How Does Complementary Logic Redefine Mathematical Infinity?

[matt grime]

That isn't a definition of complete. First it doesn't say what things might be complete.

It should start: A set S is complete iff then some criteria


you've just said complete is a property, but not what tha property is. As every set S contains all its elements trivially every set has this notional complete property. And seeing as every set with at least two elements is the union of two non-empty sets, you are also being inconsistet by demanding 'only one'


So you now admit that all this time you've been arguing with Hurkly and me, you've never known what it was that you were arguing about? Right, I'm off, that really is the final straw. To be accused of not understanding your ideas when you now admit you don't know what they are...? You don't think that a little too much?

 

[Organic]

Matt,

Your definition is: "contains all the things that it ought to"

Is this better then?

"A property that depends on the existence of all its elements in one and only one collection."

 

[Organic]

Also By my definition we can clearly understand why infinitely many elements cannot be completed.

By your definiton N is complete.

By my definition N is not complete because there is no such a thing all infinitely many elements.


I have to go so, bye bye for today.

 

[matt grime]

Oh dear I seem to be trapped.

My working defintion for complete was one that was a best attempt to put *some* mathematical meaning there to answer your challenge to give an infinite complete set. I don't claim that that is a mathematical definition. I wouldn't chose to use complete in this sense as a mathematical defintion. And as you didn't give any definition it was the best I could come up with.

If you were to press me, I would say that in the case we are talking about, some list L of elements of a set S is a 'complete' list if it enumerates all the elements of S, that is gives a bijection between N (or some subset), the natural numbers, and S.

But I still can't decide what complete might mean for a set.

Personally I would not choose to use the word complete like this in a rigorous mathemmatical context, except to give it its usual meaning in the English language - which is what my 'contains all that it ought to' was.The best off the top of my head definition of complete as an adjective - not lacking any components.


I don't see how a set cannot be complete in this sense. What element of the natural numbers is missing from N?

And, yes I would say it was better than yours as it says what complete means, yours just says complete is a property of some objects that are ambiguously defined.

 

[matt grime]

Originally posted by Organic
Also By my definition we can clearly understand why infinitely many elements cannot be completed.

By your definiton N is complete.

By my definition N is not complete because there is no such a thing all infinitely many elements.


I have to go so, bye bye for today.

But you've not defined complete!

"A property that depends on the existence of all its elements in one and only one collection"

does not define complete.

Continuity depends on the existence of an epsilon, but that doesn't say what continuous means, does it?

So, simply rephrase that,

A set is complete if and only if... WHAT?!?

it has all its elements in one and only one set?

Obviously garabage.


The set N has all its elements in the set (collection!) N, so why isn't that complete, whatever the hell that means.

"no such a thing all infinitely many elements"


EH? so all sets are finite?


I've never met anyone as obstinate about refusing to understand the bleedin' obvious. Have a medal.


I notice you are still refusing to answer any of the criticisms of your argument.

 

[matt grime]

Summary of the arguments so far:

The following terms have been introduced by you as mathematical terms without definiton despite being asked repeatedly:

opposite; non-linearity; complementary; symmetry-degree; fading transition; structural quantitative; information point; uncertainty; redundancy; {__}; approach; closeness; mutation.

We also have the axiom of infinity induction that is a mystery to everyone as you are the only person to have ever used this phrase (except for those asking what it means).
Then there are tautologies that are false; logical operations of 'and' etc 'defined' on inputs that are not statements that are true of false; the claim that an inequality is EQUAL to a set.
Not to mention this long argument about what complete means in the rigorous sense that you can't define, but use freely. Your *opinions* about it don't count for anything mathematically, it is only what you can prove from the existing mathematics that counts in this argument, you must use the existing definitions and conventions, you cannot attempt to claim it is inconsistent if the inconsistencies you claim are not based on statements within the system. You cannot say that current mathematical thinking is flawed because it does not agree with something you've just made up on the spot without knowing anything about the maths you claim to be talking about.
Fine, if you want to claim it doesn't do what you want then add to it, use some other set theory (ZF v ZFC) but don't tell us is wrong internally, when the only way you can do so is to use external objects.

You did define entropy as a partition of a number into equal parts.

And we've not sorted out what it is that you mean by cardinality not being defined for infinite sets. It is, at best you could argue that you think all infinite sets have the same cardinality. You've proved that to be false.
And then there is the misuse of lists...

 

[Hurkyl]

Can you say in simple English what is Cardinality and Ordinality to you?

Sure. I intuit Cardinality as being (a generalization of) the size of a set, and Ordinality to be (a generalization of) counting numbers.


When it comes down to rigorous work, though, I take cardinality to be something that has the property that:

Set A has cardinality no less than set B iff there is a function from A onto B. (written |A| >= |B| or |B| <= |A|, also said as "set A has cardinality at least as great as set B", or a variety of other ways)

Sets A and B have the same cardinality (written |A| = |B|) if |A| <= |B| and |B| <= |A|.

This is an equivalence relation, so we can form equivalence classes, and the things we use to represent these classes we choose to call cardinal numbers.


I'll get to ordinality later.

 

[Hurkyl]

Lemme take another angle at pointing things out...

Cantor's diagonal argument requires there to be a map from the set of columns onto the set of rows. (Actually, the index set for the columns and the rows is supposed to be the same set, but you can tweak the argument a bit to make it a bit more powerful)

(If you want, I can go through the argument and point out exactly where this is required)

e.g.

If we have a list that is 3 wide, and 3 tall, then there is indeed map from the set of columns (of size 3) onto the set of rows (of size 3), so Cantor's diagonal argument is guaranteed to give a sequence not in the list.

However, if the list is 3 wide and 4 tall, no such onto function exists. Thus, Cantor's diagonal argument is no longer guaranteed to give a sequence not in the list. (though it may get lucky!)


You don't account for this when you try to apply Cantor's diagonal argument to the list of all binary sequences. The columns are indexed by N, and the rows are indexed by P(N). Until you can prove that there is a function from N onto P(N), it is fallacious to apply Cantor's diagonal argument to this list.


I've kind of lost track of what you're trying to assert these days; if you're still trying to claim |N| = |P(N)|, then your argument is circular, because you're using Cantor's diagonal argument to do your proof, but Cantor's diagonal argument can't be used until you've proved it!

 

[Organic]

Matt,

Let us start form this definition.

A set is complete if and only if all its objects can be found.

exapmle: S={1,2,3,...} is not a complete set.

 

[matt grime]

Originally posted by Organic
Matt,

Let us start form this definition.

A set is complete if and only if all its objects can be found.

exapmle: S={1,2,3,...} is not a complete set.

Ah, but what do you mean by found? You give me any natural number and I bet I can 'find' it on that list, you see it's the order in which things are given, you take it to mean that the list must terminate, I take it to mean that given the list and any number (which is then fixed and finite call it n) that I can find it in approximately n steps - look at the n'th term. It changes as n changes but so what?

Your hand wavy arguments do not show that the set of natural numbers and its power set have the same cardinality as you want to claim. That claim is errant nonsense and shows you don't understand things you claim to use accurately. So what that there is no finite time algorithm for listing ALL the natural numbers, that isn't important. As with so many cranks you're confusing what can be constructed from what can be defined.


Just associating some 'property' to something does not a priori contradict anything, you would need to show that this 'property' meant that constructions used (in the Cantor argument here) would fail to work. But they don't; they DO work. Just because you don't understand the convention
that it is possible to specify an infinite set or string doesn't make you right.

 

[Organic]

One by One is not all.

 

[Organic]

A set is complete if and only if all its objects can be found.

It means that S have first an last objects.

exapmle: S={1,2,3,...} is not a complete set.

 

[matt grime]

Eh? You've just given me a list of the Natural numbers and challenged me to 'find' all the natural numbers. I can 'find' them simply by saying there they are, that set there you've just written down, by YOUR definition contains all the natural numbers. My usual challenge stands, tell me which natural number is not on that list, go on, just one. Or define 'find' better.

I mean, I could ask you to prove in your next post that you're not the Dalai Lama, and even tell you I@ve a simple rule that if you write X, I'll believe, you, only I won't tell you what X is, and nothing else will suffice.

You're plumbing new depths here, even by your standards.

 

[matt grime]

Originally posted by Organic
A set is complete if and only if all its objects can be found.

It means that S have first an last objects.

exapmle: S={1,2,3,...} is not a complete set.

So we conclude that a set is complete if and only if it is finite. AND?

 

[matt grime]

incidentally, what if the notion of first and last is non-sensical? I mean is the openinterval (0,1) complete in your opinion?

 

[Organic]

No the open interval (0,1) is not complete.

 

[matt grime]

What about [0,1] the closed interval, that has a first and last element in some sense.

 

[Organic]

In this case [0,1] is not (0,1)+ 0 and 1, but some finite collection between 0 and 1, + 0 and 1 included.

 

[matt grime]

Originally posted by Organic
In this case [0,1] is not (0,1)+0 and 1 but some finite collection between 0 AND 1, + 0 AND 1 included.

Good, you're at last beginning to understand things you use.

So, a set is complete iff it is finite.

Now kindly explain what the hell that has to do with Cantor's Argument?

 

[Organic]

Matt,

Now kindly explain what the hell that has to do with Cantor's Argument?

I answer only to polite persons.

 

[matt grime]

This is mathematics, not a personality contest, if you think me rude AND wrong, then what better way to correct me than to demonstrate why it is that because N is not finite that Cantor's argument is wrong.

Note, 1. Youve proved it is correct, so this is going to be a test of your consistency in mathematics.

Note, 2. I imagine you're about to wheel out the 'it makes no sense to use the word 'all' for things that aren't complete [finite]'


For 2. Why is it not permissible to say N is the set of all natural numbers? I mean if it's only because the set is not finite then I maintain, as would any mathematician, that you are not using 'all' in the context of that sentence correctly. The inequality n>n-1 is clearly true for ALL natural numbers.

N is be definition the set of ALL natural numbers. It contains all its elements. The way to show 'not all' is to find one counter example, so show me some natural number not in the set of all natural numbers.

We can talk about the set of all natural numbers, the set of all complex numbers... we cannot talk about the set of all sets obviously.

 

[Organic]

Dear Hurkyl,

x=2

Let us examine x and P(x)

x:

21
^^
||
vv
01<-->1
10<-->2

P(x):

 1 0
2 2
^ ^
| |
v v
0 0 <--> 1
0 1 <--> 2
1 0 <--> 3
1 1 <--> 4

====================================================================================================

x=3

Let us examine x and P(x)

x:

321
^^^
|||
vvv
001<-->1
010<-->2
011<-->3

P(x):

 2 1 0
2 2 2
^ ^ ^
| | |
v v v
0 0 0 <--> 1
0 0 1 <--> 2
0 1 0 <--> 3
0 1 1 <--> 4
1 0 0 <--> 5
1 0 1 <--> 6
1 1 0 <--> 7
1 1 1 <--> 8

====================================================================================================


x=|N|

Let us examine x and P(x)

x:

   4321
   ^^^^
   ||||
   vvvv
...0001 <--> 1
...0010 <--> 2
...0011 <--> 3
...0100 <--> 4
...

If we use the ZF axiom of infinity induction on the power_value
of each column then:

P(x):

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

As we can see, we get an ordered collection of infinitely many 01 unique sequences with Width aleph0 and Length 2^aleph0.


Conclusions:

1) Both Width and Length are enumerable.

2) Length > Width.


Cantor's diagonal method problems:

1) It was examined on arbitrary unordered collection of 01 sequences.

2) No collection of 2^aleph0 was examined.

3) Any "missing" 01 sequence is already in the 2^aleph0 collection,
therefore it MUST NOT be added to the collection.

4) Any complete collection cannot be but a finite collection.

5) Infinitely many elements cannot be completed, therefore the cardinality of infinitely many elements is unknown and cannot be used to establish the transfinite system.

 

[matt grime]

Edited:


Ah, so you have stopped answering me.

Anyway, you're claiming 'the axiom of infinity induction' again.

Still you are the only person to have done this, what is it?


Clearly the object you construct has both countable rows and columns because you list them, and, ergo, they are enumerable.

At least you are consistent in your nonsense, shame you really don't understand the elementary explanations given to you as to why what you're claiming is wrong.

As is eminently clear every row in your array is eventually zero by construction, so how can it possibly have cardinality 2^aleph-0. In fact you are ASSUMING alpeh-0 equals 2^aleph-0 to prove they are equal and derive a contradiction... Some one might think you didn't have the slightest idea about mathematics. They'd be right seeing as for the last God know's how long you've been insisting that 'no infinite set is complete'. In post 34 in this thread approx. you were asked to define that or at least explain what you meant. For the next 130 posts you were unable to provide any answer to that. Eventually after much prompting we were able to decide 'complete' means finite!

Now, how do we get you to see this claim of yours is wrong?

 

[Organic]

Matt,

Please read carefully what I write here.

You wrote:

This is mathematics, not a personality contest

1) For me the quality of communication between persons comes before any subject.

2) By "the quality of communication" I mean that no person in this process, using any aggressive or fanatic approach related to the examined subject.

3) I think that fruitful communication is based on cool head ,open mind, and positive approach that first of all tries to understand the other person.

4) If you can accept these basic terms of "the quality of communication" than please write your posts without “wtf”, “what the hell”, ”utter b***”, and I’ll be glad to communicate.


Yours,

Organic

 

[matt grime]

Fanatic? You are not fanatically pushing a completely stupid idea? Sorry, Organic, you got the polite 'please elaborate' posts. You didn't, just obfuscated and said that Hurkly and I didn't understand anything and yet you couldn't even post a defintion of 'complete' for 130 posts, and even then only with severe prompting and a lot of help.

Is it any wonder you've managed to frustrate me into these outbursts? Some people have just got fed up with even dealing with you nonsensical 'proofs'.

So, a set is complete iff finite, what has that to do with Cantor's argument that you find so wrong? Rememeber it's the maths that counts, not your opinion of what should and shouldn't be done with infinite sets.

 

[Organic]

Dear Matt,

A simple example to your fanatic approach.

From the beginning you did not read what I wrote about "complete" and
"all" that can be found here for example:
http://www.geocities.com/complementarytheory/GIF.pdf

More then that, by your fanatic approach you wrote:

"I am not going to read your pdf files"

If you don't read and try to understand what I write, then the only voice that you hear is your own voice.

You are in some internal war between you and your fanatic approach
that does not give you any chance to hear other person's voice, which is different from your conventions.

Let us see for the last time if you can put aside your fanatic attitude, read the two parts of:

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

And give a non-emotional response to what is written there, by writing page and paragraph number to every detailed response of yours.

Thank you,

Organic

 

[matt grime]

Slight edit:

Let us try this:

New diagonal:

page 3. Let x=|N| [:=aleph-0]

the bit in square brackets is my addition, and it is a defintion. It is unclear what you mean here, as |N| is not a number, so the previous arguments don't apply.

'axiom of infinity induction'

please explain what this is

we know what the axiom of infinity is, we know what induction is, but what is the 'axiom of infinity induction'

why are you justified in writing 'length 2^aleph-0'
by construction the cardinality of the columns is aleph-0

You appear to claim that since the 'finite' diagrams (my term) are of width n and length 2^n that it is permissible to write aleph-0 and 2^aleph-0. This is not true. As you prove later. Look at how the finite diagrams are constructed, now rigorously define the limit of this process, the best you can come up with is that there is a natural inclusion of the diagram for x=1 into x=2 into x=3 and so on.

So the limit will be something akin to the nested subset of this. Simple exercise, show that everything so constructed is countable.

Edit: THIS IS THE MOST IMPORTANT POINT, please deal with this if nothing else in the list of criticisms.
##################$# In your case the construction shows clearly that if z is the element of the power set corresponding to position n in the list, then there are only finitnely many non-zero entries in the row. z was an abritrary element of the power set. This is a contradiction. I'm not sure how more clear we can make that.$#######################


Anyway let us ASSUME that the diagram for the infinite case does list all the elements of the power set, then you prove this assumption is wrong. That is the end of the argument. You've just shown aleph-0 is not equal to 2^aleph-0.

So in no way is it justified to CONCLUDE that there are 2^aleph-0 rows.


PAage 4.

problems with cantor's argument.

1. is false. cantor's argument is defined with any ordered subset of the power set.

2. is not relevant.

3. the enumerated rows in your diagrma clearly only have finitely many elements being non-zero so that is incorrect.

4. irrelevant, as at no point in cantor's argument do we invoke the notion of complete as you define it.

5. ditto. Demonstrate that the notion of complete as you've now defined it is in anyway relevant ot the classical cantor argument, which we will reproduce at the bottom of this post.


you present a proof on page 5 of something. you claim on page 6 t prove the opposite. where is the mathematical error on page 5's proof?


page 7 'out of the scope of cantor's definition' please explain what that means, cos you seem to imply it is to do with your inability to accept that the word all has some mathematical meaning. Why is N not the set of all natural numbers. All means in maths that there is no exception.

for all, there exists, do you know what these quantifiers are?

it is your *opinion* that all is not acceptable for infinite setes, there is no basis in mathematics for that statement as far as i can tell. or indeed any other mathematician that I've met.


**Proof of cantor's argument.

Let N be the set of natural numbers, P(N) its power set, |N|:= aleph-0, we show there is no bijection from N to P(N), ie that 2^aleph-0 is not aleph-0, if you wish.


Suppose f is an *injection* f:N-->P(N)
then list the elements of the image by their preimage

f(1), f(2),f(3)...

define a set by S by n is in S iff n is not in f(n)

S is an element of P(N), by construction S is none of the f(i), so we conclude that no injection from N to P(N) can be a bijection, hence the cardinality of N is strictly less than that of P(N)**

no mention of the word all, no proof by contradiction, sets defined simply, so why is that wrong.

for what it's worth I don't like reading your pdfs because downloading takes up bandwidth, and down loading a file from a source I don't trust (in the computing sense) isn't something I encourage people to do.

In fact, you can ignore all my criticisms of your paper, and just tell me where my maths is wrong in my proof between the **'s

 

[Organic]

First paper:

Let us go step by step, and write first the most important things.

1) When we have and axiom saying that: If somthing then somthing+1
then please tell me what is its result?

 

[matt grime]

Originally posted by Organic
First paper:

Let us go step by step, and write first the most important things.

1) When we have and axiom saying that: If somthing then somthing+1
then please tell me what is its result?

If you mean, suppose i hav a family of statements labelled P(1), P(2).. and that I know that P(n) implies P(n+1) is true and that there is some r with P(r) true, then the statements P(n) are true for all n=>r

then the answer is mathematical induction.

But, as you cite the axiom of infinity induction, you presumably must know what it is, so just tell me. and the question was for you to cite it.


preferably though, just explain why my proof of cantor is wrong, and for good measure, explain why in your construction if z is in the power set P(N) it must then correspond to row n for some n, yet clearly the row n is eventaully always zero, and thus z must have only finitely many elements in it, but z was arbitrary, contradiction.

 

[Organic]

Matt,

Please hold your horses, I said we go step by step so again at this stage we are talking on my first paper.

In one of your posts you wrote: ( https://www.physicsforums.com/showthread.php?s=&threadid=12942&perpage=12&pagenumber=8 )

Here is a more mathematical definition.


There is a set W that contains the empty set and if any set y is in W then the set containing the union of y and the set containing y is also in . By induction contains every finite integer.


from:

http://www.mtnmath.com/book/node53.html

So as you see ZF axiom of infinity is an induction that define aleph0.

Isn't it?

 

[matt grime]

No, induction is used along with the axiom of infinity to construct a set which we label the integers. Aleph-0 is then defined to be the cardinality of this set. The existence of an inductive set AND induction mean that our model must contain the integers (or somethingl like it) to satisfy the ZF axioms.

This doesn't address the issue that you used

'the axiom of infinity induction'

the last word and the first 4 are understood, we just don't know what you mean whe you put the together.

In all honesty there are bigger issues - the ones I drew your attention to above.

secondly, whatever the axiom of infintiy induction is it would appeart to allow us to state, that because

n>n-1 that aleph-0>aleph-0

that should tell you that since aleph-0 is NOT and integer you can conclude nothing about aleph-0 statements based purely on induction on the integers. induction only tells you truth or otherwise about the statement P(n) for n an integer. P(aleph-0) makes no sense.

 

[Organic]

|N| is the cardinality of N where N is a collection of infinitely many n's.

Cardinality simply answer the question: "How many?"

A collection of infinitely many elements has no end by definition, therefore the cardinality of such a collection cannot be found.

Modern Math tries to hold the stick in both hands:

1) Using the name "Cardinality"
2) Instead of "How many?" using "Whet magnitude?"

By this double-definition "Cardinality" is clearly meaningless.

Cardinality is known only when we dealing with a finite collection.

I do not accept Cantor's idea about the cardinality of infinitely many
objects because it is simply a self contradiction that forcing infinitely many objects to have "well-defined" cardinality.

This poor double-definition "Cardinality" is a schizophrenic creature
that can jump beyond its own head.

The only reasonable infinity defined exactly and simply by ZF axiom of infinity.

We use it and get infinitely many unique columns but then their cardinality (notated as "aleph0") is a simple and healthy creature which its exact value is unknown, and to be unknown it is OK.

Also by the same axiom we get infinitely many unique rows but then their cardinality (notated as "2^aleph0") is a simple and healthy creature which its exact value is unknown, and again to be unknown it is OK.

By this approach we avoiding the schizophrenic state of Modern Math
about infinity, can get simple and reasonable results, that does
not put aside uncertainty.

 

[matt grime]

So we can conclude that you cannot answer mathematically any of the charges laid against you, I would suggest.

There is no problem with defining cardinalities, except in your *opinion*. Calling cardinality size (in the sense of counting its elements) is just a figurative way to express the concept.

Again you're presuming to say that the "limit" of 2^n as n goes to infinity is 2^aleph-0 because you are trying to treat natural numbers and cardinals as the same thing.

2^aleph-0 is the cardinality of the power set of the Naturals, it is not a number which you can treat in this cavalier fashion.

It seems it is your understanding of conventions again that is lacking. Just because you can write it doesn't make it so.

In fact you can entirely stop invoking the axiom because we merely need the natural numbers, and that they are not finite.

The axiom of infinity does not define infinity, it gives the existence of an infinite set, a set which does not have a finite number of elements. You are thinking that infinity is a unique object; it is a concept. Actually, striclty speaking it means that our model must contain the integers, or a set that behaves like them.

Seeing as you do not or cannot dispute the errors in your 'proof' or provide the slightest mathematical refutation of Cantor's argument within the axiomatic ZF world shall we say the issue is closed?

 

[Organic]

How "nice" is your model of infinity, it is so "nice" that you don't need infinitely many elements to define it.

Actually you don't need the existence of n's in N to define infinity because infinity is a concept.

More then that, you can use a function between no-input1 to no-input2 and find meaningful results for your highly sophisticated abstract Math, that only smart mathematicians like you can understand.

So let me tell you who am I in this story.

I am the little boy who cries: "THE KING IS NAKED".

And why the king is naked?

Let us examine a better model then you give:

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

No map can be found without input, the input in the above model is infinitely many intersections that define N Q and R collections.

No one of these collections is a finite collection, therefore their cardinality is unknown, but even though we can find differences between their sizes, and we don't need any transfinite system for this.

The idea that aleph0 is beyond n's only pushing the system to no-intersections state, and in this state (please look at my model) you have no input for any mathematical system.

So if you can understand my model then:

1) You can clearly see that "transfinite" system is too powerful for any mathematical system.

2) If you read the first paper in http://www.geocities.com/complementarytheory/NewDiagonalView.pdf
after you understand my model, then you can see that :

1) I do not use N members to define aleph0, so as you see I have a general view on cardinality.

2) length > width

3) length and width are both enumerable.

4) we don't need any transfinite system to define (2) and (3).

---------------------------------------------------------------------------
Another way to show that 2^aleph0 > aleph0 is the hierarchy of the building-blocks dependency of R objects in Q objects.

This dependency can be clearly shown here:

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


By the way, the reason that |N| = |Q| is trivial because:

(1/1)(1/2)(1/3)(1/4)...
    \          
(2/1)(2/2)(2/3)(2/4)
          \
(3/1)(3/2)(3/3)(3/4)
              \
(4/1)(4/2)(4/3)(4/4)
.                  \ 
.

that can be written as:

1 <--> 1 = (1/1)
2 <--> 1 = (1/2)*(2/1)
3 <--> 1 = (1/3)*(3/1)
4 <--> 1 = (2/2)
5 <--> 1 = (1/4)*(4/1)
6 <--> 1 = (2/3)*(3/2)
.
.

 

[matt grime]

a function without an input is not a function. Learn the maths.

i thought you wanted to deal with one point at a time in the refutation i gave. now you don't wish to deal with any of them. or learn any maths. just say it's wrong and your right.

 

[Organic]

"A function without an input is not a function."

I'll cherish this clever sentence for the rest of my life.

So, can you see what I see or not? (before we go to the next paper).

 

[matt grime]

Look, I've had a bad day, and your idiocy is no concern to me any more. Check it out, a function has a domain, it is a set, because the functions we are talking about are functions of sets.

exlpain why the proof i gave of cantor's argument is wrong. explain what the axiom of infinity induction is.

 

[Organic]

Dear Matt,

Go take a good night sleep, and we shall continue tomorrow.

 

[phoenixthoth]

matt and organic,
please read page six, paragraph 3 of
http://www.alephnulldimension.net/matharticles/tuzfcver9.pdf

cheers
phoenix

 

[Organic]

Matt,

**Proof of cantor's argument.

Let N be the set of natural numbers, P(N) its power set, |N|:= aleph-0, we show there is no bijection from N to P(N), ie that 2^aleph-0 is not aleph-0, if you wish.


Suppose f is an *injection* f:N-->P(N)
then list the elements of the image by their preimage

f(1), f(2),f(3)...

define a set by S by n is in S iff n is not in f(n)

S is an element of P(N), by construction S is none of the f(i), so we conclude that no injection from N to P(N) can be a bijection, hence the cardinality of N is strictly less than that of P(N)**

If you want to see what I have to say about it you can find it in the second paper here:

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

 

[matt grime]

It isn't clear that your observations apply to my proof because mine isn't a proof by contradiction, the one you include is, and isn't the proof as cantor gave it, in fact mine isn't his orginal proof, it is the second one he gave. modern convention means that peopleare taught the proof by contradiction method. try relating what you think to the constructive proof above.

as for what you have written

"All members which included in S , are different from each other.
Any member of Z* can be mapped with some member of P(Z*), once and only once.
Therefore t is different from each member in S, therefore t MUST BE INCLUDED in S."

page 8

is incoherent in its English and its mathematical conclusions make no sense. It is just nonsense, sorry.

 

[matt grime]

Sorry to double post but moz's form filling attributes just went bezerk.

The set S is I think defined - I don't know where you got that proof from of Cantor's argument but if I were the person who wrote it and put it out in the public domain I'd be ashamed of it. I don't mean the ideas behind it but the presentation is awful.

Once more it is you not understanding how to interpret all. The set of ALL n such that n is not an element of f(n) is a set - it is a clearly defined subset of the set of natural numbers, it is the complement of the set of m with m in f(m), again a clearly defined set. I'm sorry that the level of sophistication of your mathematical abilities isn't able to cope with these things, but that doesn't stop it being true. Or do you have some bizarre set theory in mind?

"If we want to keep S as an existing member, we MUST NOT INCLUDE t in S ."

What does this mean? keep S as a member of what?

You may wish to do your 'maths' in some bizarre model of some set theory where those things are not sets, but that is not an issue to do with the correctness of the proof in everyone else's model of their set theory.


The proof of Cantor's argument is perfectly logical.

m

 

[Organic]

Hi Matt,

My proof is very simple.

I clearly show that subset S that defined as:

S = {z in Z* such that z is not in f(z)} cannot exist in P(Z*) because by option 1 t must included in S, and by option 2 t must not included in S.

Cantor's point of view is problematic because it forces S definition to be the "checker" of t existence, which is simply non-logic because t can exist without S but S cannot exist without t.

As I wrote, S existence depends on objects like t, therefore we have to check S existence by t and not t existence by S, as Cantor did.

Please read again about the hierarchy of dependency in my second paper page 2.

Shortly speaking, S can exist only if t is "out of its scope", which means that S can exist iff t is not forced to be included in it.

My proof holds iff you understand and accept this point of view.

 

[matt grime]

it would appear that you don't understand the idea of proof by contradction. it is exactly because of the problems that happen when you define S in this way that we conclude that the function cannot be a bijection. we show there is no such S, which there would be if there were a bijection. therefore there is no bijection. end of story, I'm sorry you cannot see this but the notions of scope ond order you introduce are neither here nor there. if you don't like the contradiction, just assume f is an injection, not a bijection conclude S isn't in the image and conclude it cannot be a bijection. end. done. no issues.

 

[Organic]

we show there is no such S, which there would be if there were a bijection.

This is exactly my proof, which shows that S definition cannot exist without any connection to any mapping result between Z* and P(Z*).

Therefore it cannot be used to conclude anything.

 

[matt grime]

yes it can, it is used to conlude that as we assumed a bijection, and this led to a contradiction (a paradox, a statement that is true and false simultaneously), hence the ASSUMPTION is invalid. Find out what proof by contradiction means. this is one.

as my proof demonstrates, if you don't like contradiction, you don't have to use it. many proofs by contradiction are unnecessary just like this is.

 

[Organic]

S definition cennot exist because P(Z*) does not exist without Z*.

Simple as that.

 

[matt grime]

"I cannot comprehend the lack of understanding that would lead anyone to think that"

I think that was Babbage's response to the question asked of his calculating machine, but if you put the wrong numbers in do you get the right answer anyway?

If you don't like contradictions, then take my proof which is not a proof by contradiction.

 

[Organic]

Matt,

In both cases S definition cennot exist because P(Z*) cannot exist without Z*.

Simple as that.

 

[matt grime]

So now the integers don't exist? In that case you are not in a model for ZF, so any conclusions you draw about ZF are wrong, or at least completely unproven.

Keep digging.


Oh, you're not about to do the 'all' has no application in infinite sets thing again are you?

Round and round we go.




You didn't respond to all of the comments on your new new diagonal argument, btw.

Especially this one from where you magic up the infinity axiom of induction nonsense again. (Note, ZF does not DEFINE aleph-0, it requires there is a set of cardinality aleph-0 in your model.)

By your construction labelling the alleged 2^aleph-0 rows that you claim represent the power set of N with N we can see

##################$# In your case the construction shows clearly that if z is the element of the power set corresponding to position n in the list, then there are only finitely many non-zero entries in the row. z was an abritrary element of the power set. This is a contradiction. I'm not sure how more clear we can make that.$#######################