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Deeper than primes

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"Beyond the matrix II" game

Sure I can...and quite easily. Cantor did it already.
And he did it by using at least a collection of different <0,1> forms, which according to it, there are <0,1> forms with the same number of bits (what you call Y objects) both in S AND 2^S, such that the inverse of the diagonal of some particular S different objects' arrangment (which is actually a X^2 <0,1> matrix, where the number of bits of each different object, and the number of objects, have the same cardinality) is also an object of X bits, even if the considered collection is P(S) or 2^S (where the number of different <0,1> forms is greater than the number of the bits of each form).

The fact that Y objects (objects of X bits each) are defines both in S and P(S), enables use to define a bijection between N and P(N) objects, by simply defining a mapping between each natural number and each different case of some inverse object of the diagonal of that particular X^2 matrix.

Furthermore, S < 2^S is false, where |S| is finite or not, because it is clearly shown that there is a bijection between each natural number and each given case of different inverse object of the diagonal of some S with different X^2 matrix's arrangement.

Actually, there is S ↔ ...P(P(S))... and since the cardinality of the collection of all ...P(P(S))... is unsatisfied, then the cardinality of S is unsatisfied (incomplete).

No, S continues to be complete either way.
It can't, because its cardinality is unsatisfied.

Here is an example (based on "Beyond the matrix II" game http://www.internationalskeptics.com/forums/showpost.php?p=6907490&postcount=14351 ):

Here is some collection of P({apple, orange, lemon}) different objects, that were translated to <0,1>^|{apple, orange, lemon}| different objects:

{
000 ↔ {}
001 ↔ {apple}
010 ↔ {orange}
011 ↔ {lemon}
100 ↔ {apple,orange}
101 ↔ {apple,lemon}
110 ↔ {orange,lemon}
111 ↔ {apple, orange, lemon}
}

As can be seen the generalized <0,1>^3 forms have the same structural principle, whether they are used as P(X) or X objects, for example:

P(X)=
{
000,
001,
010,
011,
100,
101,
110,
111
}

X=
(
{
100,
110,
111
} → 000 not in the range of this particular arrangement of different <0,1> forms (where each particular arrangement is X^2 of <0,1> bits)
or
{
101,
010,
000
} → 001 not in the range of this particular arrangement of different <0,1> forms (where each particular arrangement is X^2 of <0,1> bits)
Etc…
)

Our mapping is done between each natural number and each inverse of the diagonal of each particular arrangement is X^2 of <0,1> bits, as follows:

1 ↔ 000
2 ↔ 001
...

etc. and it is shown that there is a bijection between natural numbers and any different object of 2^S collection.

Furthermore, since the diagonal method is invariant for ...2^(2^(2^S)))..., whether |S| is finite or not, there is a bijection between natural numbers and the inverse of the diagonals of ...2^(2^(2^S)))... and since the cardinality of the collection of all ....2^(2^(2^S)))... is unsatisfied, then the cardinality of natural numbers is unsatisfied (incomplete).

----------------

The same principle is shown by using Cantor's theorem ( http://en.wikipedia.org/wiki/Cantor's_theorem ).

By this theorem, cantors explicitly provides a 2^N object, called D, that is not in the range of N objects. But as we have shown above, if the bijunction is done by using such an explicitly provided 2^N object (D object) and some natural number, then there is a bijection between each natural number and each 2^N D object, where Cantor's theorem is some single case that provides an explicit 2^N D object that can be put in one-to-one correspondence with some distinct natural number.

Again, Cantor's theorem is actually a consistent method to provide some explicit 2^N D object that can be put be put in one-to-one correspondence with some distinct natural number.

Cantor's mistake is based on the fact that he concluded general conclusions by using only a one round of his theorem.
 
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..., the successor of any member of the set is also a member of the set.
This is nothing but the current agreement about the concept of successor.

By OM, the concept of Successor is simply the parmanently next different object of any given collection (notatad by "+1").

The number of the objects of the set of natural numbers is unsatisfied exactly because of the "+1" parmanently next different object for any given n, notated as n+1.
 
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Sure I can...and quite easily. Cantor did it already.
And he did it by using at least a collection of different <0,1> forms, which according to it, there are <0,1> forms with the same number of bits (what you call Y objects) both in S AND 2^S, such that the inverse of the diagonal of some particular S different objects' arrangment (which is actually a X^2 <0,1> matrix, where the number of bits of each different object, and the number of objects, have the same cardinality) is also an object of X bits, even if the considered collection is P(S) or 2^S (where the number of different <0,1> forms is greater than the number of the bits of each form).

There is so much wrong in that gibberish it is difficult to know where to start.

Cantor did not use bit maps as you claim. Cantor did not use the expression 2S as you claim. Cantor did not use numbers at all.

What Cantor did do to was show that assuming |S| and |P(S)| were the same leads to a contradiction.

Doron, since you continue to assume |S| and |P(S)| are the same, your results go beyond wrong; they are meaningless.

...
No, S continues to be complete either way.
It can't, because its cardinality is unsatisfied.

This has been a significant problem within the Catholic priesthood; I had no idea it extended to higher levels of the Church leadership.

No matter, though, since the sexual frustrations of religious celibates does not enter into set theory. Sets remain complete despite any crank's attempts to redefine terms, even in the name of the Holy Father.

...
X=
(
{
100,
110,
111
}

Case in point of yet a different version of you redefining things, this time midway through the example. You started with X = {apple, orange, lemon}, and then you insisted on introducing bit maps for no advantage other than creating confusion. Under your bit map notation X = 111, yet here you change it to something else.

X was a set of three things, and from that the power set and bit maps followed. Now you want X to be a set of three subsets, entirely different from the starting point. This change of definition in mid-stream pulls the foundation out from under your bit maps. In short, you contradict yourself.
 
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This is nothing but the current agreement about the concept of successor.

By OM, the concept of Successor is simply the parmanently next different object of any given collection (notatad by "+1").

The number of the objects of the set of natural numbers is unsatisfied exactly because of the "+1" parmanently next different object for any given n, notated as n+1.


I never said anything about the set of natural numbers being closed under an operation of succession. 2.567 as well as 2.734 are both numbers that are successors to the natural number 2 in the set of real numbers but are not included in the set of natural numbers as they are not integers and thus not natural numbers. “n+1” only denotes the next ordinal in an ordered set, although that does explicitly assert that such a next element of that ordered set is a successor to the pervious. It does not in anyway exclude the possibility of a successor to any element of that set that is not in the set. The set of all films or all numbers does exclude such possibility. Learn some set theory, please.


“The number of the objects of the set of natural numbers is” not “unsatisfied”. You are just unsatisfied because you still simply can’t (and evidently deliberately don’t want to) distinguish between a list and a set.

Again what natural number is not a natural number Doron? Within the set of all integers the set of natural numbers is closed under an operation of succession. There is no integer number that is not itself a natural number and a successor to any natural number. Any set is closed under an operation of succession within itself but may not be in some superset. Within the set of real numbers there are successors that are not natural numbers to any natural number. Just as in the set of all films there are successors that are not Fellini films to any Fellini film. This is the concept you keep missing yet try to erroneous employ for your ‘member that is not a member’ incomplete nonsense. Again the set of all Fellini films is closed to a succeeding Fellini film. If one is found tomorrow it is still a Fellini film and a member of that set of all Fellini films. Yes the cardinality of the known set of all Fellini films does change but the set is still closed under an operation of succession (not closed under your “cardinality” as you like to pretend). The problem again with your ‘interpretation’ and example is that there are no natural numbers that are not already known to be natural numbers, while there could be an unknown Fellini film that someone could find. No one is creating or finding new unknown natural numbers, Doron. So when you do dig up a natural that was not already a natural number before, please let us know. You do understand that with your “n+1” you are using natural numbers for your ordinals of the natural numbers, don’t you? So you need to find an “n+1” that is not an “n+1”. Good luck with that.
 
X was a set of three things, and from that the power set and bit maps followed. Now you want X to be a set of three subsets, entirely different from the starting point. This change of definition in mid-stream pulls the foundation out from under your bit maps. In short, you contradict yourself.

Hey, that's cross-contexting for ya. Deal with it AND without it (cause non-locally you have to do both).


Coming soon (AND before) to a cross-context theater far away, but near you (non-locally that is).
 
What Cantor did do to was show that assuming |S| and |P(S)| were the same leads to a contradiction.
He explicitly defined an object of P(S), which is not in the range of S (this object is, however, equivalent to the inverse of the diagonal, which is not in the range of X^2, where X is the cardinality (finite or infinite) of a collection of different <0,1> forms, such that the number of bits of each object is the same as the number of the objects of the considered collection).

But as I explained, first we define an explicit object, which is not in the range of X^2 matrix, and only then we define a mapping between this explicit object and some natural number.

We are doing it 2^X times and get a mapping between any 2^S object and any given natural number.

This 2^X repetition was omitted from Cantor's theorem, and as a result he missed the ability to define the bijection between S and 2^S.

Actually this bijection can be shown by using 2^S objects, without translating them to <0,1> forms (although the <0,1> forms helps to understand how to define a bijection with the natural numbers, beyond any X^2 matrix).

For example, by using Cantors argument, if the natural numbers are mapped only with P(N) member, where each mapped P(N) member contains an image of the natural number that is mapped with it, then the result is {}, which is some P(N) member.

In this case there is no problem to define a mapping between natural number 1 and {}.

Also by using Cantors argument, if the natural numbers are mapped only with P(N) members, where each mapped P(N) member does not contain an image of the natural number that is mapped with it, then the result is {1,2,3,…}, which is some P(N) member.

In this case there is no problem to define a mapping between natural number 2 and {1,2,3,…}.

By using the same technique, we can define any explicit P(N) member between {} and {1,2,3,…}, and define a mapping between this explicit P(N) member and some natural number.

We are doing it 2^X times and get a mapping between any P(N) member and any given natural number.
 
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I never said anything about the set of natural numbers being closed under an operation of succession. 2.567 as well as 2.734 are both numbers that are successors to the natural number 2 in the set of real numbers but are not included in the set of natural numbers as they are not integers and thus not natural numbers.
The Man, you demonstrate again only a context-deepened reasoning that ignores cross-contexts reasoning, which according to it, any given set is a collection of different objects, no matter if natural numbers, rational number or irrational numbers are considered.

n+1” only denotes the next ordinal in an ordered set, although that does explicitly assert that such a next element of that ordered set is a successor to the pervious. It does not in anyway exclude the possibility of a successor to any element of that set that is not in the set. The set of all films or all numbers does exclude such possibility. Learn some set theory, please.
By using also a cross-contexts reasoning, it is shown that no collection of different objects is complete, where "+1" is the exact notation of this fact.

Furthermore, in http://www.internationalskeptics.com/forums/showpost.php?p=6916077&postcount=14361 and http://www.internationalskeptics.com/forums/showpost.php?p=6918253&postcount=14366 I explicitly show that there is a bijection between N and P(N) ( or more general, there is a bijection between any S and any P(S) ), and since the set of all powersets does not exist, then any given set is incomplete.
 
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X was a set of three things, and from that the power set and bit maps followed.
Again only a context-dependent reasoning is used by you.
By using also a cross-contexts reasoning one enables to understand that {1,2,3} and {000,101,110} are some cases of a set of three things.

Now you want X to be a set of three subsets, entirely different from the starting point. This change of definition in mid-stream pulls the foundation out from under your bit maps. In short, you contradict yourself.
In short, you are unable to understand that {1,2,3} and {000,101,110} are some cases of a set of three things.
 
There is so much wrong in that gibberish it is difficult to know where to start.

Cantor did not use bit maps as you claim. Cantor did not use the expression 2S as you claim. Cantor did not use numbers at all.
That saved me some typing. Doron doesn't have the slightest idea what Cantor was actually doing and why. Perhaps it is coz of Doron's "context-independent reasoning."
 
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Again only a context-dependent reasoning is used by you.

Context-independent reasoning that you use is very convenient, such as in this case: 2x + 3 = 11.

The solution can be any real number, coz your inquiry is context-independent, focuses only on the unknown variable x and leaves the rest of the equation for the birds.
 
He explicitly defined an object of P(S)

No, Cantor's proof identifies a set, D, whose members all have a certain characteristic. The proof then shows that D happens to be a subset of the original set, S, and therefore a member of P(S).

...which is not in the range of S

No. As I said, D is a subset of S and therefore a member of P(S).

...(this object is, however, equivalent to the inverse of the diagonal, which is not in the range of X^2, where X is the cardinality (finite or infinite) of a collection of different <0,1> forms, such that the number of bits of each object is the same as the number of the objects of the considered collection).

No, it isn't. Cantor's proof made no use of numbers whatsoever. You seem stuck on them. Moreover, if X = |S| then any subset of S can be coded as an X-bit bit map, true enough, but there are 2X possible such encodings. The vast majority of bit maps don't participate in your diagonal construction.

But as I explained, first we define an explicit object, which is not in the range of X^2 matrix

So what? Your matrix is really X bits wide by 2X bit maps deep. Your "explicit object" may not be in the X by X beginning of the full matrix, but it appears further down.

...and only then we define a mapping between this explicit object and some natural number.

You are assuming things again. Incorrect things at that. And the rest of your post is just or of the same gibberish crossed with false assumptions crossed with bogus logic.
 
Again what natural number is not a natural number Doron? Within the set of all integers the set of natural numbers is closed under an operation of succession. There is no integer number that is not itself a natural number and a successor to any natural number. Any set is closed under an operation of succession within itself but may not be in some superset. Within the set of real numbers there are successors that are not natural numbers to any natural number. Just as in the set of all films there are successors that are not Fellini films to any Fellini film. This is the concept you keep missing yet try to erroneous employ for your ‘member that is not a member’ incomplete nonsense. Again the set of all Fellini films is closed to a succeeding Fellini film. If one is found tomorrow it is still a Fellini film and a member of that set of all Fellini films. Yes the cardinality of the known set of all Fellini films does change but the set is still closed under an operation of succession (not closed under your “cardinality” as you like to pretend). The problem again with your ‘interpretation’ and example is that there are no natural numbers that are not already known to be natural numbers, while there could be an unknown Fellini film that someone could find. No one is creating or finding new unknown natural numbers, Doron. So when you do dig up a natural that was not already a natural number before, please let us know. You do understand that with your “n+1” you are using natural numbers for your ordinals of the natural numbers, don’t you? So you need to find an “n+1” that is not an “n+1”. Good luck with that.
I think you are wasting your time, coz I've already tried to sneak into Doron's head the finite set of the ten digits of the decimal system with reference to the millions of known digits of number pi. No worky, coz Doron switches to his "context-independent reasoning," which is a synonym to "sheer ignorance." In other words, he knows how to protect his toys:


Definition of FINITE

3a : less than an arbitrary positive integer and greater than the negative of that integer b : having a finite number of elements <a finite set>


For some LOL reason, Merriam Webster encloses examples the same way that Doron does with his invincible <0,1> argument.
 
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No, Cantor's proof identifies a set, D, whose members all have a certain characteristic. The proof then shows that D happens to be a subset of the original set, S, and therefore a member of P(S).
No. As I said, D is a subset of S and therefore a member of P(S).
By using Cantor's theorem 2^X times one enables to define a mapping between each natural number and each D of P(S).

No, it isn't. Cantor's proof made no use of numbers whatsoever. You seem stuck on them. Moreover, if X = |S| then any subset of S can be coded as an X-bit bit map, true enough, but there are 2X possible such encodings. The vast majority of bit maps don't participate in your diagonal construction.
So what? Your matrix is really X bits wide by 2X bit maps deep. Your "explicit object" may not be in the X by X beginning of the full matrix, but it appears further down.
So what? By using Cantor's theorem 2^X times one enables to define a mapping between each natural number and each D of P(S). You seem stuck on X^2.

You are assuming things again. Incorrect things at that. And the rest of your post is just or of the same gibberish crossed with false assumptions crossed with bogus logic.
You are closed under X^2 matrix. And the rest of your post is just or of the same gibberish crossed with false assumptions crossed with bogus logic.
 
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That saved me some typing. Doron doesn't have the slightest idea what Cantor was actually doing and why. Perhaps it is coz of Doron's "context-independent reasoning."
Like it or not, 2^S and P(S) are equivalent.
 
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Again what natural number is not a natural number Doron?
We are talking about a complete collection of natural numbers.

But as a show in http://www.internationalskeptics.com/forums/showpost.php?p=6916077&postcount=14361, http://www.internationalskeptics.com/forums/showpost.php?p=6918253&postcount=14366 and http://www.internationalskeptics.com/forums/showpost.php?p=6919671&postcount=14373, such a collection is ever increasing, and therefore its cardinality is unsatisfied (or in other words, it is incomplete).
 
Cantor's proof made no use of numbers whatsoever.
Yes he does. He uses number 1 for 1 round of his theorem, and as a result he misses the abilitiy to show that
there is a bijection between S and P(S).

I corrected it by using his theorem 2^|S| times.

Let us demonstrate my technique of using Cantor's theorem on P({1,2,3}) case.

Round 1 particular case:
-----------------------
1 ↔ {1} |
2 ↔ {2} | Provides {} ↔ 1
3 ↔ {3} |
-----------------------

Round 2 particular case:
-----------------------
1 ↔ { } |
2 ↔ {2} | Provides {1} ↔ 2
3 ↔ {3} |
-----------------------

Round 3 particular case:
-----------------------
1 ↔ {1} |
2 ↔ { } | Provides {2} ↔ 3
3 ↔ {3} |
-----------------------

Round 4 particular case:
-----------------------
1 ↔ {1} |
2 ↔ {2} | Provides {3} ↔ 4
3 ↔ { } |
-----------------------

Round 5 particular case:
-----------------------
1 ↔ { } |
2 ↔ {3} | Provides {1,2} ↔ 5
3 ↔ {1,3} |
-----------------------

Round 6 particular case:
-----------------------
1 ↔ { } |
2 ↔ {2} | Provides {1,3} ↔ 6
3 ↔ {1,2} |
-----------------------

Round 7 particular case:
-----------------------
1 ↔ {1} |
2 ↔ {2} | Provides {2,3} ↔ 7
3 ↔ {1,2} |
-----------------------

Round 8 particular case:
-----------------------
1 ↔ { } |
2 ↔ {1} | Provides {1,2,3} ↔ 8
3 ↔ {2} |
-----------------------
 
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By using Cantor's theorem 2^X times one enables to define a mapping between each natural number and each D of P(S).

Using Cantor's theorem exactly once proves you wrong.

The basic mathematical fact remains, |S| < |P(S)| for any set S.
 
Yes he does. He uses number 1 for 1 round of his theorem, and as a result he misses the abilitiy to show that
there is a bijection between S and P(S).

There are no "rounds" in Cantor's theorem.

It with an assumption that |S| = |P(S)|, relying on mapping functions to define the ordering relations for cardinality. It then shows that given that some mapping function f() must exist, it establishes the existence of a set D based on f(). D is shown to be a subset of S and therefore a member of P(S), and that ends up leading to a contradiction.

Just a straight sequence of steps, from assumption to contradiction. No rounds anywhere.
 
We are talking about a complete collection of natural numbers.

But as a show in http://www.internationalskeptics.com/forums/showpost.php?p=6916077&postcount=14361, http://www.internationalskeptics.com/forums/showpost.php?p=6918253&postcount=14366 and http://www.internationalskeptics.com/forums/showpost.php?p=6919671&postcount=14373, such a collection is ever increasing, and therefore its cardinality is unsatisfied (or in other words, it is incomplete).

"Increasing"? Again what natural number was not a natural number before you 'increased' your set of natural numbers? You keep deliberately portending that a set is a list.
 
I think you are wasting your time, coz I've already tried to sneak into Doron's head the finite set of the ten digits of the decimal system with reference to the millions of known digits of number pi. No worky, coz Doron switches to his "context-independent reasoning," which is a synonym to "sheer ignorance." In other words, he knows how to protect his toys:


Definition of FINITE

3a : less than an arbitrary positive integer and greater than the negative of that integer b : having a finite number of elements <a finite set>


For some LOL reason, Merriam Webster encloses examples the same way that Doron does with his invincible <0,1> argument.

That's OK, I've got time to waste.
 
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