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Deeper than primes

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As can be seen the same <0,1> also proves the incompleteness of P(P(N)):

{
000000000... ↔ {},
010000000... ↔ {{1}},
001000000... ↔ {{2}},
000100000... ↔ {{3}},
110000000... ↔ {{},{1}},
101000000... ↔ {{},{2}},
101000000... ↔ {{},{3}},
111111110... ↔ {{},{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3}}
...
}

and in this case the complement <0,1> form starts with 10001110...

So by induction (and by the fact that the set of all powersets does not exist) we conclude that ...P(P(P(N)))... is incomplete.
 
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doronshadmi said:
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.
Click to expand...

So, then it should be no problem for you to prove.

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:
Click to expand...

It is very simple, Doron. You have a terrible time expressing complete thoughts. So, anytime you post something, everyone else has to first translate some gibberish then add in all sorts of assumptions to make what you post even begin to make sense.

If you want to apply a diagonal method proof to something, then -- even though you didn't say it -- you must have a 1-to-1 correspondence between list members and the natural numbers. In that case, it is a valid assumption to include that as an unstated part of your gibberish.

Now that you deny the correspondence explicitly, other assumptions now become perfectly reasonable to attempt to make sense of your Doronetic effluence.

If the list is in 1-to-1 correspondence with the natural numbers -- an explicit requirement to use the diagonal proof method -- then an element of the power set can be found that is not in the list. Therefore, the list isn't the power set.

If the list does represent the entire power set, then it cannot be put in 1-to-1 correspondence with the natural numbers. Therefore, the diagonal proof method is inapplicable.

So which is it, doron. You cannot pick and choose from both lists. Either way, your proof, your conclusions, and your logic are all completely bogus.
 
doronshadmi said:
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.
Click to expand...

No, you've got me stumped this time. What on earth did you mean to say instead of denationalization?
 
doronshadmi said:
<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.
Click to expand...
If N really stands for natural numbers then the cardinality of the power set is |P(N)| = 2 א‎0 (2 to the power of aleph null.) How do you express it in your notation?

Btw, the diagonal argument wasn't the proof; it was just a tool that entered the proof that real numbers couldn't be put in 1-to-1 correspondence with natural numbers.
 
jsfisher said:
-- an explicit requirement to use the diagonal proof method --
Click to expand...
No, there is no such an explicit requirement, jsfisher.

The diagonal method works whether the collection of distinct objects is finite or not or a powerset or not.

Distinction is the common concept here, and in the following examples we are using <0,1> as the general form for both P(N) and P(P(N)) as follows:

The power set of N ( notated as P(N) ) includes {},{1,2,3,...} and any finite or infinite subset between {} and {1,2,3,...} and It is obvious that any given N subset (whether it is finite or not) is translatable to <0,1> form, for example:

{
00000000000... ↔ { },
11000000000... ↔ {1,2},
10000000000... ↔ {1},
10101010100... ↔ odd numbers {1,3,5,...},
10100000000... ↔ {1,3},
01010101010... ↔ even numbers {2,4,6,...},
01000000000... ↔ {2},
01100000000... ↔ {2,3},
00100000000... ↔ {3},
11111111110... ↔ N numbers {1,2,3,...},
...
}

where the form that is not in the range of P(N) starts with 1011101110..., in this case.

The same <0,1> also proves the incompleteness of P(P(N)):

{
0000000000... ↔ {},
1000000000... ↔ {{}},
0100000000... ↔ {{1}},
0010000000... ↔ {{2}},
0001000000... ↔ {{3}},
1100000000... ↔ {{},{1}},
1010000000... ↔ {{},{2}},
1001000000... ↔ {{},{3}},
1111111100... ↔ {{},{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3}},
...
}

and in this case the complement <0,1> form starts with 111111111...

So by induction (and by the fact that the set of all powersets does not exist) we conclude that ...P(P(P(N)))... is incomplete.
 
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epix said:
Btw, the diagonal argument wasn't the proof; it was just a tool that entered the proof that real numbers couldn't be put in 1-to-1 correspondence with natural numbers.
Click to expand...
No epix, the diagonal argument is exactly a proof that any collection of finite or infinite distinct objects (whether the collection is powerset or not) is incomplete, because by using it one explicitly defines an object that has the same form of the distinct objects of a given collection, but it is not in the range of the given collection.
 
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Furthermore, by translate the natural numbers to <0,1> form we actually prove that there is no 1-to-1 correspondence between the natural numbers and some proper subset of the natural numbers.

For example, let us use the 1-to-1 mapping between the natural numbers and the proper subset of even numbers, by using the <0,1> form:

Code: 
1 1 1 1 1 1 1 1 1 ... ↔ {1,2,3,4,5,6,7,8,9,...}
↕ ↕ ↕ ↕ ↕ ↕ ↕ ↕ ↕  
0 1 0 1 0 1 0 1 0 ... ↔ {2,4,6,8,...}

As can clearly be seen, there is no 1-to-1 correspondence between the natural numbers and the even numbers.


As for the mapping between odd and even numbers, both of them have the same 1-to-0 mapping,
Code: 
1 0 1 0 1 0 1 0 ... ↔ {1,3,5,7,...}
↕ ↕ ↕ ↕ ↕ ↕ ↕ ↕   
0 1 0 1 0 1 0 1 ... ↔ {2,4,6,8,...}

such that they can be considered as collections of the same magnitude.
 
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doronshadmi said:
No, there is no such an explicit requirement, jsfisher.

The diagonal method works whether the collection of distinct objects is finite or not or a powerset or not.
Click to expand...

Nope, but feel free to prove your bogus assertion.

Heck, for that matter, just prove that the element generated by the diagonal method isn't in the list. No fair indexing the list, though, since you have declared it unindexed.
 
Enjoy the incompleteness of Q :

FareyPoincareMovie.gif


http://www-bcf.usc.edu/~fbonahon/Images/FareyPoincareMovie.gif
 
doronshadmi said:
As can clearly be seen, there is no 1-to-1 correspondence between the natural numbers and the even numbers.
Click to expand...

OMG!!! Arithmetic has suddenly stopped working. The natural correspondence between the natural numbers and the positive even numbers (i.e. f(i) = 2i) is now broken. What will we do?? What will we do?!!

Doron, please save us!
 
jsfisher said:
OMG!!! Arithmetic has suddenly stopped working. The natural correspondence between the natural numbers and the positive even numbers (i.e. f(i) = 2i) is now broken. What will we do?? What will we do?!!

Doron, please save us!
Click to expand...
Save yourself from the illusion of completeness at the level of Collections.
 
doronshadmi said:
No epix, the diagonal argument is exactly a proof that any collection of finite or infinite distinct objects (whether the collection is powerset or not) is incomplete, because by using it one explicitly defines an object that has the same form of the distinct objects of a given collection, but it is not in the range of the given collection.
Click to expand...


istockphoto_1415913-three-apples-in-a-basket.jpg


There are three apples in the basket. Which one is missing?


(Clue: Try your diagonal method tailor-made for the purpose.)
 
doronshadmi said:
No epix, the diagonal argument is exactly a proof that any collection of finite or infinite distinct objects (whether the collection is powerset or not) is incomplete, because by using it one explicitly defines an object that has the same form of the distinct objects of a given collection, but it is not in the range of the given collection.
Click to expand...

Mammals exist in two categories: the male and the female category.

Example:

11589_port_man_woman_520.jpg


According to your version of the diagonal argument, the two categories are not complete. Which category is missing?

Enjoy the incompleteness of Q -- until the letter "I" shows up. But atheists say there are no miracles.
 
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epix said:
Mammals exist in two categories: the male and the female category.

Example:

[qimg]http://www.archives.gov.on.ca/english/on-line-exhibits/connon/pics/11589_port_man_woman_520.jpg[/qimg]

According to your version of the diagonal argument, the two categories are not complete. Which category is missing?

Enjoy the incompleteness of Q -- until the letter "I" shows up. But atheists say there are no miracles.
Click to expand...
Where is the child?
 
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