What is the Collatz Problem and how can it be solved?

[matt grime]

Originally posted by Organic
Matt,

This is theory development forum, where I can define a tree in my way.

but you didn't define tree though. i asked you about it repeatedly but you never acutally described it properly. undoubtedly it made sense in your head, but you didn't explain it to anyone else. in fact anyone who actually looked in your article would see that you drew a tree as is understood in graph theory, albeit with an infinite vertex set. now you claim the 'tree' is the cantor set. yet the tree is the natural numbers, therefore you're talking crap again! as even allowing for your inconsistent notation you've said it is countable and uncountable, an impossible dichotomy.

 

[Organic]

Matt,

You don't remember the last version of my work on Cantor's diagonal methed, so please look at it again:

http://www.geocities.com/complementarytheory/NewDiagonalView.pdf

And this time please read all of it, if you want to understand it.

 

[matt grime]

ah lovely, that heap of garbage again. still using the axiom of infinity of induction despite there being no such thing, still claiming the number of rows is 2^aleph-0 because of the finite case. still wrong despite the number of revisions you've undertaken.

there is not justification for claiming there are 2^aleph-0 rows. there aren't. you are wrong and there really is no simpler way of saying this. so why are there 2^aleph-0 rows? go on pleae state here and now in mathematical terms why there are 2^aleph-0 rows which are enumerable.here is the counter proof to your assertion

the list you produce is enumerable and is alleged to be the power set of N. Let z be in the power set of N. it is in the list at some point, n(z). by construction though the element at n(z) has only finitely many non-zero entries, therefore as z was arbitrary we have a contradiction.

you've still not managed to refute that counter example to your unfounded assertion.

 

[Hurkyl]

Allow me to reemphasize my conclusion:

If we assume the real numbers are countable, we can find a set whose total length is 1, yet this set contains every point of the entire real line!


Are you actually comfortable with the implication that the entire real line a length no greater than 1?

 

[Organic]

Matt,

by construction though the element at n(z) has only finitely many non-zero entries

By what construction?

Please give a detailed example of this construction.

 

[Organic]

Hurkyl,

That’s exactly the Idea, only a solid line (which means no points in it) has length 1.

No collection of infinitely many points can use the model of a solid line.

Fullness = Solid line = {__} content = Mathematics language strong limit.

For better understanding please look once more at(please pay attention to the Continuity that stands in the basis of empty or full(green) triangles):

http://www.geocities.com/complementarytheory/4BPM.pdf


Emptiness = {} content = Mathematics language weak limit.


Mathematics language is already aware to {} content.


It is the time to fulfill the symmetry by being aware to {__} content.

 

[matt grime]

Originally posted by Organic
Matt,By what construction?

Please give a detailed example of this construction.

]]

by your construction. column 1 goes 010101010...

column two goes 001100110011...column 3 goes 000111000111...

you cycles based on 2^n remember.therefore the n'th column starts with n zeros. Te resulting infinite matrix you write down is thus strictly lower tringular - just look at the first few diagrams you've drawn in that newdiagonal.pdf

for any row, row r say, reading right to left, all the entries become 0 after the r'th place (if not sooner), thus there are only finitely many non-zero terms in the r'th row (at most r of them).

I've told you this on at least 5 occasions and you've never managed to disprove it. you can't because it's clearly true.

In fact the thing you constructed precisely enumerates the 'finite' power set - the set of finite subsets of N which is countable.

 

[Hurkyl]

That’s exactly the Idea, only a solid line (which means no points in it) has length 1.

But, by definition of length, [0, 2] has length 2. And [0, 2] is part of the real line, so the real line has to have length no less than 2.


And 1 isn't special; allow me to modify my proof a little:

Suppose the real numbers is countable. Choose any enumeration of them.

Create a countable collection of intervals such that the i-th interval contains the i-th real number, and has length 1/2^(i+1).

This collection of intervals contains every real number, however, the total length of all of the intervals is 1/2.

So now I've proven the length of the real line is no greater than 1/2.


(In fact, I can prove the length of the real line is equal to zero, with an addition to this argument)

 

[Organic]

Matt,

for any row, row r say, reading right to left, all the entries become 0 after the r'th place (if not sooner), thus there are only finitely many non-zero terms in the r'th row (at most r of them).

There is no r'th place where after it you know exactly what is the next notation (depends on the base value for example: in base 2 the notation can be 0 XOR 1, in base 3 the notation can be 0 XOR 1 XOR 2, in base 4 the notation can be 0 XOR 1 XOR 2 XOR 3, and so on).

Which means that when we dealing with fractalic(=a^b) subsets where b is non-finite, probability enters to the picture and can't be ignored, as i clearly show here:

http://www.geocities.com/complementarytheory/PTree.pdf

This is one of the fundamental mistakes that Cantor did when he researched infinity, and he made this mistake because in his time information theory and fundamental concepts like redundancy and uncertainty were not “must have” concepts of infinity research.

 

[matt grime]

Your argument doesn't hold water - you fixed base 2. that is how you constructed this object, this 'list' of strings of 0s and 1s. To say that you might have 2s or 3s in the strings of 0s and 1s is frankly misleading, if not a down right attempt to change the subject away from something where you are wrong. You're off again on an unrelated topic.

Jus look at the construction YOU gave, the t'th column starts with t zeroes! You can cleary see that on the r'th row, every entry after the r'th column must be zero. Look at you'ure own diagram where you can see the pattern that all the numbers above the diagonal are 0 - it is trivial to show that this pattern continues in the 'list' as I've just proven.

Can we make it a bit clearer? the t'th entry in row r is the entry from column t, if t>r (and r is fixed remember) then as the t'th column starts with t zeroes and r<t it must be that the r'th entry in that column is 0 becuase all the entries from 1,2,...r,..,t are zero. (This is your construction, yet you do not even understand this simple observation.) So after the r'th place in row r all the entries are zero. thus the r'th row has only a finite number of non-zero entries, (at most r). Thus the corresponding element in the power set is a finite set.

 

[Organic]

Hurkyl,

This is exactly the beautiful thing in {_} content, for example:

.__. = Finite line = [__]

__ = Infinite line = (__)

.__ = Infinite line = [__)

|{.__.}| = 1

|{.____.}| = 1

|{.________.}| = 1

|{.__}| = 1

|{__}| = 1

Shortly speaking, __ is the essence of an invareant self similarity over scales.

 

[Organic]

Matt,

To say that you might have 2s or 3s in the strings of 0s and 1s is frankly misleading,

I don't believe that you as a mathematician say such things.

In base 2 the notations are 0,1

In base 3 the notations are 0,1,2

In base 4 the notations are 0,1,2,3

And so on ... (the fixed base value is only in your head)

In my model the important thing is the power_value, where base value
can be any finite natural number.

Because you can't understand the idea of probability in my system, let us look at it without using probability.

by construction though the element at n(z) has only finitely many non-zero entries

Each row is a unique combination of infinitely long sequence of notations (depends on base value) no more no less.

If power value is non finite then “left” side can be (when base value = 2) …000…, …111… or any sub-combination of 01 notations.

 

[matt grime]

you fixed base 2, don't start changing it now. these things you write down are the indicator functions of sets and thus only take values 0 and 1. Where did you start to bring base 3 or different? what would base 3 even mean. don't answer that as it isn't important and just let's you wriggle out of justifying anything again.

look at your own paper the colums are labelled with powers of 2, that's the key thing.

and as the specific case of 2 here is false, the general case, whatever that might be, cannot be true.

by construction you do not produce any strings with infinitely many 1s in them if you are claiming to be able to enumerate them as you do.

Pick any element in the list, look we've proved it has only a finite number of non-zero entries if your claimed enumeration is true.

therefore your claimed enumeration is false, not mathematics.

and moreover the claimed enumeration sends a string to the binary expansion it denotes, and thus can only be defined for strings with finitely many 1s in.how is it bad mathematics to state a string of 0s and 1s cannot have 3s or 4s in it? i'd have thought that was bloody well obvious.

 

[ahrkron]

Originally posted by Organic
Please look at my paper:
http://www.geocities.com/complementarytheory/ET.pdf

I read it.

There is nothing there that applies to the current discussion.

However, there are many comments I can make about its contents.

1. You define "equation trees" in a barely acceptable way (though *very* informal), but then you start using undefined notations and properties. Examples: "symmetry-degree", "information's clarity-degree", the "xor" notation in page 2, the parentheses notation on page 3.

2. On page 3, you say that these trees may be used to "construct and explore complex relations between elements". With "elements", do you mean natural numbers? If so, your assertion is wrong, since natural numbers are well defined already. Your "equation trees" are based on them, and cannot produce further understanding of the numbers they are buit upon!

3. The diagrams on pages 4 and 5 are simple exercises in combinatorics. No new info is gained from looking at them explicitly.

4. On page 6, the product of ET's is never defined. Furthermore, in order to decide if the system is commutative or not, you need to define: a) how to obtain the product, and b) when ET's are equal (the drawings you show there are indeed different, but they could perfectly be two different "symbols" for the same trees).

5. Diagrams in page seven, again, are never defined. They seem entertaining to make, but they also seem useless. The "operations" marked in blue on them also use an undefined notation.

6. Page 9: You never defined what "information point" is supposed to mean.

7. The drawings that occupy the entirety of pages 9 and 10 give no new information, neither do they illustrate any idea from the text. They seem, again, as simple exercises on combinatorics (... related to an unspecified problem).

I'm sorry to say this, Doron, but your "complementarytheory" is really far from providing any insight into natural numbers, and has nothing to do with the issue at hand.

However, you are putting a lot of effort into this. Why not taking a "vacation" from defending your theories and spend the time learning (with an open mind) how things are done in "standard" math? You could gain much from it. I mean it, I'm not trying to patronize you or be condescending.

 

[matt grime]

in addition to all that, ahrkron, it has come to light that when he says tree he doesn't mean tree as you or i would understand it, but actually a cantor set. apparently it's ok to do this because we're in theory development. as this thread was started by Organic in a math forum and as he never offered a definition of his 'tree', I don't find that a remotely compelling argument, don't know about anyone else. There's also the issue that post facto redefining extant terms to suit yourself is frowned upon: how do we even know that what organic means by cantor set is what the rest of the world means? He didn't even know what a bijection was until it was explained to him, but that hadn't stopped him talking about them before, so the chances of him knowing what a cantor set really is are quite slim.

 

[Organic]

ahrkron,

All you demonstrate is that at this point you don't understand my work, no less no more.

By the way, because you are a professional mathematician with maybe a lot of knowledge please show me some mathematical branch where multiplication and addition are complementary operations, for example:
http://www.geocities.com/complementarytheory/ASPIRATING.pdf

You are invited to visit my web site: http://www.geocities.com/complementarytheory/CATpage.html

And maybe if you will let your self to be opened to another point of view on Math language, we will be able to communicate with each other in the near future.

Yours,

Organic

 

[matt grime]

Originally posted by Organic By the way, because you are a professional mathematician with maybe a lot of knowledge please show me some mathematical branch where multiplication and addition are complementary operations, for example:
http://www.geocities.com/complementarytheory/ASPIRATING.pdf

if i go to that article will it define what it means for two binary operations to be complementary? Let's see...

oh look no it doesn't! so simply here and now define what it means for two binary operations to be complementary. Acutally, could you even justify why they are operations on N? each element of N is a collection of partitions of a set with n elements. you mulitply 2*3 and get _some_ of the partitions for 6, but you don't acutally get a number do you, you don't get an element of N. You get a subset of an element of N. How can that define a binary operation from NxN to N? so it is a map from NxN to something that isn't N, even with your definitions of numbers.

I can see an easy way of coming up with some analogy using operads and turning them into, perhaps, some groupoid in a second way, which is after all all we need to do -- your multiplication isn't defined on all partitions, only some of them, at least that is what you appear to say (that bit about structure on page 2)

 

[ahrkron]

Originally posted by Organic
All you demonstrate is that at this point you don't understand my work, no less no more.

I understand your desire to create on math, and some of your intent while defining the ET's, but I also understand clearly that there are many flaws on your attempt.

It is not necessary to understand the totality of your work in order to see the problems I pointed out in my previous post.

And maybe if you will let your self to be opened to another point of view on Math language

I am very much open to new math concepts and notations. I frequently use quite bizarre notations myself, but it is important to do it in a self consistent way.

Also, if you are to claim that you have uncovered anything about natural numbers, you need to make sure that you translate properly between your own symbols and those used by other people. Otherwise, you need to say "I have discovered fascinating properties about what I call natNumis"; you will surely have less of an audience that way.

Finally, if you allow me, I'd like to advise you not to disregard criticism made on your ideas by just saying "you didn't understand me", since then you will not gain anything from the interaction, and you risk loosing an oportunity to correct perceived or real errors on your work. Very often, good science is done by giving up on ideas that seemed beautiful and that were close to our hearts, but were incorrect nonetheless.

[Edit: added last 4 words]

 

[Organic]

Dear Matt,

0,1 (base 2) is just some example that can be translated to any fractal which is based on some finite n>1.

A fractal, as you know has a non-linear property and in the case of base 2 we gat a Binary tree, that can be represented in non-compressed way by an ordered matrix of aleph0 width on 2^aleph0 length 01 notations.

 

[Organic]

Dear ahrkron

Finally, if you allow me, I'd like to advise you not to disregard criticism made on your ideas by just saying "you didn't understand me", since then you will not gain anything from the interaction, and you risk loosing an oportunity to correct perceived or real errors on your work.

First thank you for your gentle and posivite attitude, but from criticism I have have learned that you simply don't understand my work,therefore the detailed remarks that you gave in the previous posts, cannot help not to you an not to me, at this stage.

If you agree let me start step by step and ask you again this question:

Because you are a professional mathematician with maybe a lot of knowledge please show me some mathematical branch where multiplication and addition are complementary operations, for example:
http://www.geocities.com/complementarytheory/ASPIRATING.pdf

 

[matt grime]

but it is you assertion that the set rows simultaneously countable, uncountable, and obtained by some undefined induction that is not true.

forget fractals, forget probability.

you cannot, have not, and evidently will not prove anything about the rows. I have, on the other hand, proved you are wrong and you haven't managed to contradict the proof I've offered. Perhaps because you yourself do not even know how to produce this array 'inductively'?

Do you agree that the first column (on the right) is the sequence 01010101... that the second column is 00110011... etc

that is that by construction there rows are countable - their are as many rows as there entries in a sequence and the sequence is indexed by N.

Now why do you insist that there must be 2^aleph-0 of them when my proof demonstrates that under this construction there are no infinite subsets of N in the construction?

 

[Organic]

Matt,

Why you repeat yourself?

You proved nothing.

My matrix of 01 notations is:

<---width=aleph0 ---0
                    | 
                    |
                    |
                    |
                    |                    
                    |
                    |
                    |
              length=2^aleph0  
                    |
                    |
                    |
                    |                    
                    |
                    |
                    |
                    V

and both width and length are countable and any row or column is a non-finite unique sequence of 01 notations.

By the way it is constructed the result cannot be but an aleph0 x 2^aleph0 01 matrix, where aleph0 cannot be beyond the infinitly many notations from one hand, and also to be their cardinal on the other hand.

The reason is very simple and can be demostrated in this model:

http://www.geocities.com/complementarytheory/RiemannsLimits.pdf

Therefore the the idea of the transfinite universes is a conceptual mistake.

Your "proof" that takes any ... as ...000... simply demonstrates that you closed under this conceptual mistake.

 

[matt grime]

I repeat myself because I am correct. We have agreed on how to construct this infinite array, yes?

the first column is on the right, then going to the left we number the columns accordingly.

the first column is 'based on 2^1' and goes 0101010101...

the second colummn reads, downwards, 00110011...

and so on for each column.

the r^th column starts with r zeroes.
then r ones then r zeroes...
Yes?

Let t be any row, the entry in the s'th place (reading right to left) is the entry from the s'th column.

whenever s >t this entry must be zero.

therefore any row is eventually all zeroes, and every element you enumerate in the power set is finite.

Read that argument carefully and don't dismiss it simply because it is me. as you keep telling people they must be open to new ideas, well, so must you.

so go through it and at each point tell me what you think is wrong with the deduction there.

 

[Organic]

Matt,

I have no choice bu to say it again,

By the way it is constructed the result cannot be but an aleph0 x 2^aleph0 01 matrix, where aleph0 cannot be beyond the infinitly many notations from one hand, and also to be their cardinal on the other hand.

The reason is very simple and can be demostrated in this model:

http://www.geocities.com/complementarytheory/RiemannsLimits.pdf

Therefore the the idea of the transfinite universes is a conceptual mistake.

Your "proof" that takes any ... as ...000... simply demonstrates that you closed under this conceptual mistake.

Please try carefully to understand my model of infinity.

 

[matt grime]

which ... am i taking to be alll zero? I'm not doing any such thing.

so take my proof point by point and say what's wrong at each point.

anyone else reading this care to tell me where it goes wrong?

(the link you point to is of no importance here)does the rth column not go 000..01111..11100..0001111.111...

where the are r zeroes and r ones in each block?

look at the diagrams you've drawn. look at the 1st row. doesn't it go 00000 all the way across (i'm reading the numbers backwards here)

doesn't the second row go 1000000...

doesn't the third go 0100000... thence all zeroes?

doesn't the 4th go 1010000000...thence all zeroes

look at the 5th rowtake the 6th entry from the right. that is the 5th entry in the 6th column, the 6th column starts with 6 zeroes.

look at the 7th entry in the 5th row, it comes from the 5th entry in the 7th column, the 7th colum starts with 7 zeroes

look at the rth column where r is bigger than 5. it starts with r zeroes, so the 5th one must be a zero.now i can repeat that for any row, and see that the t'th row is all zeroes after the t'th entry by the way you've defined the columns.

now which step is wrong in that?

now, you claim that the rows enumerate the power set of N, by the standard indicator function argument, I've just shown you every row has a finite number of 1's in it.
here's a little test. Let S_n be a set with n elements. is the union over n in N countable? yes. suppose now S_n has 2^n elements in it. is the union over n in N countable? yes, but you seem to think it isn't because of this axiom of infinity misconception you have lying around.

anyway, take the proof rewritten there including an illustration of the method for the specific exanples of t=1,2..5 and explain where it is wrong.

 

[Organic]

Matt,

Also please look at this:

    3 2 1 0
   2 2 2 2
   ^ ^ ^ ^
   | | | |
   v v v v
...1 1 1 1 <--> 1
...1 1 1 0 <--> 2
...1 1 0 1 <--> 3
...1 1 0 0 <--> 4
...1 0 1 1 <--> 5
...1 0 1 0 <--> 6
...1 0 0 1 <--> 7
...1 0 0 0 <--> 8
...0 1 1 1 <--> 9
...0 1 1 0 <--> 10
...0 1 0 1 <--> 11
...0 1 0 0 <--> 12
...0 0 1 1 <--> 13
...0 0 1 0 <--> 14
...0 0 0 1 <--> 15
...0 0 0 0 <--> 16
...

 

[matt grime]

and how are we defining this this time? what makes you think the rows you've now defined are included in the original construction? you say they must be without offering any proof, other than asserting something that is unjustified (because of you misconception about the axiom of infinity) proof, and i prove they aren't.

it doesn't matter if you even permute rows because i can just unpermute them back to the form you have in the article.

it looks like you've just put 1s in every place above the diagonal, and thus you've exactly got the elements of the power set whose complement contains only finitely many elements, these are called cofinite and are also countable.so, go through the water tight proof I've offered you twice now in the last 5 posts or so alone (and many times before that) point be point and say where you think it is wrong, and why, and give a counter example based upon the construction you offer in the article newdiagonl.pdf which it has to be said is now worse because you've taken out any mention of how you actually *might* construct the infinite array.

 

[Organic]

Matt,

What is the result of 2^aleph0 - aleph0?

 

[matt grime]

if by that what you mean what is the cardinality of a set of cardinality 2^aleph-0 after removing aleph-0 elements, then the answer is that the cardinality of this set is strictly greater then aleph-0. (it is 2^aleph-0 as we will show)

simple example: there is a bijection between between R and R\Z

defein the map piecewise

on [0,1)

send 0 to 1/2, 1/2 to 1/3, 1/3 to 1/4 etc call this map f and extend to the rest of [0,1) by setting it to be the identity

define the map analogously on each interval [n,n-1)alternatively, for R\N, say, it is still infinite, let x(n) be any sequence in R\N, send

n to x(2n) and x(2n) to x(2n+1)

that do you?

 

[Organic]

Thank you Matt,

So if you look now at this:

    3 2 1 0                   3 2 1 0
   2 2 2 2                   2 2 2 2
   ^ ^ ^ ^                   ^ ^ ^ ^
   | | | |                   | | | |
   v v v v                   v v v v
...0 0 0 0 <--> 1    or   ...1 1 1 1 <--> 1
...0 0 0 1 <--> 2    or   ...1 1 1 0 <--> 2
...0 0 1 0 <--> 3    or   ...1 1 0 1 <--> 3    
...0 0 1 1 <--> 4    or   ...1 1 0 0 <--> 4    
...0 1 0 0 <--> 5    or   ...1 0 1 1 <--> 5   
...0 1 0 1 <--> 6    or   ...1 0 1 0 <--> 6    
...0 1 1 0 <--> 7    or   ...1 0 0 1 <--> 7    
...0 1 1 1 <--> 8    or   ...1 0 0 0 <--> 8   
...1 0 0 0 <--> 9    or   ...0 1 1 1 <--> 9   
...1 0 0 1 <--> 10   or   ...0 1 1 0 <--> 10  
...1 0 1 0 <--> 11   or   ...0 1 0 1 <--> 11  
...1 0 1 1 <--> 12   or   ...0 1 0 0 <--> 12  
...1 1 0 0 <--> 13   or   ...0 0 1 1 <--> 13  
...1 1 0 1 <--> 14   or   ...0 0 1 0 <--> 14
...1 1 1 0 <--> 15   or   ...0 0 0 1 <--> 15  
...1 1 1 1 <--> 16   or   ...0 0 0 0 <--> 16
...                  or   ...

or a mixing of them

has a length of 2^aleph0 unique 01 sequences.

 

[matt grime]

and what? why won't you even attempt to refute the proof offered that you're wrong? if you're so sure of you position it should be quite easy for you.i'm not sure where you're going but the two lists you've written enumerate the finte and cofinite elements of the power set. there are more sets in the power set than that. in fact it is a trivial exercise to show that the countable union of countable sets is countable, and you're not going to get anywhere with this idea.

come on organic, you think i don't understand infinity, and you're the expert. what's wrong with the proof offered to you?

it is perfectly mathematically sound. you're not the first person to have made this mistake and you won't be the last.

 

[Hurkyl]

Hurkyl,

This is exactly the beautiful thing in {_} content, for example:

.__. = Finite line = [__]

__ = Infinite line = (__)

.__ = Infinite line = [__)

|{.__.}| = 1

|{.____.}| = 1

|{.________.}| = 1

|{.__}| = 1

|{__}| = 1

Shortly speaking, __ is the essence of an invareant self similarity over scales.

So how do you reconsile this with the fact that, if I assume the real numbers are countable, I can prove the real numbers have a length less than 1/2?



Anyways, one of your major problems is that you seem to confuse the order of the quantifiers in the statements we make.


For example, consider these two statements:

For any (non-bald) person P you pick, I can choose a color C, such that person P's hair is color C.

I can choose a color C, such that for any (non-bald) person P you pick, that person P's hair is color C.


One of these claims is very easy, and one of these is impossible! I hope this demonstrates why the order of these operations is important.


Sometimes, doing things procedurally helps understanding. You fulfill each quantifier one step at a time before moving onto the next one.

For instance, I claim this is possible:

Step 1: You choose any person.
Step 2: I choose a single color.
Fact: The person you chose has hair with the color I chose.

Proof: At step 2, I can look at the person you chose, and select his hair color.

I claim this is not always possible.

Step 1: I choose a single color.
Step 2: You choose any person
Fact: The person you chose has hair with the color I chose.

Proof: At step 2, you know what color I chose, and you can choose a person with a different hair color.


The problem at hand (enumerating the list of binary sequences)

There exists a list L such that for any binary sequence S, S is on the list L.

So you have to do this in steps:

Step 1: You have to choose a list.
Step 2: I choose a binary sequence.
Query: Is the sequence I chose on the list you choose?

Now, the kicker is that step 2 doesn't happen until you've completely specified your list L. Once I've chosen a binary sequence, you can't go back and change your list.

Let me say this again.
You have to specify everything relevant about your list before we start choosing binary sequences.

Let me give an example.

Suppose you give us a specification for a list.
We mention a binary sequence.
You come back with a new specification for a list.

In this example, you have failed. The list must be completely specified before we start choosing real numbers.



And, incidentally, for your latest attempt, I choose the binary sequence:
...010101

that is, the sequence $\{x_n\}$ where $x_i = (1 + (-1)^i)/2 (i \in \mathbb{N})$, or equivalently, $x_i = 1$ iff $x_i$ is even.

 

[Organic]

Matt,

You Wrote:

I've just shown you every row has a finite number of 1's in it.

If this is your proof then it does not hold on this:

   3 2 1 0
   2 2 2 2
   ^ ^ ^ ^
   | | | |
   v v v v
...1 1 1 1 <--> 1
...1 1 1 0 <--> 2
...1 1 0 1 <--> 3
...1 1 0 0 <--> 4
...1 0 1 1 <--> 5
...1 0 1 0 <--> 6
...1 0 0 1 <--> 7
...1 0 0 0 <--> 8
...0 1 1 1 <--> 9
...0 1 1 0 <--> 10
...0 1 0 1 <--> 11
...0 1 0 0 <--> 12
...0 0 1 1 <--> 13
...0 0 1 0 <--> 14
...0 0 0 1 <--> 15
...0 0 0 0 <--> 16
...

because now the next notation in ... is always 1.

You conceptual mistake is this:

You clime that each column is constructed from finite number of 0 or 1 notations, therefore the list is a collection of infinitely many finite structures and therefore it is a countable list.

Also you clime that each length of 0 or 1 sequences along each column must be infinitely long, which means that if we start from 0 notation we will never get 1 notation or if we start in 1 notation (in the above example) we will never get 0 notation, therefore we can conclude that the length of each column must be countable and finite.

My answers are:

1) by using this trick 0 0' 1 1' 2 2' 3 3' 4 4' ( please see http://home.ican.net/~arandall/abelard/math12/Cantor.html )
we can build this list:

    3 2 1 0
   2 2 2 2
   ^ ^ ^ ^
   | | | |
   v v v v
...1 1 1 1 <--> 1 (1)
...0 0 0 0 <--> 2 (1') 
...1 1 1 0 <--> 3 (2)
...0 0 0 1 <--> 4 (2')
...1 1 0 1 <--> 5 (3)
...0 0 1 0 <--> 6 (3')
...1 1 0 0 <--> 7 (4)
...0 0 1 1 <--> 8 (4') 
...

which is a mixed list of positive-negative sequences of 2^aleph0 length.

2) aleph0 as some "transfinite" object beyond n in N CANNOT EXIST because when we go beyond n in N we are in no information state for any mathematical research (in this case no base value can exists therefore no_base_value^aleph0 is meaningless), as clearly can be shown in this model:
http://www.geocities.com/complementarytheory/RiemannsLimits.pdf

It goes like this:

No base_value --> no notations --> on information --> no Math (--> no "transfinite" universes).

Shortly speaking, the "transfinite" universes do not hold water.

 

[matt grime]

but you're cheating by writing that, aren't you? that isn't the array you had in the article that you've been using. that array, if it is defined in the obvious way lists the cofinite sets and I can prove by the same argument but switching a couple of details that not all the sets in the power set are there, indeed only a countable number of them are.

I proved that none of the elements on the new list youve written down is on the old one.

You haven't properly specified how to generate that list.Read Hurkyl's last post about how you fix the selection of elements in the power set (by the construction in the paper where the 1st column is 010101... the second is 00110011..) and THEN we find something not on the list (any infinite element of the power set).Now you introduce a new list with no finite element of the power set on it.

not every element of the power set is on the two lists; it would take an uncountable set of lists like this to do so.

why do you think you can just create a new list like this?

go back, quote my proof, and after each step say whether you agree or disagree with the deduction and why, offering counter example or counter proof if you think you can.

my proof had nothing to do with the new list you've produced, but it didnt' have to deal with it.it is only your belief that you can continue to enumerate the rows AND have them be the power set because you think the finite case implies the infinite. that is not true, as this argument proves, and as you yourself prove, but then disregard becuase it doesn't suit your argument. you cannot prove what you want because it isn't true, and your only argument against this is 'well i think it ought to be'.

 

[matt grime]

just seen your edited post. it contains many more errorsat no point to i state that if the column starts with a 0 it stays 0, in fact i state it is an alternating block of r 0s, then r 1s then r0s and so on, for the rth column.

read the proof, defined for the array as you wrote it in the original form.your new array interleaves finite and cofinites sets. the set of even integers is not on the list.