How Does Complementary Logic Redefine Mathematical Infinity?

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In summary: Complimentary logic is not one of those fields. In summary, this conversation does not provide a detailed or precise explanation of how complimentary logic works, or what its potential applications might be.
  • #71
No Matt,

S = {} or S does not exist by your definition.

If it is not empty then by your definition no object is uninteresting.

Therefore S (by your definition) does not exist, because even if one member of it is interesting then we can find the relation of each member in S to this interesting member, so all members are interesting or S has no members.

Shortly speaking no set can include its negation and survive.
 
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  • #72
Do you mean to this part?
--------------------------------------------------------------------------
We can get a combinations list of infinitely many places, by using the ZF Axiom of infinity induction, on the left side of our combinations list, by using the induction on the power_value of each column, for example:

2^0, 2^1, 2^2, 2^3, ...
--------------------------------------------------------------------------
 
  • #73
yes i mean that. the axiom of infinity merely tells us that there is something akin to the natural numbers in our model. how does that allow you to produce these strings, and what does it tell you about the strings so produced, and how, for that matter, are you inducting?
 
  • #74
ZF axiom of infinty simply says: If n exists then n+1 exists.


Do you agree with that?
 
  • #75
and what's that got to do with what you're doing?

incidentally, in immediate prelude to that there is a mistake (well, there are many...}

you say the right column is 'based on 2^0' which is questionable, because it is 01010101. check, i think you'll find it's 11001010
 
  • #76
Matt,

Now I see that you simply don't understand my method.


2^0=1, and 1 tells us after how many times to switch from 0 to 1 or from 1 to 0, therefore the result is 0101010101...

2^1=2, therefore the result is 001100110011...

2^2=4, therefore the result is 0000111100001111...

and so on.
 
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  • #77
firstly, i think you'll find that the first column in your paper IS NOT 0101010...

now you've explained that, let us examine what this means - given any row it is immediate that after some point, reading right to left, that every entry is zero. This means that the list you construct has only strings that have finitely many non-zero elements in the. Thus it is some sublist of ALL strings. Clearly it is countable.

If you'd explained your method clearly this would be unimportant.

What IS important is you assertion that the list produced contains every element, that is that the diagonal argument produces something that is NOT added to the list because it is already there. Clearly that is not true. Even if only because befoer that you've misused the diagonal argument.

Incidentally, in what way have you USED the axiom of infinity INDUCTION to do anything?
 
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  • #78
Matt.

01010101... 001100110011... 0000111100001111... are columns not rows.

Now I clearly see that you simply don't read what I write in my papers.
 
  • #79
look at page 2, the array for n=3 and all the combinations there. you claim that the right hand column is based on 2^1 or something, that is that it ought to be 01010101. it isn't it is 1100101 how can we see that is what you say it is when it clearly isn't?

i know they're columns, where first is the rightmost. look at the rightmost column on page 2 for n=3.
 
  • #80
so let us assume that you've picked some ordering of the rows where you've got this pattern - which you CAN do for n=3, as you've got on page 1 and page 3, but not page 2.

So, we can produce a doubly infinite array (to the left and down) where each column is periodic ok, getting there. this is not using the infinity axiom of induction. whatever that might be - you'vr not explained that either, btw, axiom of infinity yes, but not the axiom of infinity induction.

SO? Reading across in each column gives us an element of the combination list, doesn't it? Well, as I stated before, that tells us that each element on the list, that is each row, has only finitely many non-zero entries.

How are you concluding that this list has all the elements you want on it? Cantor's diagonal argument produces something not on that list, it has infinitely many non-zero entries, you claim that this need not be added to the list because it is already on it. No it isn't - this is the basis for you deciding 2^aleph-0=aleph-0
 
  • #81
Again you show us that you don't fully read what I write, therefore don't understand what you see.

Go back to page 1 read all of it and then read cerfully the first lines in the top of page 2.
 
  • #82
Have done, realized what was going on, so see the last post which i reprint here:
so let us assume that you've picked some ordering of the rows where you've got this pattern - which you CAN do for n=3, as you've got on page 1 and page 3, but not page 2.

So, we can produce a doubly infinite array (to the left and down) where each column is periodic ok, getting there. this is not using the infinity axiom of induction. whatever that might be - you'vr not explained that either, btw, axiom of infinity yes, but not the axiom of infinity induction.

SO? Reading across in each column gives us an element of the combination list, doesn't it? Well, as I stated before, that tells us that each element on the list, that is each row, has only finitely many non-zero entries.

How are you concluding that this list has all the elements you want on it? Cantor's diagonal argument produces something not on that list, it has infinitely many non-zero entries, you claim that this need not be added to the list because it is already on it. No it isn't - this is the basis for you deciding 2^aleph-0=aleph-0
 
  • #83
Be aware that we have an ordered collection of infinitely many 01 sequences by using the ZF axiom of infinity built-in induction on the power level of 2^power_level.

Again ZF axiom of infinity: If n then n+1.

Therefore the power_level = aleph0 and we have an ordered collection of 2^aleph0 elements.
 
  • #84
you see, now this is where you go horribly wrong, you pass from a finite cardinal to an infinite one. YOU CANNOT DO THIS! It is not valid.

AND, you've still not said what you mean by

the ZF axiom of infinity in-built induction. What the hell is this?? On the power level 2^powerlevel, what the hell is power level??

And the axiom of infinity does not state if n then n+1
 
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  • #85
Please write the ZF axiom of infinity in English words.
 
  • #86
The simple interpretation is that there is an inductive set, that is we can produce a set that behaves like the natural numbers. It is bad english to say that n impies n+1. The maths states that given a set which we can label by n we can form another set which we can label n+1, by taking the union of the set labelled n with the set containing the empty set. It is your english that is most at fault, and virtue of that then your maths. It is the existence of the set indexed by n, with the set containing the empty set that implies there is a set labelled n+1, the labels being the cardinalities.


All you are doing is saying that there infinitely many natural numbers, OK.

let the r'th column of a doubly infinite array be the string of 2^r 0s, then 2^r 1s then 2^r 0s and so on. There is no need to use any axiom of infinity induction (which is...?0 merely that the naturals are infinite! There is no induction! we are not using the s'th level to define the s+1'st level, which is what an induction would be!
 
  • #87
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
 
  • #88
My language is Hebrew.

Now Zf axiom of infinity is:

There is a set Omega that contains the empty set and if any set y is in Omega then the set containing the union of y and the set containing y, is also in Omega.

By induction Omega contains every integer.

So as you see, we are talking about Omega = {1,2,3,...}
and Omega is aleph0.

Therefore our collection is 2^aleph0 collection.
 
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  • #89
No, aleph-0 is not a set. It is the cardinality of a set. Your omega is the set of natural numbers, its cardinality is Aleph-0.

The last line:

Therefore our collection is 2^aleph0 collection.


Makes little sense to me.
 
  • #91
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.
 
  • #92
No Matt,

By using the induction of ZF axiom of infinity on 2^power_value (by mistake i wrote power_level) power_value = aleph0.
 
  • #93
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!
 
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  • #95
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!
 
  • #97
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.
 
  • #98
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).
 
  • #99
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.
 
  • #100
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.
 
  • #101
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 defintion! 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.
 
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  • #102
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
 
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  • #103
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?
 
  • #104
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
...
...
 
  • #105
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 becuase, 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.
 

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