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Deeper than primes

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zooterkin said:

Ok, so now I'm sure you don't have a clue what it means. What next?
Click to expand...
LOL. No one really knows what it means, coz part of the reference provided by Doron shows the door that opens for the uncountable set to enter.
http://en.wikipedia.org/wiki/File:Diagonal_argument_2.svg

Note the indexing E1, E2, E3... with 1, 2, 3... being counting numbers. The vectors that made the whole set are therefore partially countable with one vector unaccounted for (the swapped diagonal), similar to the case where one member of a family is missing from the get-together family photo -- the one who made the pic. Give it to Doron as a birthday present to play with.
 
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zooterkin said:

Ok, so now I'm sure you don't have a clue what it means. What next?
Click to expand...
Yes zooterkin I'm sure that you are unable to understand my novel use of the diagonal argument, in order to prove that P(S) is incomplete, so?
 
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epix said:
LOL. No one really knows what it means, coz part of the reference provided by Doron shows the door that opens for the uncountable set to enter.
Click to expand...
Wrong, counable or uncountable is just an illusion because given any collection (finite or not) it is naturally incomplete exactly because it does not have the totality of Emptiness (that has no predecessor) or Fullness (that has no successor).

epix said:
Note the indexing E1, E2, E3... with 1, 2, 3... being counting numbers.
Click to expand...

EDIT:

Note that my novel use of the diagonal number is equivalent to the members of P(N),

{
0000000... ↔ { },
1100000... ↔ {1,2},
1000000... ↔ {1},
1010000... ↔ {1,3},
0100000... ↔ {2},
0110000... ↔ {2,3},
0010000... ↔ {3},
...
}

where the form that is not in the range of P(N) starts with 1011111..., in this case.
 
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doronshadmi said:
Yes zooterkin I'm sure that you are unable to understand my novel use of the diagonal argument, in order to prove that P(S) is incomplete, so?
Click to expand...
Doron, according to your definition of completeness and incompleteness, is number Pi in expressed in its base10 numerical form complete or incomplete?
 
To remind you what you claimed:
doronshadmi said:
It is very useful. By this generalization we are able to understand that incompleteness is an essential requirement of the consistency of any non-empty collection, whether it is finite or not.
Click to expand...

Do you think you could leave aside infinite sets, and show how your novel use works for the set {x,y,z} and its powerset?

• { } = 000
• {x} = 100
• {y} = 010
• {z} = 001
• {x, y} = 110
• {x, z} = 101
• {y, z} = 011
• {x, y, z} = 111

Which set is missing from the powerset?
 
doronshadmi said:
Wrong, counable or uncountable is just an illusion . . .
Click to expand...

. . . especially when your silly argument below limps due to the typo that's a big no-no when you try to prove something.

0000000... ↔ { },
1100000... ↔ {1,2},
1000000... ↔ {1},
1010000... ↔ {1,3},
0100000... ↔ {2},
0110000... ↔ {2,3},
0010000... ↔ {2},
...
}
Click to expand...
 
doronshadmi said:
Note that my novel use of the diagonal number is equivalent to the members of P(N),

{
0000000... ↔ { },
1100000... ↔ {1,2},
1000000... ↔ {1},
1010000... ↔ {1,3},
0100000... ↔ {2},
0110000... ↔ {2,3},
0010000... ↔ {2}, (typo; should read 3)
...
}

where the form that is not in the range of P(N) starts with 1011111..., in this case.
Click to expand...

Doron, you can assign a set of binary strings to any set of objects that are either finite or infinite.

{
1000000... ↔ 1
0100000... ↔ 2
0010000... ↔ 3
0001000... ↔ 4
0000100... ↔ 5
0000010... ↔ 6
0000001... ↔ 7
...
}

Reverse diagonal = 0000000 and is not included in the set.

In this case 0000000=8, which is not in the set. If the set above is not bound, then the reverse diagonal equals n+1 where n is a natural number and lim n → ∞.

Keep discovering . . .
 
epix said:
Doron, you can assign a set of binary strings to any set of objects that are either finite or infinite.
Click to expand...
I know, this is a proof of the incompleteness of any given set, whether it is finite or not.

epix said:
{
1000000... ↔ 1
0100000... ↔ 2
0010000... ↔ 3
0001000... ↔ 4
0000100... ↔ 5
0000010... ↔ 6
0000001... ↔ 7
...
}

Reverse diagonal = 0000000 and is not included in the set.

In this case 0000000=8, which is not in the set. If the set above is not bound, then the reverse diagonal equals n+1 where n is a natural number and lim n → ∞.

Keep discovering . . .
Click to expand...

{
10000000... ↔ 1
01000000... ↔ 2
00100000... ↔ 3
00010000... ↔ 4
00001000... ↔ 5
00000100... ↔ 6
00000010... ↔ 7
00000000... ↔ 8
...
}
Now we have 00000001 that is not in the set etc ... ad infinitum ... , so?
 
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zooterkin said:
To remind you what you claimed:


Do you think you could leave aside infinite sets, and show how your novel use works for the set {x,y,z} and its powerset?

• { } = 000
• {x} = 100
• {y} = 010
• {z} = 001
• {x, y} = 110
• {x, z} = 101
• {y, z} = 011
• {x, y, z} = 111

Which set is missing from the powerset?
Click to expand...
You can see it if you compare some collection of distinct <0,1>^3 with P(<0,1>^3)=<0,1>^8, as follows:

00001111
00110011
01010101
---------
00000000
10000000
01000000
11000000
00100000
10100000
01100000
11100000
00010000
10010000
01010000
11010000
00110000
10110000
01110000
11110000
00001000
10001000
01001000
11001000
00101000
10101000
01101000
11101000
00011000
10011000
01011000
11011000
00111000
10111000
01111000
11111000
00000100
10000100
01000100
11000100
00100100
10100100
01100100
11100100
00010100
10010100
01010100
11010100
00110100
10110100
01110100
11110100
00001100
10001100
01001100
11001100
00101100
10101100
01101100
11101100
00011100
10011100
01011100
11011100
00111100
10111100
01111100
11111100
00000010
10000010
01000010
11000010
00100010
10100010
01100010
11100010
00010010
10010010
01010010
11010010
00110010
10110010
01110010
11110010
00001010
10001010
01001010
11001010
00101010
10101010
01101010
11101010
00011010
10011010
01011010
11011010
00111010
10111010
01111010
11111010
00000110
10000110
01000110
11000110
00100110
10100110
01100110
11100110
00010110
10010110
01010110
11010110
00110110
10110110
01110110
11110110
00001110
10001110
01001110
11001110
00101110
10101110
01101110
11101110
00011110
10011110
01011110
11011110
00111110
10111110
01111110
11111110
00000001
10000001
01000001
11000001
00100001
10100001
01100001
11100001
00010001
10010001
01010001
11010001
00110001
10110001
01110001
11110001
00001001
10001001
01001001
11001001
00101001
10101001
01101001
11101001
00011001
10011001
01011001
11011001
00111001
10111001
01111001
11111001
00000101
10000101
01000101
11000101
00100101
10100101
01100101
11100101
00010101
10010101
01010101
11010101
00110101
10110101
01110101
11110101
00001101
10001101
01001101
11001101
00101101
10101101
01101101
11101101
00011101
10011101
01011101
11011101
00111101
10111101
01111101
11111101
00000011
10000011
01000011
11000011
00100011
10100011
01100011
11100011
00010011
10010011
01010011
11010011
00110011
10110011
01110011
11110011
00001011
10001011
01001011
11001011
00101011
10101011
01101011
11101011
00011011
10011011
01011011
11011011
00111011
10111011
01111011
11111011
00000111
10000111
01000111
11000111
00100111
10100111
01100111
11100111
00010111
10010111
01010111
11010111
00110111
10110111
01110111
11110111
00001111
10001111
01001111
11001111
00101111
10101111
01101111
11101111
00011111
10011111
01011111
11011111
00111111
10111111
01111111
11111111

{
00100100,
10100100,
01100100,
11100100,
00010100,
10010100,
01010100,
11010100
}
where the form out of the range of <0,1>^3 is 11011011 , in this case (and again, the set of all powersets does not exist).
 
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zooterkin said:
Stick to the 3 element case, please. Where is the incompleteness? Which set is missing from the powerset?
Click to expand...
It is a 8 elements case because P(S)=<0,1>^3 = 23, so in order to see the missing diagonal we need 8 symbols for each distinct form of the 8 elements case.
 
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zooterkin said:
Irrelevant. Please show the missing set from the 3 element case.
Click to expand...
It is relevant, you still do not understand the diagonal method and how <0,1>^k form is used for both S and P(S) (where the set of all powersets does not exist).
 
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doronshadmi said:
I know, this is a proof of the incompleteness of any given set, whether it is finite or not.



{
10000000... ↔ 1
01000000... ↔ 2
00100000... ↔ 3
00010000... ↔ 4
00001000... ↔ 5
00000100... ↔ 6
00000010... ↔ 7
00000000... ↔ 8
...
}
Now we have 00000001 that is not in the set etc ... ad infinitum ... , so?
Click to expand...
It just tells you that your "proof" is nonsense. That's all.
 
doronshadmi said:
I know, this is a proof of the incompleteness of any given set, whether it is finite or not.



{
10000000... ↔ 1
01000000... ↔ 2
00100000... ↔ 3
00010000... ↔ 4
00001000... ↔ 5
00000100... ↔ 6
00000010... ↔ 7
00000000... ↔ 8
...
}
Now we have 00000001 that is not in the set etc ... ad infinitum ... , so?
Click to expand...
That's another demonstration of you not being able to intercept inconsistency.

Look again at what I concluded with

A.

1000000... ↔ 1
0100000... ↔ 2
0010000... ↔ 3
0001000... ↔ 4
0000100... ↔ 5
0000010... ↔ 6
0000001... ↔ 7

The reverse diagonal 0000000 is not included in the set and {0000000} = 8.

But tha's hardly possible, coz 8 = {00000001} when the vectors of the matrix are being extended by one element.

B.

10000000... ↔ 1
01000000... ↔ 2
00100000... ↔ 3
00010000... ↔ 4
00001000... ↔ 5
00000100... ↔ 6
00000010... ↔ 7
00000001... ↔ 8

Which natural number is {0000000} assigned to in the case A., Doron? It can't be 8.

Obviously, {0000000} cannot be assigned to a natural number. But at the same time all those binary vectors are assigned to the natural numbers.

There is an argument called "proof by contradiction." It's a very handy tool. Euclid used it as much as he could; Cantor too. So what the proof by contradiction has to say to the above?
 
epix said:
Which natural number is {0000000} assigned to in the case A., Doron? It can't be 8.
Click to expand...

It grows like this:

1000000... ↔ 1
0100000... ↔ 2
0010000... ↔ 3
0001000... ↔ 4
0000100... ↔ 5
0000010... ↔ 6
0000001... ↔ 7

Now we add the reverse diagonal 0000000 to the list and add an extra 0 at the right side of each <0,1> form in order to deal with 8 elements, as follows:

10000000... ↔ 1
01000000... ↔ 2
00100000... ↔ 3
00010000... ↔ 4
00001000... ↔ 5
00000100... ↔ 6
00000010... ↔ 7
00000000... ↔ 8

Now we add the reverse diagonal 00000001 to the list and add an extra 0 at the right side of each <0,1> form in order to deal with 9 elements, as follows:

100000000... ↔ 1
010000000... ↔ 2
001000000... ↔ 3
000100000... ↔ 4
000010000... ↔ 5
000001000... ↔ 6
000000100... ↔ 7
000000000... ↔ 8
000000010... ↔ 9

Now we add the reverse diagonal 000000011 to the list and add an extra 0 at the right side of each <0,1> form in order to deal with 10 elements, etc. ... ad infinitum.
 
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