The goal of transfinite theory is to create a dichotomy between the rational and real numbers and to use this dichotomy as the foundation for a definition of the continuous line.

Transfinite theorist claim to have accomplished this goal with the diagonal method. However, I have found the diagonal proof lacking. For example, using the diagonal method, I am able to show that the set of namable numbers is both denumerable and non-denumerable. It appears that the diagonal method itself is not sufficient to establish the claimed distinction between rational and reals, and I am unable to even begin a study of transfinite theory.

Transfinite theorists claim that the dichotomy that exists between the rational and the real numbers also exists between the set of ordered pairs and the power set of the integer. Exploring these sets might give me better answers to the nature of the transfinite.

The "power set" is the set of all sets that you can create from a
given set. For example, the power set of {a,b} is { {},{a},{b},{a,b} }. Using
simple combinatorics, we know that the power set of a set with n elements has
2^{n} elements. The distribution of the elements follow the binomial
pattern (Pascal's Triangle):

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

...

Using Cantor's cross section method, it is simple to create an enumeration of the power set of an arbitrarily large set. This exercise, should give a clearer understanding of the issue addressed by the theory.

Before launching into this project, I thought I should make a brief comment about order. In set theory, order doesn't matter, the set {a,b} is equal to {b,a}. In the process of creating tables, it is easier to leave the duplicates. Instead of working with sets, I will work with ordered groupings and treat [a,b] as a different object than [b,a]. I will also treat [a] as a different object than [a,a]. I invite the reader to do the same exercise dropping the duplicates. The basic mathematics is still the same.

The set of permutations with one integer is denumerable by default. It can be listed as follows:

1 2 3 4 5 6 ... [1] [2] [3] [4] [5] [6] ...

Note, this set of groupings of one integer seems to be pretty much the same size as the set of integers.

I can list the set of ordered pairs by putting them in a table, and taking cross sections of the table. To make the table, I put my grouping of single integers in the the header row. In the first column of my table, I list all integers starting with one. The cell in postion (m,n) contains the grouping from the top row and the integer listed at the beginning of the nth row.

[1] [2] [3] [4] [5] [6] ... 1 |[1,1][2,1][3,1][4,1][5,1][6,1]... 2 |[1,2][2,2][3,2][4,2][5,2][6,2]... 3 |[1,3][2,3][3,3][4,3][5,3][6,3]... 4 |[1,4][2,4][3,4][4,4][5,4]... 5 |[1,5][2,5][3,5][4,5]... 6 |[1,6][2,6]...

I hope my methodology is clear. To create this table, I put the ordered collection of integers in the top row, then list all possible integers in the first column. Again, I am ignoring order, but mark the duplicates as magenta. [1,2] is black, [2,1] is magenta.

The table containing the set of the the ordered pairs created with the first n integers will have n^{2} elements.

I now take cross sections of the table. I will go from the left then upwards right. This produces the listing:

1 2 3 4 5 6 7 8 ... [1,1] [1,2] [2,1] [1,3] [2,2] [3,1] [1,4] [2.3] ...

NOTE: You can create a list of the rational numbers by treating the first entry as the numerator and the second as the denominator, and removing duplicates.

To make the set of ordered triplets, I simply put the set of ordered pairs in the header row. In the first column of the table I list the integers. The cell in position (n,m) contains the ordered pair from the header row and the integer at the beginning of the row.

[1,1] [1,2] [2,1] [1,3] [2,2] [3,1] [1,4]... 1 |[1,1,1][1,2,1][2,1,1][1,3,1][2,2,1][3,1,1][1,4,1]... 2 |[1,1,2][1,2,2][2,1,2][1,3,2][2,2,2][3,1,2][1,4,2]... 3 |[1,1,3][1,2,3][2,1,3][1,3,3][2,2,3][3,1,3][1,4,3]... 4 |[1,1,4][1,2,4][2,1,4][1,3,4][2,2,4][3,1,4][1,4,4]... 5 |[1,1,5][1,2,5][2,1,5][1,3,5][2,2,5][3,1,5][1,4,5]... 6 |[1,1,6][1,2,6]...

The set of ordered triplets has n^{3} members. Again, I leave it to
the reader to adjust from the set of ordered triplets to the set of three
numbers. There are quite a few duplicates, but the process is easy. I like using
the ordered triplets because the total size of the set is easier to calculate.

Taking cross sections of the list produces the enumeration. You can drop out the duplicates...(hint, they are magenta.):

1 2 3 4 5 6 ... [1,1,1][1,1,2][1,2,1][1,1,3][1,2,2][2,1,1] ...

To make my set of ordered quintuplets, I simply take the results of the
ordered triplets, put them in the headers, and list the integers in the columns
of my new row. I will leave this exercise to the student. Note that this table
with have n^{4} rows.

I can repeat this process until they lock me away. I can go even further than
that if they are kind enough to lock me away with a pen and paper. The set of
ordered n-lets will have n^{n} members.

The power permutation is akin to the power set, the big difference is that the power permutation has repeats. If you drop the repeats out of the power permutations, you will have the power set. Anyway, I will make the complete enchilada as follows:

I will start with the null set. Being a Java programmer, I think I will assign the null set the number 0 instead of the number 1. I think it looks prettier that way. I will then create one last table. In the first row of this table, I will put the integers from 1.1. In the second row, I put the collection of ordered pairs from 1.2 and so on, as follows:

null | [] 1.1 | [1] [2] [3] [4] [5] [6] ... 1.2 |[1,1] [1,2] [2,1] [1,3] [2,2] [3,1] ... 1.3 |[1,1,1][1,1,2][1,2,1][1,1,3][1,2,2][2,1,1] ... 1.4 |[1,1,1,1]... ... 1.n |[1,1,...]...

I can now take cross sections of my power collection. This will produce the result:

0 1 2 4 5 6 7 8 9 ... [] [1] [1,1] [2] [1,1,1] [1,2] [3] [1,1,1,1] [1,1,2] ...

I now have a listing of the power permutations. Set theorists can drop the duplicates to produce the power set. (You can tell a number is a duplicate if it is not in numerical order. For example [2,4,5,2] is a duplicate because 5 > 2. [78,90,105] is not a duplicate because the numbers are in order.

The size of this table would be 1 + n + n^{2} + n^{3} + n^{4}
+...n^{n. This is a massive collection of data. The power collection of a
set with 10 integers would have 11,111,111,111 elements. NOTE: The power set
of the first ten integers is also quite large. It has 210 elements.}

I successfully put the power set of the integers into a table. This process is similar to the process used to show that the rational numbers are "denumerable." Something is amiss. Either the power set of the integers is denumerable, or the process of tabulating and taking the cross sections of a table is not sufficient to establish the Cantorian dichotomy. I will be generous and assume the latter. However, the fact that we can put the power set in a table shows yet another gaping hole in the standard presentation of transfinite theory.

(Actually, what I should say is that we can start a list of the power set of the integers as easily as we can start a listing of the rational numbers. Since both sets are infinite, we will never be able to finish either set.)

Of course, there are interesting differences between the set of ordered pairs and the power set. An ordered pair has two and only two elements. There are no restriction on the number of elements in a given member of the power set. Each set in the power set will have between zero and infinity elements.

I suspect that the actual goal of the diagonal method is to show the difference between sets where the elements are finite and those where the elements are infinite.

To illustrate: In my listing of the power set, I can find the position of any finite grouping of integers. However, it is more difficult to find the position of an infinite grouping of numbers...like the set of all even numbers. The set of all even numbers is a member of power set of integers, but is also infinite.

In my listing, I can calculate the finite position of the set of the first n even integers, but do not have a definite position for the entire set of even numbers.

This is not surprising. The set of even numbers numbers does not have a definite number of members. Since the set does not have a definite number of members, I cannot find a definite position for the listing. An unknown input leads to unknown output.

Since I am unable to comprehend a completed infinity, I find it much easier to ask the same question about arbitrarily large sets. An arbitrarily large set is as big as I desire, but is still finite in size. It has a fixed upper limit. So, instead of speaking of the power set of all integers, I will speak about the power set of the first n integers.

We can ask and directly answer some very interesting questions about the set of the first n integers. For example, we could ask, what is the smallest rational number that we can create using one of these integers as the numerator and another as a denumerator. The answer would be 1/n. We can ask: what is the largest ordered pair? The answer would be [n,n]. Likewise, we can ask and find the position of the set of all even numbers less than n in our power set.

If I increased the value of n. The size of the smallest rational rational number would decrease, likewise, the position of the set of all even numbers less than n would change. I could write a computer algorithm that would determine the position of the set of all even numbers less than n based on the value of n. This function would be pretty much like all of the other functions in mathematics.

Of course, transfinite theorists are not interested in something as pedantic as arbitrarily large sets. They are interested in infinite sets. Nothing less than everything will do. So, while I sit here with a finite brain and stare at the equation, 1/n, I see that it gets smaller as n approaches infinity. Transfinitists, with infinite brains claim that 1/n actually equals zero when n=infinity. So, here we can finally see the basis for Cantor's dichotomy between the denumerable, and non-denumerable sets: It all has to do with speculations about what happens when the set of integers is infinite.

The transfinists claim that, at the completed infinity, the number 1/n equals
zero. Imagine my large set of integers. We discovered that we can make (n^{2}
- dups) rational numbers from the first n integers. At infinity, the laws change.
Just as 1/n disappeared, the rest of the rogue rationals will take a hike. At
infinity: n^{2}=n. The fact that the integers are a proper subset of
the rationals is inconsequential. Likewise, the set of ordered pairs, and all
of the algebraic numbers fall into the same black hole of reasoning.

Please don't mind that the integers are a proper sub set of the rationals. At infinity, one to many relations become one to one correspondenses. There are as many atoms as molecules. The same thing happens with the set of ordered pairs, and algebraic numbers. The rogue ordered pairs at the end of the list fall into the black hole of the transfinite's thoughts.

The power set, however, has infinities within infinities. Although it's childs
play to talk away the rogue rationals, when you look at the power set of the
integers, you have definable entities like the set of even numbers, the set
of threes, the set of squares that are not easy to dispose of. These chicken
bones of sets get stuck in the transfinitist's craw. While it is easy to say
that n^{2}=n, and that set of rational numbers is the same size as the
set of integers, transfinite theory draws a line at 2^{n}. and say that
2^{n} is greater than n at infinity.

It is all based on this infinity in an infinity idea. Sets that have just one layer of infinity are called denumerable. Sets with two layers of infinities are called non-denumerable. Cantor gave the "denumerable" sets the name aleph-0, and the second he gave the name aleph-1. It is all based upon infinities inside of infinities.

So, what happens if you take the power set of the power set. Well now, you have infinities inside of infinities inside of infinities. This multiplicity of infinities deserves a name. Cantor gave it the name aleph-2. Each time you layer in a new bag of infinities you get a new "cardinal" number. You end up with a whole new arithematic which is pretty much logically consistent. (Well, it is logically consisitent until you have an infinite number of infinities embedded in infinities. At that point you end up with an enigma.)

One of the main arguments in favor of transfinite theory is that, to a very large extent, the mathematics performed with the cardinal numbers appears to be consistent. However, if you really think about how sets and numbers behave (group theory) this really is not surprising. Let's jump back to our observations about the arbitrarily large sets. In this conversation we realized that multiplying a large set by a number had a bigger impact than addition. Doubling Bill Gate's fortune is a more impressive task than adding pennies to his billions.

Taking 2 to nth power is a more powerful operation than addition or multiplication. So if you began with a large set of 2n members, where n is an arbitrarily large number. You will end up with a number system that looks an awful like Cantor's cardinal numbers. The whole transfinite number system is not opening up or creating a new world.

On to the conclusion....

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