Deeper than primes - Continuation

Status
Not open for further replies.
jsfisher said:
Doron makes an incompetent post. Doron attempts to pass gibberish as intelligence. Doron defends it by blaming everyone else.
Click to expand...

That about sums it up.
 
Some traditional mathematician says: :"That would be the form you tried to demonstrate and failed, miserably."

Argument:

Given a non-empty set with a given form, there is a power set of these forms, such that there is an explicit form that is a member of the power set, but it is not a member of the given set.

As a result the given set is incomplete, because not all the given forms are its members.



Proof by construction:

The form is based on 0;1 symbols.

S={0}

P(S)={0,1} where form 1 is an explicit member of P(S) but it is not a member of S.


S={1}

P(S)={0,1} where form 0 is an explicit member of P(S) but it is not a member of S.


By using Diagonalization among the forms that are based on 0;1 symbols, we always are able construct an explicit form that is a member of the power set, but it is not a member of the given set.

S=
{
10,
11
}

P(S)={00,01,10,11} where form 00 is an explicit member of P(S) that is not a member of S.


S=
{
10,
00
}

P(S)={00,01,10,11} where form 01 is an explicit member of P(S) that is not a member of S.


S=
{
00,
11
}

P(S)={00,01,10,11} where form 10 is an explicit member of P(S) that is not a member of S.


S=
{
01,
10
}

P(S)={00,01,10,11} where form 11 is an explicit member of P(S) that is not a member of S.


...


Things do not change even if the 0;1 forms are infinite, as shown by Cantor.

Since the largest P(S) does not exist, we can conclude that no collection with distinct forms is complete.

Q.E.D
 
Last edited:
doronshadmi said:
Argument:

Given a non-empty set with a given form, there is a power set of these forms, such that there is an explicit form that is a member of the power set, but it is not a member of the given set.

As a result the given set is incomplete, because not all the given forms are its members.
Click to expand...

So, a set is Doron-incomplete if it doesn't contain things that are not a member of it? Gee, that makes the Doron-incomplete concept fundamentally useless.

Proof by construction:

The form is based on 0;1 symbols.

S={0}

P(S)={0,1} where form 1 is an explicit member of P(S) but it is not a member of S.
Click to expand...

No. If S = {0} then P(S) = {{}, {0}}. Moreover, neither member of P(S) is a member of S.

Doron, until such time as you stop blending your thoughts and uttering confused gibberish, your nonsense hits a brick wall.

By the way, this ass-backwards excursion into binary numbers adds a level of confusion (for you) and adds no clarity (for us).
 
"No. If S = {0} then P(S) = {{}, {0}}." is a good example of a mathematician that can't get any notion beyond the traditional notation.

Say no more.
 
jsfisher said:
Not much more needs to be said. Doron makes an incompetent post. Doron attempts to pass gibberish as intelligence. Doron defends it by blaming everyone else.
Click to expand...

Q.E.D.
 
doronshadmi said:
zooterkin said:
And what if

S=
{
00,
01,
10,
11
}


? What element is missing then?
Click to expand...
First please construct its power set, in order to explicitly construct the missing member.
Click to expand...

Ok, I see that you still do not get the generality of 0;1 form.

Please be aware that in this particular case we are using here distinct members under powersets with cardinality ...2^(2^(2^(2^1)))... and the number of 0;1 symbols must grow in order to save their distinction w.r.t each other.

By using it, the powerset of S with cardinality 4 is:
Code: 
0 0 1 1
0,1,0,1
-------
P(S)=
{
0 0 0 0,
1 0 0 0,
0 1 0 0,
1 1 0 0,
0 0 1 0,
1 0 1 0,
0 1 1 0,
1 1 1 0,
0 0 0 1,
1 0 0 1,
0 1 0 1,
1 1 0 1,
0 0 1 1,
1 0 1 1,
0 1 1 1,
1 1 1 1,
}

such that each member is made of 4 0;1 symbols and the cardinality of S is 4, where the cardinality of P(S) is 2^4=16

Now please use Diagonalization among the 4 forms that are based on 0;1 symbols as follows:

S=
{
1001,
0111,
0111,
0001,
}

such that 0000 is an explicit member of P(S) that is not a member of S, etc., as shown in http://www.internationalskeptics.com/forums/showpost.php?p=8144371&postcount=942.

In other words, even if the cardinality of P(S) 2^∞, there is S with ∞ members and both S AND P(S) have members with 0;1 form, where each member has ∞ symbols.

Since (P(S) and S have the same construction of forms, but there are members of this construction in P(S), but not in S) AND (the largest P(S) does not exist) we can conclude that no collection with distinct members is complete (whether the cardinality > 0 is finite or not).
 
Last edited:
The traditional mathematician here does not understand yet the result of the general use of 0;1 from among cardinality 2^k(where k = 1 to ∞) whether the cardinality grows by ...2^(2^(2^(2^1)))... or not, where by this general use S and P(S) have forms with the same number of 0;1 symbols.

The reason: he can't get any notion beyond the traditional notation.

<snip> is its best.
 
Last edited:
doronshadmi said:
The traditional mathematician here does not understand yet the result of the general use of 0;1 from among cardinality 2^k(where k = 1 to ∞) whether the cardinality grows by ...2^(2^(2^(2^1)))... or not, where by this general use S and P(S) have forms with the same number of 0;1 symbols.

The reason: he can't get any notion beyond the traditional notation.

<snip> is its best.
Click to expand...

The ''traditional'' notation works. Your made-up notation does not work.
 
jsfisher said:
Doron, why does it surprise you so that a set of 16 elements will have some elements that do not appear in a set of only 4?
Click to expand...

Doron, in your dismissive fit you neglected to address the question posed to you. I've quoted it above in full for your convenience.

In due time, too, we can explore your unfounded belief that 2^|S| behaves as you think for non-finite sets.
 
Sorry, haven't tuned in in quite a while. Just thought I ask: hey Doron, anything published yet? Any results at all? Anything? Or still just wanking on internet forums over the attention some bored folks give you there?
 
He had some sidekicks, then he managed to put them off as well. Now he is with the old stick of:

- If you get it, you get it.
- If you don't get it, you don't get it.
- Only Doron gets it.
 
Let's simplify the argument.

Given a collection of distinct elements that are constructed by 0;1 symbols, such that the size of each element is the same as the size of the collection, there is always an explicit element of 0;1 symbols that is constructed by using Diagonalization, which is not an element of the given collection, no matter if the considered size is finite or infinite.

By following this fact, we can conclude that no such collection of distinct elements is complete.
 
Last edited:
doronshadmi said:
Let's simplify the argument.

Given a collection of distinct elements that are constructed by 0;1 symbols, such that the size of each element is the same as the size of the collection, there is always an explicit element of 0;1 symbols that is constructed by using Diagonalization, which is not an element of the given collection, no matter if the considered size is finite or infinite.

By following this fact, we can conclude that no such collection of distinct elements is complete.
Click to expand...

So what? I think that Cantor mentioned that.
 
doronshadmi said:
Let's simplify the argument....
Click to expand...

No need for all that binary nonsense nor pretending things to be power sets when they really aren't. It is just a very simple idea you seem hell-bent on presenting in the most convoluted manner possible. The question remains, though: Why does it surprise you so that a set of 16 elements will have some elements that do not appear in a set of only 4?
 
Status
Not open for further replies.

Share this page

ISF - Join now!

Every member here is approved by hand. No bots, no spam, just people who care about evidence and honest debate.

Membership is free!

Create your free account

Back
Top Bottom