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Dimensional Equation Tested

I previously posted an equation on the Dimensionism group that perhaps I should attribute to my brother Brian.

In my two- to three- year history of intermittent searching for better expressions of the data, only recently was I able to re-orient myself towards something close to a correct formulation for multiple types of sets. This was due in large part to a hint (or perhaps, if I don't remember, much more than a hint...) which was embarrassingly "2 X 2" "2 X 2"... It's unclear whether what he meant at the time was that he had solved the problem (probably he had, since he's such a smart person), or whether 2 X 2 was simply a notion of how the two categories crossed in the simplest version of a quadra arrangement.

At any rate, the resulting equation was:  2 * {2^ (M root of C) - 1}.

Some initial testing early this morning suggests that this equation is highly effective at pruning out trivial cases, something that might be noted may be characteristic of my brother's intelligence rather than mine.

Here is the equation originally posted at:  http://www.facebook.com/dimensionism):

This equation works for odd numbers of categories:
2 * {2^ (M root of C) - 1}

And, this equation works for even numbers:

2 ^ (M root of C).

An approximation can be reached for quadra sets by taking the square root of the number of categories.

[Pending new results with penta, some of the figures are being revised. Some of the following is theoretical, as the methodology may be under dispute. The very viability of categorical deductions has been scrutinized in the past].


QUADRA TEST [even number]
2 ^ (4 root of 4)
= 2, correct.

2 ^ (4 root of 16)
= 4, correct, in the non-unity method.

2 ^ (4 root of 64)
= 8, correct so far as I know.


DUALISTIC TEST

2 ^ {2 root of 2}
= 2, correct.

2 ^ {2 root of 4}
= 4, correct.

2 ^ {2 root of 8}
= 8, correct.


TRINITARY TEST

2 * {2 ^ (3 root of 3) - 1}
= 2, correct! (forwards and backwards, instead of combinations of two).

2 * {2 ^ (3 root of 9) - 1}
= 6, correct! (3 * 2 * 1 combinations).

2 * {2 ^ (3 root of 27) - 1}
= 14. ? I predict greater efficiency.

The more general estimate gives 2 ^ (3 root of 27) = 8 deductions for 27 categories.


PENTA TEST

2 * {2 ^ (5 root of 5) - 1}
= 2, correct! (2 rotations of the diagram, no symmetric altercations).

2 * (2 ^ {5 root of 25} - 1)
= 6, (3 opposites * 2 opposites * 1 opposite, and no more opposites!)

Disproof of Cantor's Diagonal Argument

A series of sets of infinite sequence is said to resemble the following:

s1 = (0, 0, 0, 0, 0, 0, 0, ...)
s2 = (1, 1, 1, 1, 1, 1, 1, ...)
s3 = (0, 1, 0, 1, 0, 1, 0, ...)
s4 = (1, 0, 1, 0, 1, 0, 1, ...)
s5 = (1, 1, 0, 1, 0, 1, 1, ...)
s6 = (0, 0, 1, 1, 0, 1, 1, ...)
s7 = (1, 0, 0, 0, 1, 0, 0, ...)

It is claimed that there exists a set which is not included in the list, which is simply the abversion of the nth-most digit in each set (where n resembles the set number).

What has not been considered is that the sets are not value-ordered as stated. In other words, what has been used is the most random form of sequence for the sets. The abversion then represents a sequence which is provable in the infinite, but not in the finite.

What is difficult about an infinite set when it is organized? Such a set is said to have infinite value without exception. Or perhaps the concern is the 'volume' of the set? Isn't it possible that Cantor is confusing one infinite with another? Infinite volume may not by countable, but infinite volume does not assume quantity at all. Perhaps it is about fundamental issues, rather than mathematics.

Here is a more adequate order:

0000
0101
1010
1111

Still more accurate is the following

0000
0001
0011
0111
1111

But, in an infinite set, why wouldn't the components consist of fractions?

For example,

010 could become
0, 1/2, 1/2, 0
0,1/4, 1/4, 1/4, 1/4, 0 etc.?

By assuming the numbers have no space, Cantor may be assuming that the numbers have no quantity.

Why would the boundary be a set quantity when there is no difference in the values? In other words, why wouldn't some parts of the set be larger than others? After all, that is already how the set is defined. That is the micro problem. But there is also a macro problem. In the case of infinite sets, there is a kind of Achilles and Tortoise problem in addressing not the adequate number of sets, but rather the contents of those sets. This may cause confusion if it is not understood, e.g. if the order of the sets is arbitrary, then we must conclude that the contents are arbitrary. But if the order of the sets is ignored, the conclusion is that an arbitrary result is rational, and explicative. But that is not the case.

In the case of an infinite set these are uncountable sets, because they must be counted from last to first to ascertain value. This is what is called the entity problem in mathematics. In other words, several conflicts emerge: [1] There is a difference between a decimal set (a 0-dimensional value) and a first-in-sequence set, the second type being more geometric and less arbitrary, [2] The notion of set is affected by the concept of set, e.g. separability versus repetition and potentially still further concepts, in the case that the numbers are conceptualized, [3] The notion of set may admit to areas of numbers more than sequence of numbers, unless these sequences are seen to be 0-dimensional, or to describe multiple sets simultaneously.

The general observation is that there is a degree of arbitrariness in considering sets as though they exist in a categoric sequence without making efforts to address the quantity property of such categoric organization.

Again, it seems that the order given in Cantor's diagonal argument, at least within wikipedia, is too arbitrary to constitute a categorical set. This may be called the categorical criticism.

And, if an actual categorical organization can be conceptualized, how is it uncountable? It must simply be determined that it has volume when it is infinite.