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Deeper than primes

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doronshadmi said:
By this particular order

00000000 ↔ { }
00100000 ↔ {z}
01000000 ↔ {y}
01100000 ↔ {y,z}
10000000 ↔ {x}
10100000 ↔ {x,z}
11000000 ↔ {x,y}
11100000 ↔ {x,y,z}

one of the elements that has <0,1> form, which is not in the range of the list above, is 11111111 which belongs to the list in http://www.internationalskeptics.com/forums/showpost.php?p=6797324&postcount=14030, where also that list is incomplete exactly because the set of all powersets does not exist.
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I want to ask about the above finite power set. Each subset, such as {z}, for example, corresponds with a binary string that has 8 elements. The reverse diagonal has 8 elements as well. It follows that the reverse diagonal 11111111 corresponds with some subset of the set that is not listed. I'm curious about which subset is missing. If |S|=3, then |P(S)|=23=8. It seems to me that all subsets including the empty set have been accounted for, but 11111111 indicates otherwise.
 
jsfisher said:
No. You did that wrong. In order to apply a diagonal method proof, you need to put the elements of your list into 1-to-1 correspondence with the natural numbers.
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Wrong.

This proof is perfectly done by using <0,1> form with the members of P(N), and also in this case there is <0,1> form that is not in the range of P(N).

So the right conclusion of the diagonal method is the incompleteness of any given collection of distinct forms, whether they are finite or not, powersets or not.

Your limited reasoning simply can't comprehend the generalization of <0,1> form, and how it is used as a rigorous proof of the incompleteness of any collection of distinct objects.

The notion of enumerable\non-enumerable collections is based on essential misunderstanding of the diagonal method, exactly because the <0,1> form was not used by Traditional Math, in this fine case.
 
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epix said:
The above set is finite. So why did you chose to bold the left-right diagonal and not the right-left one?
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You can choose the right-left one, but it does not change the fact that the complement <0,1> form is not in the list.
 
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jsfisher said:
Ok, let's accept that as proven, even though you cannot. I'll give you that this bit map representation for sets is suitable for all finite and countably infinite sets. Anything beyond that, no.
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Wrong jsfisher, your reasoning simply can't comprehend the generalization of <0,1> form and the use of the diagonal method on this form.

<0,1> form is in 1-to-1 correspondence with P(N) members, as follows:

{
0000000000... ↔ { },
1100000000... ↔ {1,2},
1000000000... ↔ {1},
1010101010... ↔ odd numbers {1,3,5,...}
1010000000... ↔ {1,3},
0101010101... ↔ even numbers {2,4,6,...}
0100000000... ↔ {2},
0110000000... ↔ {2,3},
0010000000... ↔ {3},
1111111111... ↔ N numbers {1,2,3,...}
...
}

where (by using the diagonal method on <0,1> form) the form that is not in the range of P(N) starts with 1011101110..., in this case.

In other words, we have proved that P(N) is incomplete and since the set of all powersets does not exist, this proof holds for any given powerset.
 
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epix said:
I'm curious about which subset is missing.
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It is not about a missing subset but it is about the ability to use the subsets (that were translated to <0,1> form) in order to explicitly define another <0,1> form that is not in the range of the original list of <0,1> distinct forms.

This general fact is not changed even if the order or the amount (finite or not) of the <0,1> distinct forms is changed.

In other words, the incompleteness of any given collection of distinct objects is universal.
 
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doronshadmi said:
It is not about a missing subset but it is about the ability to use the subsets (that were translated to <0,1> form) in order to explicitly define another <0,1> form that is not in the range of the original list of <0,1> distinct forms.
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And of what use is that, exactly? In what way does it mean the powerset is incomplete?

This general fact is not changed even if the order or the amount (finite or not) of the <0,1> distinct forms is changed.

In other words, the incompleteness of any given collection of distinct objects is universal.
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So, you've defined something meaningless and assigned it a misleading name?
 
zooterkin said:
And of what use is that, exactly?
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Please pay attention to the fact that the complement that is not in the range of the given thing, is the ability of any given thing to be developed beyond its current boundaries. Furthermore, this ability is universal.

zooterkin said:
So, you've defined something meaningless and assigned it a misleading name?
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Since when the ability of everlasting development is meaningless?
 
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doronshadmi said:
Please pay attention to the fact that the complement that is not in the range of the given thing, is the ability of any given thing to be developed beyond its current boundaries. Furthermore, this ability is universal.
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I didn't order word salad, I asked what use your latest 'discovery' was.
Since when the ability of everlasting development is meaningless?
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QFT. :rolleyes:
 
doronshadmi said:
...
where (by using the diagonal method on <0,1> form) the form that is not in the range of P(N) starts with 1011101110..., in this case.
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Great! Now all you have to do is prove 1011101110... doesn't appear in that list. Go ahead. We'll wait.
 
zooterkin said:
I didn't order word salad, I asked what use your latest 'discovery' was.
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Tsk! Have you learned nothing? it can be applied wherever Doron's version of Organic Mathstm is useful, though you may need Direct Perceptiontm to see how it applies... :D
 
doronshadmi said:
In other words, we have proved that P(N) is incomplete and since the set of all powersets does not exist, this proof holds for any given powerset.
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Ignoring for a moment that you haven't proven anything except your own inability to prove things, and that the lack of a set of all power sets is irrelevant to the discussion, just how did you jump to that bogus conclusion?
 
jsfisher said:
Great! Now all you have to do is prove 1011101110... doesn't appear in that list. Go ahead. We'll wait.
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This is the principle of being the complement of a given denationalization, the complement <0,1> form is different by at least one bit from any given form in the list.

Actually you had no problem to accept that the complement <0,1> form is not in the list, when you used it in order to conclude that no N member is in 1-to-1 correspondence with the complement form.

Now, when I use the same denationalization method in order to prove that P(N) is incomplete you suddenly ask about a proof that the complement <0,1> form is not in the list.

Why is that jsfisher? :rolleyes:
 
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jsfisher said:
Ignoring for a moment that you haven't proven anything except your own inability to prove things,
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Ignoring for a moment that you haven't got the novel use of the denationalization, which proves the incompleteness of any given collection of distinct forms, what else is left to discuss with you on this fine subject?
 
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