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.
  • #106
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)?
 
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  • #107
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.
 
  • #108
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, ...
 
  • #109
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.
 
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  • #110
Matt,

Yes ...01010101 is in T but not in L.
 
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  • #111
Hurkyl,

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

Please look at the axiom of infinity.
 
  • #112
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.
 
  • #113
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?
 
  • #114
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.
 
  • #115
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.
 
  • #116
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.
 
  • #117
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"
 
  • #118
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.
 
  • #119
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
 
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  • #120
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.
 
  • #121
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 it product is clearly Omega(=aleph 0)?

What does this have to do with

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

?


(P.S. I'm content with simply saying &omega; = N)
 
  • #123
Originally posted by 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.

Priceless! So, your infinite list has two ends?

What is clear is that reading the list



...00000
...00001
...00010
...00111
...01100


that every string on the list has only finitely many elements that are non-zero. My proof holds here: let x be the string on row r, as 2^r>r for all r, it follows that after the r'th postion on the string (right to left) that every subsequent entry is zero as by definition the s'th coumn starts with 2^s zeroes.



Now the second list reading the orignal list from bottom to top (which now implies that the list is finite)

...1111
...1110
...1001
...1000


the same proof demonstrates that any string on the list only has a finite number of 0's in it. Now you are saying these lists are the same? Where is the string ...1111 on th first list? I only ask because in you new diagonal paper you say it isn't on the first list. But it is on the second, and the lists emnumerate the same elements,. namely the thing you call T?




And you say if n then n+1

n is not a statement that is true or false. It is evident you don't understand the slightest thing abuot logic and its conventions. What does it mean for n to be true?
 
  • #124
Theorem:

If S is a set, and subsets of S have both smallset and largest elements, then S is finite.


As a special case:

If L is a list, and each sublist of L has both a first and a last entry, then L is a finite list.
 
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  • #125
Matt,

My proof is not about |T|=|L| but |T|>=|L|.

in T there are:

...000...1111
...000...1110
...000...1001
...000...1000

and also there are:

...111...1111
...111...1110
...111...1001
...111...1000

You still don't understand that there is no such a thing like cardinality of infinitly many objects, because any collection of infinitly many objecsts is
a non-complete collection.

Therefore its cardinality cannot be found.
 
  • #126
Not every collection can be represented by a list.
 
  • #127
Originally posted by Organic
Matt,

My proof is not about |T|=|L| but |T|>=|L|.

in T there are:

...000...1111
...000...1110
...000...1001
...000...1000

and also there are:

...111...1111
...111...1110
...111...1001
...111...1000

You still don't understand that there is no such a thing like cardinality of infinitly many objects, because any collection of infinitly many objecsts is
a non-complete collection.

Therefore its cardinality cannot be found.

The wholepoint of this is that you ARE claiming |T|=|L|, how do you do this? by claiming the list for T has cardinality 2^aleph-0, that it is a complete list. It isn't as you know because of th diagonal argument.

Explain why N is not complete, in your view; cardinality does make sense for infinite sets; answer any of the fatal errors in your posts today.
 
  • #128
Hurkyl,

I am talking about (...111,...000] XOR [...111,...000)
XOR (...101010,...111] XOR (...010101,...000] XOR ...
 
  • #129
Matt,

The fatal error is the cardinality of infinitly many elements.


Please show me how can infinitely mant elements can be complete.
 
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  • #130
(...111,...000] XOR [...111,...000)
XOR (...101010,...111] XOR (...010101,...000] XOR ...

(a) We've told you over and over again that this notation is nonsensical; it doesn't even come close to the standard mathematical definition, and you refuse to define it (and consistently claim that you are using it in a standard way)

(b) You're trying to make a list. What does exclusive or have to do with anything?

(c) You do realize that if A xor B xor C xor ... does NOT mean "Exactly one of A, B, C, ... are true"; it means "An odd number of the A, B, C, ... are true".

e.g.: if A, B, and C are all true, then A xor B xor C is true.
 
  • #131
Originally posted by Organic
Hurkyl,

I am talking about (...111,...000] XOR [...111,...000)
XOR (...101010,...111] XOR (...010101,...000] XOR ...


Oh, look, more misuse of logic. XOR is a logical predicate, its inputs are things that are eithert true or false. what does it mean of (...1111,...000] to be true. Hell, for that matter, what does it mean? I guess you'll post about complementary logic, even though it isn't clear why you'd do that because all the questions are about proper mathematical objects.
 
  • #132
Please show me how can infinitely mant elements can be complete.

(I'm guessing at what you mean by complete)


If C is a collection, then C = C, right?

Then, C is clearly a complete collection of the elements of C.

N is clearly a complete collection of the elements of N.

R is clearly a complete collection of the elements of R.
 
  • #133
Ok Hurkyl,



I am talking about (...111,...000] OR [...111,...000)
OR (...101010,...111] OR (...010101,...000] OR ...
 
  • #134
A or B or C or ... means "At least one of the A, B, C, ... are true"

A and B and C and ... means "All of the A, B, C, ... are true"


Oh, I should also note that logical expressions may only have a finite number of terms.


That still doesn't address the issue that

(...111,...000]

has not been given a meaning comprehensible to anyone but yourself.


And it still doesn't address the issue that this seems to have absolutely nothing to do with lists. Based on your earlier posts, I might guess you're trying to say "Do something that corresponds to (...111,...000], then the something that corresponds to [...111,...000), then ..."

But I know neither what (...111,...000] is, nor what that something corresponding to (...111,...000] is.
 
  • #135
Originally posted by Organic
Matt,

The fatal error is the cardinality of infinitly many elements.


Please show me how can infinitely mant elements can be complete.


Let N be the collection of all natural numbers, it is complete in the sense that it contains all the natural numbers, and its existence is not contrary to the ZF axioms (in fact it is required by the axiom of infinity). It is a set.


Let Z be the ring defined by formally adding inverses to elements in Z and including 0. Clearly it is still a set, and is 'complete' in any reasonable sense - that is the opereations of addition and subtraction do not take one out of the set. It doesn't contain any pink elephants, but then its existence doesn't require pink elephants.


Form the fraction field, clearly this is still a set as the construction of itself proves (there are countably many elements in any equivalence class, and countably many equivalence classes. It is Q, and it is again complete in terms of algebraic operations..

perhaps you should define what you mean by complete?




Let Q be
 
  • #136
Hurkyl,

If a set has infinitely many elements we cannot use the word "complete" from a quantitative point of view, because no quantity can be captured and notated by one symbol, and then can be used as a meaningful input for some mathematical system.

When infinitely many elements are forced to be notated as one concept
we are dealing with actual infinity, and no theory can use actual infinity as input, and cannot explore it.

Please see: http://www.geocities.com/complementarytheory/RiemannsLimits.pdf
 
  • #137
Originally posted by Organic
Hurkyl,

If a set has infinitely many elements we cannot use the word "complete" from a quantitative point of view, because no quantity can be captured and notated by one symbol, and then can be used as a meaningful input for some mathematical system.

When infinitely many elements are forced to be notated as one concept
we are dealing with actual infinity, and no theory can use actual infinity as input, and cannot explore it.

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


So it is your personal meaning for complete that no one else understands? What quantity do you mean?


Anyway, why haven't you answered the questions posed to you?

Does this mean you accept that your attempts to show |T|=|L| is non-sense?
 
  • #139
Looking back through your posts, it seems that you mean

a set C, is not complete, if given any list (enumeration) of some (possibly all) of its elements, then there is some element of C not listed.


This is exactly the defintion of uncountable.

N is 'complete' as the trivial enumeration shows, so complete is countable? Since N exists countable/complete sets exist, and clearly you accept Cantor's diagonal argument, hence there aer uncountable sets. Which in your world would be 'incomplete'. As there already exists a word 'countable' and 'complete' hasn't been well defined by you, can I suggest this as a definition, and simultaneous ask that you stop using it 'cos complete is used in lots of situations already, and why invent another name for something already known
 
  • #140
Originally posted by Organic
Matt,

Please read again pages 3 an 4 in:

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

and also look again at:

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

Thank you.


Why, what are these going to answer?

Just provide a defintion of complete here, shoulc be a matter of a few lines. We don't need to reread new diagonal because there are aleraedy enough errors explained to you here that you've failed to address, why would we add more.
 

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