of all non-selfish numbers. If we stay true to that definition, we will never find a number to pair it with, notwithstanding the fact that number 3 hasn't been paired with any subset. We break down and decide to map 3 to S. Now there is a complete bijective mapping between N and P(N) (where N and P(N) each have infinite members)--so why should anyone care that S is no longer "the set of all unselfish numbers"?

Let's now address the other proofs (including the famous diagonalization proof) that seem to support the claim that R and N don't have the same cardinality. These proofs start with a list of, say, all the real numbers. Each real number is paired with a natural number. (It is assumed that all natural numbers are listed.) A new real number is then produced that is not on the list. This supposedly contradicts the assumption that the real numbers are countable. If the real numbers are countable, the new real number would be on the list--so the argument goes. However, if the real numbers are countable, then it should be possible to produce a new counting number (natural number) that is not on the list for every new real number that is not on the list. Here's how it is possible:

Translate the natural numbers and/or Cantor ordinals to ASCII or unicode. This will produce natural numbers that will bijectively map to the real numbers. Now, use any proof that will produce a real number that is not on the list. This new real number will map to a unique natural number that is also not on the list. Below is an example of a possible translation scheme:

The following is the bijective mapping scheme:

Note that omega+1 circled in red is translated to 9694349. This is a unique natural number that is not on the list (that includes omega). All the natural numbers listed (translated to unicode) have a left-leading digit of 4 or 5. The unlisted natural number has a left-leading digit of 9, and, unlike omega (969) has a 43. It can be mapped to the unlisted real number. Also note that every order of infinity can be translated into a unique, finite natural number. One concern is the unicode numbers can become quite large compared to untranslated natural numbers. Will the larger unicode numbers reach infinity sooner than their smaller counterparts as the count increases, causing a more incomplete list? The following shows a comparison between n and much larger n^n. If n is the largest number short of infinity, is n^n infinity or beyond?

It appears that n^n will always be finite as long as n is finite and n^n won't be infinite until n is infinite. Thus the unicode natural numbers can map to the same infinite (or finite) list as the smaller untranslated natural numbers. This is good news! There is simply no finite or infinite number that can't be translated into a unique natural number. For every unlisted real number there is a corresponding unlisted natural number. To falsify this conclusion, all one needs to do is show that a Cantor ordinal or combination of Cantor ordinals and natural numbers can't be translated into a unique unicode natural number.

With all the forgoing in mind, let's review the question again: "Is there any set which has more members than the set of natural numbers (N), but fewer members than the set of real numbers (R)?" No, because R and N have the same cardinality.

ADDENDUM:

Is Infinity Real?

By definition, no number is greater than infinity, and, infinity seems unobtainable. The largest number imaginable can always be made larger. If infinity is obtainable, what is the largest finite number just short of infinity? We can agree that an arbitrary line segment has an infinite number of points. We might infer that a longer line segment has a larger infinity of points--but it too has an infinite number of points. How is that possible? If we think of the points as infinite zeros added together, these infinite zeros can add up to any finite number n. The value of n can be determined by how zero is expressed:

Clearly infinity is obtainable when the point length is zero and we are not willing to traverse one point at a time. Instead, we choose to slide past those infinite points to reach a finite number. Also, there doesn't appear to be more than one infinity, since the same infinite sum of zeros leads to any number n. On a grander scale, an infinite number of members leads to any infinite set. Like finite numbers, different infinite sets appear to be different, but involve the same infinity.

Limits and Infinite Precision

If infinity does not exist, then no precise or approximate number would exist? Wait ... approximate numbers should exist, since they don't require infinite precision, right? Wrong. Each approximation is a precise value of itself. To illustrate, suppose we want to expand the exponent e^x using a Taylor series. Let's assume there is no such thing as infinity, so the expansion only has a limited number of terms:

We have an approximation of e^x, but how can we if we don't believe in infinity? Let's set the approximation equal to k, then we shall expand k:

Without an infinite number of terms, we can't have an approximation of e^x--only an approximation of an approximation of e^x. Plus, we can carry out this insanity further with a limited expansion of the approximation of the approximation ... and so on. To claim that we have an approximation is to claim we don't have the number we were shooting for, but we do have the precise value of a number that is close. If we come up short or overshoot a precise point on a number line, we land on a different point that has its own infinite precision. When we have an error margin, that margin is bound by two numbers. Whether they are deemed precise or approximate, they are precisely what they are. That precision implies a Taylor expansion with infinite terms. Even if we insist that a number can't be precisely had, we can at least slide past it along the number line. If there is no infinity, there are no numbers.

Circling back to the question, what is the largest finite number short of infinity? Using the line-segment analogy, there are infinite points between the start and finish. If we can't pin down the exact point that qualifies as the largest finite number, we can at least slide past it.

References:

1. Cantor, Georg. Grundlagen einer allgemeinen Mannigfaltigkeitslehre (Foundations of a General Theory of Sets). jamesrmeyer.com.

2. Cantor, Georg. Uber eine elemtare Frage de Mannigfaltigketslehre (On an Elementary Question of Set Theory). jamesmeyer.com. 3. Cantor's Theorem. Wikipedia

4. 2020. SP20:Lecture 9 Diagonalization. courses.cs.cornell.edu

5. Cantor's Diagonal Argument. Wikipedia

6. Cardinality of the Continuum. Wikipedia

7. Continuum Hypothesis. Wikipedia

8. Huge List of Unicode Symbols. vertex42.com

Proof that Various Infinities Have a Finite Value

In the late 19th century, German mathematician Georg Cantor demonstrated that there are a variety of infinities of various sizes. Equations 1 and 2 below compare two different arbitrary infinities:

The infinity at equation 2 is n times bigger than the infinity at equation 1.

Srinivasa Ramanujan, an Indian mathematician (1887–1920), was cleverly able to extract finite values from various divergent infinite series. If having various infinities isn't weird enough, imagine various infinities having a finite value! Normally, the process of extracting a finite value from an infinity requires tackling that infinity with a unique set of steps. Thus it is not clear if all infinities have a finite value. Perhaps there are some that do and some that don't. Can mathematics provide any clues? Let's begin with an arbitrary function f that diverges to infinity:

What makes this particular infinity unique is not that variable N tends to infinity, but that N is multiplied by a coefficient alpha. This infinity is alpha times the size of a benchmark infinity that is simply N with an infinite limit. Below is another infinity (function s) with a coefficient of beta, where beta may not equal alpha:

The relationship between f and s is as follows:

Using some Ramanujanian algebra, we can solve s:

We can now solve f by making a substitution:

If coefficient k is finite and x doesn't equal 1, the solution to any arbitrary infinite function is finite. Otherwise it is infinite.

References:

1. Matson, John. 07/19/2007. Strange but True: Infinity Comes in Different Sizes. Scientific American

2. Dodds, Mark. 09/02/2018. The Ramanujan Summation: 1 + 2 + 3 + ⋯ + ∞ = -1/12? Cantor's Paradise

Introducing Stochastic Trigonometry for Quantum Physics and Statistical Mechanics

In the field of quantum physics, each eigenvalue has an eigenvector, and, when the eigenvector is normalized and squared, we get the probability for the eigenvalue. The normalized eigenvector is sometimes referred to as the probability amplitude.

When all the probability amplitudes are squared and added, the total should be 1. We can represent this with the Pythagorean theorem and the right triangle below:

The above diagram consists of two probability amplitudes: 'a' and 'b.' One is a wave function cos(theta) and the other is a wave function sin(theta).

Now, suppose there are more than two eigenvalues/eigenvectors? The diagram below shows that a and b can be broken up into smaller pieces or smaller and more numerous probability amplitudes. As before, when they are all squared and summed, they give us a total of 1.

It is possible to break up 'a' and 'b' into as many pieces as we like. Below we focus on amplitude 'a':

We can imagine breaking up amplitude 'a' into as many as an infinite number of sub-amplitudes. This can be done in both Euclidean and curved space. Equation 10 below shows how amplitude 'a' and its sub-amplitudes are invariant within flat or curved space.

With a little algebra, we can derive equation 14:

Equation 14 shows amplitude 'a' consists of an infinite number of eigenvalues (eta), each with its own probability (P(eta)). Without the probabilities, the etas would add up to infinity, and that would necessitate some sort of re-normalization technique. If we assume, however, that all quantum numbers have a probability, we will not get infinity; rather, we get the expectation value, i.e., the value actually observed.

What kind of probability values yield a finite result when eta increases linearly to infinity? Probability values that decrease exponentially. Below we derive such a probability function by using the natural-log function and converting eta to 'n':

At 16.2 we have a probability function that will reduce the probability exponentially. It gives us a number between 0 and 1, but we can derive a better function that gives us a number between 0 and 1, and, we can make a substitution. The end game is equation 16.9:

Equation 16.9 claims that if n = Q, the probability of Q (P(Q)) equals the definite integral of the probability function over a range from Q-1 to Q. We can further justify this claim with the diagram below which shows the relation between discrete values (in red) with continuous values (blue line).

Note how the area under the blue line, say, from Q-1 to Q is the same as the area of the red squares from Q-1 to Q. Equation 17 models the fact the the area under the blue line is the same as the area of the red squares over the entire range.

Now, to get a finite expectation value (amplitude 'a') we could combine equations 16.9 and 17, but the math would be more complicated than need be. To simply the math we will encounter later, let's first stretch the above diagram vertically:

Next, we draw a yellow line from zero N+1. This new line is going to make our lives easier and has the same area beneath it as the red line. Wouldn't it be nice if we could nix the red and substitute the yellow? Sure! But first we have to rotate the diagram:

Ah ... now we're in business! Below is the adjusted diagram and equation 18 with a new slope of N/(N+1):

The integral has a new range of zero to N+1, so we give the probability integral the same range:

Let's combine equations 18 and 19 to get 20:

If the limit of N is infinity, equation 20 will always give us the finite probability amplitude 'a.' No re-normalization required.

Using the diagram below and equations 21 and 22, we can derive a formula that finds probability densities:

What we've covered so far allows to find probabilities for integer values. This works fine if the value is, for example, the number of vertices in a Feynman diagram. Albeit, energy can have values of n+.5. Below is the math for that circumstance:

Notice if we divide both sides of equation 26 by Q+.5, and use summation signs, we arrive at equation 23, the formula for finding probability densities.

Now that we have the math the way we want it, let's put it to a test. Let's say we want to add up an infinite number of quantum numbers to get a finite value. Let's assume that the principle of least action applies: the most probable value will be the least action (e.g. least energy, least time, least distance, least resources required, etc.). The least probable value will be the action or event that requires the most resources, time, energy, etc. So we expect the probability to drop exponentially as the value of 'n' increases linearly--this will ensure a finite result.

Let's also assume that experiments confirm that probabilities change according to equation 27:

OK, now we only have to do some complicated math to find the expectation value 'a,' right? Wrong! At 28 and 29 below we convert the right side of 27 to a natural exponent function. If we look at equation 20, it becomes obvious that we can solve this problem by mere inspection. Looking at the exponent, everything to the right of -n is 1/a. Thus equation 31 is our final result.

Here's another test: What is the probability that a particle will travel a distance 'Q' along a pathway 'omega'? Equation 34 below can answer that. At 32 we assume that each pathway has the same probability if the distance traveled is constant, since the action is the same along each pathway (except for the direction, angles, curves, twists, turns, etc.).

Equation 35 gives us a definite answer if we want to know the probability density of a range of distances and pathways the particle can travel:

As you can see, stochastic trigonometry simplifies mathematics that can turn into a complicated, ugly, and infinite mess. It can also improve statistical mechanic's coarse-graining techinique:

Why use squares when you can use triangles?

Update: The following math shows both a convergent series and divergent series can yield a finite number 'q.' First, we start with a divergent series and make it convergent by using the probability function we derived above.

Next, we take a divergent series and assume the coefficients (the c's) don't add up to 1. Each could be any finite size; they could be a random series. The strategy is to factor out 'c' from the coefficients and use one of Ramanujan's techniques:

Another update: The following math generalizes the idea that a finite value can result from any arbitrary convergent or divergent series: