Deeper than primes

[zooterkin]

The 3 element case is S and not P(S), and it is already shown in http://www.internationalskeptics.com/forums/showpost.php?p=6795906&postcount=14013 .

EDIT:

000... ↔ { }
110... ↔ {x,y}
100... ↔ {x}
101... ↔ {x,z}
010... ↔ {y}
011... ↔ {y,z}
001... ↔ {z}
...

where the diagonal's first 3 symbols, in this case, is 101...

First, what are the '...' supposed to indicate?

Second, what's the significance of 101, when it's the next element in the power set in the list?

Third, why have you left the list incomplete by one item?

Fourth, why have you chosen such a peculiar order?

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

So, which one is missing? I don't see one.

 

[epix]

Fourth, why have you chosen such a peculiar order?

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

That seems to be natural way to assign numbers to the subsets, but Doron has its own mysterious ways, as seen here, where he wreaks havoc with orderly developing diagonal.
http://www.internationalskeptics.com/forums/showpost.php?p=6797645&postcount=14040

 

[jsfisher]

Just how does that make it incomplete? You keep offering an element not in a set as proof the set is incomplete. That isn't what incomplete means. You really, really should have asked for good dictionary for Hanukkah instead of those socks.

Incompleteness, in this case, is the ability to show that some collection is a part of a bigger collection.

We don't need Doronetics to show that any collection is an element of some bigger collection, and we certainly didn't need to corrupt the term incompleteness to name it.

Congratulations, Doron, you have once again elevated the trivial to new heights.

...

(1) You have never demonstrated that your set P(S) is the power set of S, and since it isn't you will be hard-pressed to proof it is, but I welcome you to try.

Wrong.
{
0000000... ↔ { },
1100000... ↔ {x,y},
1000000... ↔ {x},
1010000... ↔ {x,z},
0100000... ↔ {y},
0110000... ↔ {y,z},
0010000... ↔ {z},
...
}
is the powerset of {x,y,z,...}

Prove it. Saying it is so is not proof.

...(where there are infinitely many distinct symbols like "x","y" or "z" for example: "@", "%" , "&", "*", etc... ad infinitum), and the diagonal that is not in the range of P(S) starts by 1011111..., in this case.

Well, look at that. That would be an informal proof that this Doron special P(S) is not the power set of S. Well done, Doron. You continue to provide your own contradictory claims.

 

[jsfisher]

I know, this is a proof of the incompleteness of any given set, whether it is finite or not.

Well, now that it has been revealed that by "incompleteness" you mean "member of another set", doesn't the Axiom of Union already cover this.

Doron, not only should you give up trying to disprove definitions, you needn't bother trying to proof axioms, either.

 

[epix]

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.

Doron, just think about this for a sec. The purpose of the diagonal argument is to show that some infinite sets contain members that cannot be put in 1-to-1 correspondence with natural numbers.

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

The reverse diagonal 0000000 indicates that the set of natural/counting numbers from 1 to 7 is not complete. But it is complete. If you extend the set of natural numbers so there is no bound, the reverse diagonal 000... would indicate that there is a member in that set that has not been accounted for. But that could be hardly true.

 

[doronshadmi]

We don't need Doronetics to show that any collection is an element of some bigger collection, and we certainly didn't need to corrupt the term incompleteness to name it.

Congratulations, Doron, you have once again elevated the trivial to new heights.



Prove it. Saying it is so is not proof.



Well, look at that. That would be an informal proof that this Doron special P(S) is not the power set of S. Well done, Doron. You continue to provide your own contradictory claims.

You continue to talk nonsense.

For example, the power set of N ( notated as P(N) ) includes {},{1,2,3,...} and any finite or infinite subset between {} and {1,2,3,...}

It is obvious that the <0,1> form is equivalent to any given N subset (whether it is finite or not) , for example:

10101010... ↔ odd numbers
10000000... ↔ {1}
01010101... ↔ even numbers
11000000... ↔ {1,2}
11111111... ↔ N numbers

etc ...

My novel use of the diagonal method is equivalent to the members of P(N),

{
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 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.

 

[zooterkin]

You continue to talk nonsense.

What's that popping sound? Let me see, I think it's coming from the cupboard in the corner. Ah, it's where we keep our supplies of irony meters; I think it's time to order another batch...

 

[doronshadmi]

Doron, just think about this for a sec. The purpose of the diagonal argument is to show that some infinite sets contain members that cannot be put in 1-to-1 correspondence with natural numbers.

Wrong.

Please see http://www.internationalskeptics.com/forums/showpost.php?p=6800369&postcount=14046 .

 

[doronshadmi]

Let me see

Nobody, except yourself, prevents from you to see.

Until this very moment you have no clue what a 3 element case is ( as clealry shown in http://www.internationalskeptics.com/forums/showpost.php?p=6797324&postcount=14030 ).

 

[doronshadmi]

Doron, just think about this for a sec. The purpose of the diagonal argument is to show that some infinite sets contain members that cannot be put in 1-to-1 correspondence with natural numbers.

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

The reverse diagonal 0000000 indicates that the set of natural/counting numbers from 1 to 7 is not complete. But it is complete.

No, it is not complete because 0000000 ↔ 8 (where 0000000 is some <0,1> form) is not in the list. This is the beauty of the diagonal method, you can use it whether the list is finite or not.

 

[zooterkin]

No, it is not complete because 0000000 ↔ 8 (where 0000000 is some <0,1> form) is not in the list. This is the beauty of the diagonal method, you can use it whether the list is finite or not.

Then please use it for the 3 element ({x,y,z}) case.

 

[doronshadmi]

Then please use it for the 3 element ({x,y,z}) case.

What is a 3 element case for you? please explain in details.

 

[doronshadmi]

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.

 

[zooterkin]

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.

Oh, my.

You've listed all the members of the power set, and then you say one is missing because of the arbitrary way you've chosen to number them?

Where does the 'set of all powersets' come in, anyway? That's not what was being discussed, only the powerset of {x,y,z}.

 

[doronshadmi]

Oh, my.

You've listed all the members of the power set, and then you say one is missing because of the arbitrary way you've chosen to number them?

Where does the 'set of all powersets' come in, anyway? That's not what was being discussed, only the powerset of {x,y,z}.

What is discussed is the diagonal method, an by using it on any set (finite or not, powerset or not) we define a <0,1> distinct form that is not in the range of the original list of <0,1> distinct forms.

You simply don't get this beauty, and how it is a rigorous proof for the incompleteness of any given collection of distinct forms (whether the collection of finite or not).

 

[zooterkin]

What is discussed is the diagonal method, an by using it on any set (finite or not, powerset or not) we define a <0,1> distinct form that is not in the range of the original list of <0,1> distinct forms.

You simply don't get this beauty, and how it is a rigorous proof for the incompleteness of any given collection of distinct forms (whether the collection of finite or not).

If the powerset is incomplete, please identify which element of it is missing, or else explain, in English, what you mean by 'incomplete'?

 

[doronshadmi]

If the powerset is incomplete, please identify which element of it is missing, or else explain, in English, what you mean by 'incomplete'?

You've listed all the members of the power set, and then you say one is missing because of the arbitrary way you've chosen to number them?

The order is insignificant, given a list of distinct <0,1> forms (finite or not, powerset or not) there is always a distinct <0,1> form (the "reverse" of the diagonal <0,1> form) that is not in the range of the original list, and therefore the original list is incomplete.

 

[zooterkin]

The order is insignificant, given a list of distinct <0,1> forms (finite or not, powerset or not) there is always a distinct <0,1> form (the "reverse" of the diagonal <0,1> form) that is not in the range of the original list, and therefore the original list is incomplete.

Incomplete in what sense? No element of the powerset is missing.

 

[jsfisher]

You continue to talk nonsense.

Well, since I've simply been feeding back things you say, you'd know best, wouldn't you.

For example, the power set of N ( notated as P(N) ) includes {},{1,2,3,...} and any finite or infinite subset between {} and {1,2,3,...}

Well, if we forgive the excursion into gibberish there near the end, ok. Yes, the power set of the set of natural numbers includes all the subsets of the set of natural numbers.

It is obvious that...

Many things seem obvious, but that seldom constitutes proof.

...the <0,1> form is equivalent to any given N subset (whether it is finite or not) , for example:

10101010... ↔ odd numbers
10000000... ↔ {1}
01010101... ↔ even numbers
11000000... ↔ {1,2}
11111111... ↔ N numbers

etc ...

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.

My novel use of the diagonal method is equivalent to the members of P(N),

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

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. You didn't do that. Without that correspondence, you cannot reach your bogus result. You need to be able to show the element generated from the diagonal of the list isn't in the list, and you do that by showing it cannot be in any position n because of the disagreement at the n-th bit. No correspondence, so your assertion fails.

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.

Ignoring your unjustified redefinition of incomplete, you have proven neither. You assertion fails for any infinite set.

 

[epix]

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.

The above set is finite. So why did you chose to bold the left-right diagonal and not the right-left one?

Doron, is it true that if S = {}, then T = P(P(P...P(S)))... = {{}, {}, {}, {}...} ?