How Does Complementary Logic Redefine Mathematical Infinity?

[matt grime]

No, you do not have a collection of 2^aleph-0 elements, assuming you mean the set of 'combinations' you produce. I do not know how you are concluding this but it isn't 2^aleph_0. This is the cardinality of the power set of of the natural numbers. The list you produce is not in bijection with any set of cardinality 2^aleph-0, it is in fact obviously of cardinality aleph-0.

 

[Organic]

No Matt,

By using the induction of ZF axiom of infinity on 2^power_value (by mistake i wrote power_level) power_value = aleph0.

 

[matt grime]

edit: word power inserted at crucial point

OK, the cardinality of the power set of n-elements is 2^n, the cardinality of a POWER set of card alpeh-0 is 2^aleph-0. The list you produce has card aleph-0. The only way you could claim it were 2^aleph-0 was if it were the list of all subsets of N, but it isn't it only contains the finite subsets. This is the crux of the issue - you do not produce a 'list' of 2^aleph-0 elements, you prove it is countable yourself.

The only way it you could do otherwise would be to demonstrate the list you produce by some 'induction' that isn't an induction, contains all 'combinations'; it doesn't!

 

[Organic]

Again no Matt,

My collection is all possible 01 combinations of 2^aleph0 ordered elements finite an non-finite.

Please read again page 1 in:
http://www.geocities.com/complementarytheory/NewDiagonalView.pdf

and see for yourself that i am talking about P(aleph0).

 

[matt grime]

But it clearly isn't!

The thing you construct is the doubly infinite array, the first column (right to left) is 0101010...

the second is 0011001100...

the r'th column we now agree is 0000...00111..111000..00... with 2^r 0s and 1s in each segment.

SIMPLE proof this only gives the finite subsets - let x be some combination on the list, say it is the n'th. Since every column after the n'th starts with 2^n zeroes, and 2^n>n for n >1, then clearly every entry after the n'th, reading right to left, is zero, and thus there are only finintely many elements in that subset. THe set was arbitrary.


As you send a string to the binary expansion, to show it's countable, you can only have a finite number of 1s in the damn string!

 

[Organic]

No Matt,

It gives P(aleph0).

Again you don't read what I wrote in page 1 :

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

Your "PROOF" is not about 2^aleph0 members, because n not= aleph0.

 

[matt grime]

But your bizarre bastardization for the finite sets doesn't apply in the infinite list.

In fact as you go top left to bottom right in your diagonal argument, can I ask how you use this in the infinite case - which element in the first string (row) do you use?

I can't say it any more clearly, the list you produce, by your own admission only contains strings with a finite number of non-zero entries on it. This is 'your' proof they are countable (they are trivially countable by construction, but you don't seem to realize this).

By your own admission, the element ..1111 is not on the list, yet it ought to be if the list enumerated the power set. And yet you claim it contains all the combinations... bizarre and self contradictory.

 

[Organic]

Left-right or right-left diagonals holds only for finite P(n).

When I deal with P(aleph0) ordered list then you can see that it is
a right-left diagonal.

It is no problem to say that ...111111 is also in the list but then
we clearly deal with a finite ordered list, which is not our case.

Again, be aware to the fact that we are dealing with an ordered collection.

Also ...111111 is not just a one member but an open interval of aleph0 scales (of 2^aleph0 ordered collection).

 

[matt grime]

Originally posted by Organic
Left-right or right-left diagonals holds only for finite P(n).

When I deal with P(aleph0) ordered list then you can see that it is
a right-left diagonal.

It is no problem to say that ...111111 is also in the list but then
we clearly deal with a finite ordered list, which is not our case.

Again, be aware to the fact that we are dealing with an ordered collection.

Also ...111111 is not just a one member but an open interval of aleph0 scales (of 2^aleph0 ordered collection).

So, we should looka t the diagonal from top right to the 'bottom left'.


the scales thing is not important - ...1111 corresponds to the element in the power set that is the set N.


That doesn't answer anything important anyway.



The thing you construct is a doubly infinite array from right to left and top to bottom, it contains only strings with a finite number of non-zero elements as I've proved independently of you and as you prove yourself by writing an explicit bijection with 2-adic expansions.

Clearly the list is countable (nb, for mathematicians, lists are countable by definition), yet you insist that it contains all combinations, despite proving it doesn't yourself and repeatedly saying the string ..1111 isn't on it! Nor is ..01010101, nor is ..001100110011 etc.

You make two accurate assertions - that there is no bijection between N and P(N) and that the Finite Power set is countable. The problem is you then say they are the same thing! They are not. You prove this yourself.



And I don't understand why you seem think that N is not an element of P(N) {N is the set of natural numbers}, that is the only way I can read your statement about when ...1111 is a combination.

 

[Organic]

1) I proved that |P(N)|>=|N| iff |N| is the cardinal of ALL N members.

2) I clime that there is no such a thing the cardinal of infinitely many elements, because they cannot be completed.

3) In this case all we have is (...111,...000] XOR [...111,...000)

4) There is no shuch a thing [...111,...000]

5) I am going to sleep, so see you and have a good night.

 

[matt grime]

Originally posted by Organic
1) I proved that |P(N)|>=|N| iff |N| is the cardinal of ALL N members.

2) I clime that there is no such a thing the cardinal of infinitely many elements, because they cannot be completed.

Well, you didn't prove 1 assuming N denotes the set of natural numbers, although you assert it with a wrong proof, as you take a fininte cardinal result and put in aleph-0 and claim the answer. Well, here's a counter example:

for all finite cardinals n>n-1 hence aleph-0<aleph-0

and 2. is a definition! which incidentally you use in part 1. Besides, it doesn't matter what you believe, it matters what you can prove, or disprove. For instance, in what way is N, the set of natural numbers not complete? I mean, there meanings where it is not complete, algebraically. If I list the numbers in N in sequence, which one do I omit? Either give me an example, ro proive I must omit one.

 

[Organic]

Matt,

PROOF:

Let T be the 01 collection with cardinality of P(N).

Let L be the 01 collection with cardinality of N.

T is:

...000000
...111111
...000001
...111110
...000010
...111101
...000011
...111100
...
...

In this private case we can see that because T 01 collection is longer then L 01 collection, ...010101 is not in L but it is in T.

We can change the order of T 01 collection and then we shall find another 01 sequence which is not in L but is in T.

In this stage we can clearly say that |T|>|L|.

The width of T 01 collection = |N|

The lengh of T 01 collection = |P(N)|

Because any missing 01 sequence is already in T, we MUST NOT add it to L.

Therefore we can find a 1-1 and onto between 1,2,3,... to any T member.

But because |L|=|N| we can conclude that |L|=|T|.

Now we have |T|>=|L| which is a contradiction.

Therefore transfinite cardinality does not exist.

Q.E.D


This proof holds for finite or infinitely many objects.

Let us see it in a finite 01 collection.


T has the cardinality of P(3)

L has the cardinality of 3


T 01 collection is:

000
111
001
110
010
101
011
100

01 sequence of 101 is already in T therefore we MUST NOT add it to T.

Therefore there is 1-1 and onto between any 1,2,3,... to any T member:

000 <--> 1
111 <--> 2
001 <--> 3
110 <--> 4
010 <--> 5
101 <--> 6
011 <--> 7
100 <--> 8

 

[Hurkyl]

Let T be the 01 collection with cardinality of P(N).

Let L be the 01 collection with cardinality of N.

T is:

...000000
...111111
...000001
...111110
...000010
...111101
...000011
...111100
...
...

How do you justify writing T as a list?

 

[Organic]

Hi Hurkyl,

Please read this:

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

As you can see I can combine between:

...000000
...000001
...000010
...000011
...
...
(...111,...000]

and

...111111
...111110
...111101
...111100
...
...
[...111,...000)

and get T collection:

...000000
...111111
...000001
...111110
...000010
...111101
...000011
...111100
...
...

 

[matt grime]

I must echo Hurkly here - you are, by writing it as a list and applying the diagonal argument to it presuming it is enumerable AND contains all the 'combinations'; you prove this is not possible. END OF PROOF. But no, you carry on...


More abuses of maths are:

1. The set with cardinality |P(N)|? Don't you just mean P(N)? There are lots of sets with cardinality |P(N)|, you can't talk of THE set...

2. Same as above but with N not P(N) in there

3. you don't know what's in L because you've not said what L is


What is the bijection you claim from T to N? Please say it's not by the numbering of the row.


The finite case does not imply the infinite case. You are attempting to say that because, in the finite case we only look at the top nxn square in an nx2^n array, we can do the same in the infinite case - but the construction doesn'tdo this. THere are as many rows as columns in the infinite case, and no row is not used in the diagonal argument.

 

[Hurkyl]

Let T be the 01 collection with cardinality of P(N).

As you can see I can combine between:

...000000
...000001
...000010
...000011
...
...
(...111,...000]

and

...111111
...111110
...111101
...111100
...
...
[...111,...000)

and get T collection:

...000000
...111111
...000001
...111110
...000010
...111101
...000011
...111100
...
...

Why do you think this list has cardinality P(N)?

 

[matt grime]

Originally posted by Organic
Hi Hurkyl,

Please read this:

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

As you can see I can combine between:

...000000
...000001
...000010
...000011
...
...
(...111,...000]

and

...111111
...111110
...111101
...111100
...
...
[...111,...000)

and get T collection:

...000000
...111111
...000001
...111110
...000010
...111101
...000011
...111100
...
...

The first collection gives you those with finitely many 1s in, the second gives you the collection with finitelym any 0s. THey both have card aleph-0 so does their union. But it isn't all of the things in T, only the finite and cofinite ones. At no point is the repetitive string...010101010101 in there.

 

[Organic]

Because T is a collection of 2^Omega unique memebrs, where Omega
is the result of using the ZF axiom of infinity bulit-in induction on the power_value
of 2^0, 2^1, 2^3, ...

 

[Hurkyl]

Because T is a collection of 2^Omega unique memebrs

Why do you think that? What is omega? Is it cardinality of the natural numbers in ZF?

Omega
is the result of using the ZF axiom of infinity bulit-in induction on the power_value
of 2^0, 2^1, 2^3, ...

This sentence makes no sense.

I assume you mean "Omega is the result of applying induction to the sequence 2^0, 2^1, 2^3, ..."

But this makes no sense because this is not even close to the form of something to which one applies induction... induction is, essentially, when you compute values of something based on previous values (when the domain of the function is well-ordered)

Here, each value is given explicitly; 2^0, 2^1, 2^3, et cetera. There is no computation based on previous values.

 

[Organic]

Matt,

Yes ...01010101 is in T but not in L.

 

[Organic]

Hurkyl,

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

Please look at the axiom of infinity.

 

[matt grime]

Originally posted by Organic
Because T is a collection of 2^Omega unique memebrs, where Omega
is the result of using the ZF axiom of infinity bulit-in induction on the power_value
of 2^0, 2^1, 2^3, ...

As we established last night, you've not defined 'the axiom of infinity built-in induction'. in fact a quick google for the phrase reveals you're the only person on the entire web as google sees it that has used that phrase. There is the axiom of infinity which merely asserts our set theory has some set that contains the Natural numbers (we do: the natural numbers). There is also the principle of induction, which you don't use.

 

[matt grime]

Originally posted by Organic
Matt,

Yes ...01010101 is in T but not in L.

But you just told us how to construct T, and the ...010101 is not there. You said to add together the finite and cofinite lists to produce the T collection.


And you've still not adequately define L (or T really), but I think we know what they are. Want to make sure?

 

[Organic]

No Matt,

The pruduct of using ZF axiom on infinity on the power_value of
2^power_value is an ordered collection of 2^aleph0 unique members.

 

[Organic]

Matt,

I did not add anything to T I just changed the order of its members
to show you that your "proof" by left endless zeros does not hold.

 

[matt grime]

Originally posted by Organic
No Matt,

The pruduct of using ZF axiom on infinity on the power_value of
2^power_value is an ordered collection of 2^aleph0 unique members.

Erm, that's not what the axiom of infinity states. I see you've dropped the word induction.

The axiom assures us that there is a set conatining the natural numbers. It does not allow us to whack in an alaph-0 at will. Aleph-0isn't even part of ZF's axioms.


So, explain what you are doing to
2^0, 2^1, 2^2...

the nearest I can get is that you are taking unions of sets of size 1,2,4,8,...

the union clearly is countable.

 

[Hurkyl]

And anyways, induction doesn't require the axiom of infinity, or even much set theory. Here are some definitions:

Definition: < is a well ordering on a set S iff subsets of S have smallest elements (with respect to <)

Notation: We often abbreviate this by saying "S is a well ordered set" and leave the ordering, <, implicit.


Theorem (induction): If

(1) S is a well ordered set with smallest element s
(2) P(x) is a logical proposition
(3) P(s)
(4) (y in S, y < x, P(y)) &rarr; P(x)

Then

z in S &rarr; P(z)


This barely requires any set theory, and doesn't even require things like "for all" or "there exists"

 

[matt grime]

Originally posted by Organic
Matt,

I did not add anything to T I just changed the order of its members
to show you that your "proof" by left endless zeros does not hold.

from your original definition in the first new diagonal argument my proof is perfectly valid.


I can now prove that any string has either no 1's after some point, OR it has no 0's given your new way of constructing T by interlacing two clearly countable sets, since it came from one of two original lists where one or the other holds.

 

[Organic]

Hurkyl,

It is clear as a middle-noon sun, if n exist then n+1 exist.

How can you say that this is not an induction, which its product is clearly Omega(=aleph0)?

Please look here: http://us.metamath.org/mpegif/aleph0.html

 

[Organic]

Matt,

You still don't get it, what you call two lists is the same list with
3 different orders.


The first is top --> bot.

The second is bot. --> top

The third is one form top, one from bot., one from top, one from bot., ...

In all of the cases we have the same 2^aleph0 01 unique members.