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
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