Deeper than primes

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doronshadmi said:
No, jsfisher. You have simply ignored generalization because you can't rid of sub sets.

By generalization, both S and P(S) have the same form (no sub sets are used), and all we care is the size of S and P(S).
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No, that is not generalization. That is abandonment. To be a generalization of sets and power sets, you'd be covering a broader area, but you'd still be covering sets and power sets. Since what you claim to be a power set isn't, your so-called generalization isn't.

Once again, Doron, you struggle to disprove a definition. It does not work that way.
 
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.

This notion enables us to understand that completeness is an essential requirement of the consistency of Emptiness and Fullness.
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If incompleteness relates to non-empty collection, then the opposite -- completeness -- must relate to empty collection and Emptiness and . . . Fullness? That doesn't seem to be consistent conclusion, unless "empty" and "fullness" are listed as tight synonyms in the Thesaurus of DoroneticsOM.

This works like a red flag: When a neural system fails to correctly conclude a simple comparison based on opposites, then it means that there is a cool argument hidden in the anchor terms.

Too bad, that you are unimaginably close-minded person, otherwise you would see Emptiness and Fullness in action.
 
epix said:
Too bad, that you are unimaginably close-minded person, otherwise you would see Emptiness and Fullness in action.
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Too bad, that you are unimaginably close-minded person, otherwise you would see Emptiness and Fullness as totalities, where action is done at the intermediate level of collections that their consistency is exactly their non-total and therefore incomplete nature, which is used as the level of ever developed framework.
 
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jsfisher said:
To be a generalization of sets and power sets, you'd be covering a broader area, but you'd still be covering sets and power sets.
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No, I show the consistency of collections by discover the built-in (essential) incompleteness of the concept of collection.

It is done by using a common from for both sets and power sets by which is a broader view of their traditional understanding that is currently based on set\subset concepts.

By this broader area the consistency of collections is exactly their incompleteness.

Jsfisher, http://www.internationalskeptics.com/forums/showpost.php?p=6760345&postcount=13915 is broader than your reasoning about this fine and important subject.
 
doronshadmi said:
No, I show the consistency of collections by discover the built-in (essential) incompleteness of the concept of collection.
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No, what you did was pretend that power set meant something completely different from what it really means, then you strung together some self-inconsistent gibberish, then claimed an illogical conclusion.
 
jsfisher said:
No, what you did was pretend that power set meant something completely different from what it really means, then
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Wrong, by <0,1> both set and its power set are based on a common form, such that no sub sets are used and all we care is the sizes of sets and their power sets, which are based on the same form.

This is a generalization of sets and their power sets, which are not separated anymore by the structure of sub sets.

Your reasoning, which is based on sub sets, can't get this generalization.
 
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doronshadmi said:
Wrong, by <0,1> both set and its power set are based on a common form, such that no sub sets are used and all we care is the sizes of sets and their power sets, which are based on the same form.
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Repeating a false statement doesn't make it true. Stop trying to disprove a definition. Power sets are what they are; you don't get to redefine the term. Moreover, your bogus generalization doesn't show anything close to what you claim, especially since half way through your presentation you change one of the premises.
 
jsfisher said:
especially since half way through your presentation you change one of the premises.
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Please show it.

EDIT:

Also Let us correct some thing:

We start from the empty set {}.

The power set of {} is {0}.

The diagonal number that is not in the range of Nothing is 0, we get {0} as the power set of {},
and also in this case no sub sets are used, and < ,0> is used as a common form for nothing or member 0.

The power set of set {0} is {0,1} (no subsets are used and <0,1> is used as a common form for member 0 or member 1).

Acctually the general common form is <x,y> , where x can be a place holder for Nothing.

The diagonal number (which has a s size of a single symbol out of two possible single symbols) that is not in the range of 0, is 1 (or 0, if 1 is considered as the single symbol).

The power set of set {0,1} is
{
00,
01,
10,
11
}
and the members of S are partial collections of any possible 2 distinct members that have a common <0,1> form with P(S) distinct members, for example:

S =
{
10,
11
}
or
{
00,
10
}

etc ... , where given any S version of 2 members , there is a diagonal object that is based on <0,1> form ( which is common for both S and P(S) ) that is not in the range of S, but it is in the range of P(S).

But also P(S) is a set that has a common <0,1> form with P(P(S)), for example:

0 1 0 1
0 0 1 1
-------
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

and in this case some 4 P(S) objects ( which are partial case of P(P(S)) ) are:

P(S)=
{
0 1 1 0,
1 1 1 0,
0 0 0 1,
1 0 0 1
}
or
{
0 1 0 0,
0 0 1 1,
1 1 1 1,
0 0 1 0
}

etc ... , where given any P(S) version of 4 members , there is a diagonal object that is based on <0,1> form ( which is common for both P(S) and P(P(S)) ) that is not in the range of P(S), but it is in the range of P(P(S)).



The same reasoning works also if S has an infinite size, as follows:

S=
{
.0 1 1 0 ...,
.1 1 1 0 ...,
.0 0 0 1 ...,
.1 0 0 1 ...,
...
}
or
{
.0 1 0 0 ...,
.0 0 1 1 ...,
.1 1 1 1 ...,
.0 0 1 0 ...,
...
}

etc ... are partial cases of P(S) where the diagonal member is not in the range of any S version ( although both S and P(S) have a common <0,1> form ).

By using <0,1> ( or more generally: <x,y> ) as a common form for both sets and powersets (finite or not), we discover that the ZFC axiom of powerset is actually a "Trojan horse" , which defines some object of ZFC that have the properties of a given set (it obeys the construction rules of the given framework) but it is not in the range of the given set (but can't be proved within this framework), exactly as Godel's first incompleteness theorem demonstrates.

-----------------

Again, no subsets of S are used as the members of P(S), because both of them have members of the same form.

Only the size of S and P(S) is different.

Here is again the example of this notion:

The power set of set {0,1} is
{
00,
01,
10,
11
}
and the members of S are partial collections of any possible 2 distinct members that have a common <0,1> form with P(S) distinct members, for example:

S =
{
10,
11
}
or
{
00,
10
}


etc ... , where given any S version of 2 members , there is a diagonal object that is based on <0,1> form ( which is common for both S and P(S) ) that is not in the range of S, but it is in the range of P(S).
 
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doronshadmi said:
(you still can't get rid of subsets in order to get the generalization of S P(S) common <x,y> form, so?)
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Since that's that a power set is, the set of all subsets of a set, I am not the one having trouble with power sets and their generalization.

Please stop trying to disprove a definition, Doron. It is really a waste of time.
 
doronshadmi said:
You are right, it is really a waste of time to talk about this subject with persons that can get S P(S) only in terms of subsets.

It is about time to ignore you about this subject.
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Yeah, what a pain if can't just make stuff up as you go along. Doronetics is so flexible that way. Too bad it is useless for anything real.
 
doronshadmi said:
You are right, it is really a waste of time to talk about this subject with persons that can get S P(S) only in terms of subsets.
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But the power set loses entirely its purpose when you divorce it from the concept of sets and subsets that threw the monkey wrench into the invincible mind of early 19th century Homo sapiens.

Aah, that glorious 19th century! That was the illuminating time that gave birth to atheism -- a reason-based uprising against God -- yet, professors of mathematics still couldn't count right.

What will become the most important invention of the 19th century? Tell us, Heavenly Father.

Well, the 19th letter of the alphabet is letter S. It must be something described by the word that starts with S.

Soup?

I've already invented that.

Hmm... How about Salami?

That's been invented as well, a bit later after Stomach. Besides, I must invent something that the mind could feast on for a change. How about Set?


And so, God the Lord, Ph.D., shook the Tree of Knowledge and The Serpent fell to the ground. His Irreversible Wisdom and Omniscience, God the Lord took The Serpent and cut off its head and tail to bless the 19th century with:
Se-rpen-t.

From this point on, Man could count right.
 
By generalization, S and P(S) sizes are based on <0,1>^k form, where k = 0 to ∞.

By <0,1>^0
P(S)=
{0}
and
S=
(
{}
)

By <0,1>^1
P(S)=
{0,1}
and
S=
(
{0}
or
{1}
)

By <0,1>^2
P(S)=
{00,01,10,11}
and
S=
(
{
10,
11
} → 00
or
{
10,
00
} → 01
or
{
00,
11
} → 10
or
{
01,
10
} → 11
)

etc… ad infinitum … and as can be seen, no subsets are used.
 
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doronshadmi said:
By generalization...
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Nope. It is not a generalization if for no other reason than it does not include as a special case that which it claims to generalize. If anything, it is the opposite of a generalization; it is very restrictive as to what may be considered a set or whatever this P operator thing is supposed to be.

Really, Doron. Try to learn at least one something. Perhaps the meaning of the word, generalization, as it would apply here.
 
jsfisher said:
If anything, it is the opposite of a generalization; it is very restrictive as to what may be considered a set or whatever this P operator thing is supposed to be.
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EDIT:

P(S) is the power set of set S, and <0,1>^k, where k = 0 to ∞, is a general form for both P(S) and S.

The traditional form of P(S) and S has no general form for both P(S) and S because P(S) members are subsets of S (which is very restrictive).

Because of the lack of general form for both S and P(S), the traditional approach of this subject gets the illusionary dichotomy between enumerable and non-enumerable collections of distinct objects.

The general form enables us to understand directly (without any need of 1-to-1 mapping) that any given non-empty collection of distinct objects is consistent exactly because it is incomplete (and actually there are infinitely many incomplete infinities).

By using <0,1>^k as a general form for both sets and powersets (finite or not), we discover that the ZFC axiom of powerset is actually a "Trojan horse" , which defines some object of ZFC that have the properties of a given set (it obeys the construction rules of the given framework) but it is not in the range of the given set (but can't be proved within this framework), exactly as Godel's first incompleteness theorem demonstrates ( please see also http://www.internationalskeptics.com/forums/showpost.php?p=6756356&postcount=13911 ).

This discovery can't be achieved by the traditional approach exactly because there is no general form for both P(S) and S.
 
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doronshadmi said:
The traditional form of P(S) and S has no general form for both P(S) and S because P(S) members are subsets of S (which is very restrictive).
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That P(S) is the set of all subsets of S is not very restrictive, as you say. It is merely the definition of power set.

Why is that so hard for you to accept? Power sets are sets of subsets. You want to generalize it? Fine, but that means the subsets concept has to stay. Cast it aside, and then you aren't talking about power sets any more.

Because of the lack of general form for both S and P(S)...
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What are you on about now? Set theory provides a perfectly rational general from for both sets and their power sets. They are both sets, or had you not noticed?
 
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