Deeper than primes

[doronshadmi]

Doron you still can’t show any member of a set that is not a member of that set, so all sets are, by definition, complete.

Ok, let us do it this way:

Actually, if we are closed under a given set (finite or not) we simply can't show its incompleteness.

In order to show it, we have to show objects that have the same properties of the considered set, but they are not included in the considered set.

This is exactly what I did by translating any set to <0,1> unique (finitely or infinitely) codes.

By show that the objects a given set and its powerset have the same properties, and by using the diagonal method along these objects, I explicitly provide the inverse object of the diagonal object, which has the same properties of the set's objects, but it is not in the range of the considered set.

Conclusion: the considered set is incomplete.

An example:

Here is some collection of P({apple, orange, lemon}) unique objects, that was generalized to <0,1>^|{apple, orange, lemon}| unique objects:

{
000 ↔ {}
001 ↔ {apple}
010 ↔ {orange}
011 ↔ {lemon}
100 ↔ {apple,orange}
101 ↔ {apple,lemon}
110 ↔ {orange,lemon}
111 ↔ {apple, orange, lemon}
}

As can be seen the generalized <0,1>^3 forms have the same structural principle, whether they are used as P(X) or X objects, for example:

P(X)=
{
000,
001,
010,
011,
100,
101,
110,
111
}

X=
(
{
100,
110,
111
} → 000 is not in the range.
or
{
101,
010,
000
} → 001 is not in the range.
Etc…
)

By this generalization X and P(X) objects have the same properties (the same structural principle), and indeed we can explicitly show an object that has the same structural principle of X objects, but it is not in the range of X.

This is a finite example, but by translating any set to <0,1> unique (finitely or infinitely) codes, the incompleteness of X is shown even if it is infinite.

 

[jsfisher]

The the inverse of the diagonal that is explicitly not in the range of {000..., ... ,111...} demonstrates the incompleteness {000..., ... ,111...}, exactly because both (number of <0,1> bits)^2 list) AND (2^(number of <0,1> bits) are based on common construction method under set {000..., ... ,111...}).

Except it doesn't. The "inverse of the diagonal" you constructed is on the list. You seem to have missed that important point.

It works the same way whether the length of the bit strings is finite or any flavor of infinite: There are more elements in the collection than there are bits in each string.

 

[doronshadmi]

There are more elements in the collection than there are bits in each string.

Actually it does not matter, please read http://www.internationalskeptics.com/forums/showpost.php?p=6900668&postcount=14321 .

 

[jsfisher]

1) Each powerset is in turn a set and it also has a powerset, etc... ad infinitum.

Yes. Why do you bring this up? This point is not in contention.

2) The strings of a set and its power set, etc... ad infinitum, have the same structure.

The bit map for each higher level power set is longer than that that preceded it. You could argue they still have the same structure just different scale, but to what end? Tennis balls and planets have the same structure.

3) The set of all powersets does not exist.

Yes, and your point is?

By (1),(2),(3) {000..., ... , 111...} is universal for any collection of unique codes

Ah! Another demonstration of your out-of-context reasoning. Unfortunately for you, your conclusion is not just baseless, it is bogus.

 

[jsfisher]

Actually it does not matter

It matters a great deal. It's the salient point the assures there's nothing missing from the collection.

please read http://www.internationalskeptics.com/forums/showpost.php?p=6900668&postcount=14321

That would be the same inane dribble you have trotted out before. It's still wrong. You start out with X = a set of three specific things, recode the power set of X as a collection of bit maps, then go back and completely change the definition of X. Amidst all the bit map smoke and mirrors, you invent out of the blue a proper subsets of power set, then claim victory because it doesn't include all of the power set members.

This is all trivia, and all it shows is that a set does not include elements it does not include.

 

[zooterkin]

An example:

Here is some collection of P({apple, orange, lemon}) unique objects, that was generalized to <0,1>^|{apple, orange, lemon}| unique objects:

{
000 ↔ {}
001 ↔ {apple}
010 ↔ {orange}
011 ↔ {lemon}
100 ↔ {apple,orange}
101 ↔ {apple,lemon}
110 ↔ {orange,lemon}
111 ↔ {apple, orange, lemon}
}

As can be seen the generalized <0,1>^3 forms have the same structural principle, whether they are used as P(X) or X objects, for example:

P(X)=
{
000,
001,
010,
011,
100,
101,
110,
111
}

X=
(
{
100,
110,
111
} → 000 is not in the range.

Stop right there. Why have you now taken a subset of the power set?

 

[The Man]

Here is some collection of P({apple, orange, lemon}) unique objects, that was generalized to <0,1>^|{apple, orange, lemon}| unique objects:

{
000 ↔ {}
001 ↔ {apple}
010 ↔ {orange}
011 ↔ {lemon}
100 ↔ {apple,orange}
101 ↔ {apple,lemon}
110 ↔ {orange,lemon}
111 ↔ {apple, orange, lemon}
}

As can be seen the generalized <0,1>^3 forms have the same structure, whether they are used as P(X) or X objects, for example:

P(X)=
{
000,
001,
010,
011,
100,
101,
110,
111
}

X=
(
{
100,
110,
111
} → 000
or
{
101,
010,
000
} → 001
Etc…
)

By this generalization X and P(X) objects have the same properties (the same structural principle), P(X) and P(P(X)) objects have the same properties, ... , ...P(P(X))... and ...P(P(P(X)))... objects have the same properties, etc... ad infinitum, such that the inverse of the diagonal over {000..., ..., 111...} is a self evident truth of its incompleteness.

Still lacking the one critical property of membership for those “objects” that even you claim are not in the set. Doron this “same structural principle” as you call it is just your superfluous “<0,1>” “forms”. It is not a property those “objects” nor of the set being considered it is simply an insistence of your superfluous “<0,1>” “forms” notation. They simply take that ‘form’ because you simply insist on putting them in that ‘form’

Some correction:


Simply being <0,1> unique form that is not in the range of the collection (finite or not) of <0,1> unique forms. Again, (number of <0,1> bits)^2 list) AND (2^(number of <0,1> bits) list) are based on common construction of <0,1> unique forms.

So the “properties” you refer to specifically excludes the property of membership, no wonder you can’t show that a set is incomplete.

Also the “properties” you refer to is not a property at all and is simply your superfluous “<0,1> unique forms” notation. Color me unsurprised again.


So what you can’t show is any member of a set that is not a member of that set and all you can show is that if you ascribe one of your “<0,1> unique forms” to each element of some set and some element(s) that are not in that set then all the elements of that set and some element(s) that are not in that set will each have one of your ““<0,1> unique forms” ascribed to it.


You’ve worked on that for how many decades now?

Ok, let us do it this way:

Actually, if we are closed under a given set (finite or not) we simply can't show its incompleteness.

In order to show it, we have to show objects that have the same properties of the considered set, but they are not included in the considered set.

Once again you are specifically claiming that your “objects” do not have the property of membership “in the considered set”. So “this way” is the same as you have asserted all along. “the considered set” does not include (by your own assertion) your “objects”, it is complete without them.

This is exactly what I did by translating any set to <0,1> unique (finitely or infinitely) codes.

By show that the objects a given set and its powerset have the same properties, and by using the diagonal method along these objects, I explicitly provide the inverse object of the diagonal object, which has the same properties of the set's objects, but it is not in the range of the considered set.

Conclusion: the considered set is incomplete.

That isn’t a “Conclusion” Doron it is simply one of your preferred and baseless assertions which you have been struggling to find any way of supporting, but consistently failing. Now you simply assert that a set is incomplete because it does not include what you claim yourself is not included in the set. This simply makes your notion of “complete” meaningless as everything would be incomplete unless it self-contradictorily includes what it does not include. As always Doron you remain the staunchest opponent of just your own notions.

An example:

Here is some collection of P({apple, orange, lemon}) unique objects, that was generalized to <0,1>^|{apple, orange, lemon}| unique objects:

{
000 ↔ {}
001 ↔ {apple}
010 ↔ {orange}
011 ↔ {lemon}
100 ↔ {apple,orange}
101 ↔ {apple,lemon}
110 ↔ {orange,lemon}
111 ↔ {apple, orange, lemon}
}

As can be seen the generalized <0,1>^3 forms have the same structural principle, whether they are used as P(X) or X objects, for example:

P(X)=
{
000,
001,
010,
011,
100,
101,
110,
111
}

X=
(
{
100,
110,
111
} → 000 is not in the range.
or
{
101,
010,
000
} → 001 is not in the range.
Etc…
)

By this generalization X and P(X) objects have the same properties (the same structural principle), and indeed we can explicitly show an object that has the same structural principle of X objects, but it is not in the range of X.

This is a finite example, but by translating any set to <0,1> unique (finitely or infinitely) codes, the incompleteness of X is shown even if it is infinite.

Once again the property your “object” specifically fails to have, even by your own assertions, is membership in “the considered set”. So once again the set remains complete without your “object”.

 

[epix]

You still do not get the cross-contexts universality of set {000..., ... , 111...} which is not related to any particular number system and not to any particular <0,1> object out of the range.

{000..., ... , 111...} is a generalization of any collection of unique objects.

We are not talking about any particular object. By cross-contexts reasoning, it is simply shown that any given collection of unique objects is incomplete exactly because the universal collection of unique <0,1> forms {000..., ..., 111... } always has an unique <0,1> form that is not in the range of {000..., ..., 111... }.

Gee, I wonder if you consulted your mental atrocity with Madame Fullness. That's the chick "that has no successor." Given the incompleteness of "any given collection," she could reach and throw her tampon at you. So be gentle upon disclosing your new discovery to her.

Your invention has been discarded by Homo neanderthalensis as lacking any bit of practicality. Now we assign 1, 2, 3, ... to a collection of distinct objects where completeness or incompleteness of any collection is subordinate to definition or to common agreement. (See the story of planet Pluto.)

Not every set of items can be assigned ordinal numbers though, as Cantor showed. Nothing is missing from the set of real numbers from 0 to 1, but you can't prove it by using natural numbers.
http://www.coopertoons.com/education/diagonal/georgeanddiagonal.jpg

 

[The Man]

This is exactly the reason why you do not understand my answer, because by your "as defined" only context-dependent reasoning, this question does not exist.

Doron I asked the question so it still does exist, well, unanswered.

Again stop simply trying to posit aspects of your own failed reasoning on to others.

The Man, you are not aware of the fact that your reasoning is only context-dependent, and your "as defined" in http://www.internationalskeptics.com/forums/showpost.php?p=6893125&postcount=14275 clearly demonstrates it.

Doron you are not aware of the fact that meaning is derived from context, without it you have, and evidently so, no meaning.

Once again you are using "as defined" without the awareness that you are using this term by only context-dependent reasoning.

Doron unlike you, I’m perfectly aware of the context in which I used that phrase.

Again stop simply trying to posit aspects of your own failed reasoning on to others.

This is a typical view of a person that uses only context-dependent reasoning. He will get anything only in terms of this reasoning, by missing again and again and again ... any cross-contexts reasoning, which is universal by nature.

What, so you do think of yourself as the universe and your nonsense “cross-contexts reasoning” is “universal by nature”?

Simply being <0,1> unique form that is not in the range of the collection (finite or not) of <0,1> unique forms. Again, (number of <0,1> bits)^2 list) AND (2^(number of <0,1> bits) are based on common construction of <0,1> unique forms.

That is not a property it is simply your chosen notation. So the property your “<0,1> unique form” chosen notation still lacks is membership in the set.

Your "membership" "in question" is another demonstration of your only context-dependent reasoning, which naturally can't comprehend the universality of cross-contexts reasoning.

“Membership” Doron is also the property “in question” not your nonsense “cross-contexts reasoning” where you just note the set elements in your superfluous “<0,1> unique forms” and beguile yourself into believing that just because you can also note elements that even you claim are not in the set in your superfluous “<0,1> unique forms” that the set must be incomplete.

By understanding the universality of set {000..., ... , 111...}, it is shown (by using the diagonal method) that {000..., ... , 111...} is incomplete.

Once again you simply posit yourself as the universe as it is simply your, evidently deliberately meaningless, nonsense.

Please read also http://www.internationalskeptics.com/forums/showpost.php?p=6900038&postcount=14315 .

I already did.

1) The is no such a thing like a collection of all sets.

Who said “a collection of all sets”, other than you?

2) Your "by definition" is based only on context-dependent reasoning.

Again meaning is derived from context, without it you remain meaningless, but evidently that is simply what you want.

Again stop simply trying to posit aspects of your own failed reasoning on to others.

Unlike you, it is done by using a cross-contexts reasoning.

Yes Doron we are well aware of the fact that your reasoning is mostly meaningless.

Let us generalize it, any given collection of unique objects is incomplete exactly because the universal collection of unique <0,1> forms {000..., ..., 111... } always has an unique <0,1> form that is not in the range of {000..., ..., 111... }.

Where, identify that specific “unique <0,1> form”. By the way that is not ‘generalizing it’ you are claiming that a specific set does not include some element. As jsfisher has tried to point out before if you indentify the specific element you think is not in that set you will find that it is.

We are not talking about any particular object. By cross-contexts reasoning, it is simply shown that any given collection of unique objects is incomplete exactly because the universal collection of unique <0,1> forms {000..., ..., 111... } always has an unique <0,1> form that is not in the range of {000..., ..., 111... }.

You are specifically talking about “an unique <0,1> form that is not in the range of {000..., ..., 111... }”. The reason you don’t want to indentify any particular one is because you can’t. That is why you claim this “By cross-contexts reasoning” which simply asserts that your reasoning is intentionally meaningless.

No, the term all, if related to any collection (finite or not) of at least <0,1> unique forms, is invalid, exactly because by using the diagonal method, we are able to construct an object that is not in the list, such that X and P(X) objects have the same properties, P(X) and P(P(X)) objects have the same properties, P(P(X)) and P(P(P(X))) objects have the same properties, etc... ad infinitum.

So your invariant “inverse form of the diagonal of that collection, which is not in the range of that given collection” does vary? Again you remain the staunchest opponent of just your own notions.

Thank you for your correction, but it does not change the fact that both invariant AND variant properties are involved.

Though specifically not the way you claimed.

Furthermore, given a formula, it is considered as formula as long as it is invariant w.r.t any particular (and therefore variant) I/O of its data.

Could you please try to put that in to English?

Formulas can and do vary Doron, in spite of whatever gibberish you want to spout or whatever deliberately meaningless “cross-contexts reasoning” you want to claim you're using.

Doron, this “cross-contexts” nonsense you’ve been spouting lately may seem to you like it is your escape hatch, but it simply confirms that your assertions have no particular meaning at all. As a result you must find some meaning in some other context to get you what you want when you want it. As opposed to saving you it simply buries you in your own assertion of being deliberately meaningless.

 

[doronshadmi]

You start out with X = a set of three specific things, recode the power set of X as a collection of bit maps, then go back and completely change the definition of X.

All we care about is the number of the distinct objects in a given collection.

Theorem: Given any S collection with X different objects, such that each object is <0,1> form of X bits, S is incomplete.

Proof:

1) Let S be any collection of X different objects, such that each object is <0,1> form of X bits.

2) Each different object of any S and each different object of P(S) is <0,1> form of X bits.

3) According to (2) there are <0,1> forms of X bits each, which are not in the range of any given S.

Conclusion: S is incomplete.

Q.E.D

Furthermore, since the collection of all powersets does not exist, then any given S is incomplete.

 

[jsfisher]

All we care about is the number of the distinct objects in a given collection.

Who is this fictional "we" to which you refer?

Theorem: Given any S collection with X different objects, such that each object is <0,1> form of X bits, S is incomplete.

Define "incomplete". So far, your usage is been that incomplete means the set doesn't contain an element it doesn't contain.

 

[epix]

All we care about is the number of the distinct objects in a given collection.

Theorem: Given any S collection with X different objects, such that each object is <0,1> form of X bits, S is incomplete.

Proof:

1) Let S be any collection of X different objects, such that each object is <0,1> form of X bits.

2) Each different object of any S and each different object of P(S) is <0,1> form of X bits.

3) According to (2) there are <0,1> forms of X bits each, which are not in the range of any given S.

Conclusion: S is incomplete.

The conclusion obviously doesn't include the collection of objects bounded by Emptiness, which has no predecessor, and Fullness, which has no successor. Or did you forget about your friends from your previous voyage into insanity?

 

[epix]

It is complete only if it is not a collection.

Given any collection, if it is translatable to <0,1> unique codes, then it is incomplete.

I haven't majored in art, cardinal, but I suppose that a few Pablo Picasso's sketches might have gone unaccounted for.

Your Holiness, I'm talking about the collection of his canvas paintings. Professor Shadmi proved their quantitative incompleteness by purely mathematical means. This is not about subjective incompleteness generated by the artistic spirit of gifted individuals who are never satisfied with their accomplishments, like Pablo Picasso; this is a purely material notion that carries an implication; namely, Picasso's work is not finished -- it's being expanded!

Cardinal, as I said, I'm not an art major, but I'm socially aware of certain related facts, such as the one regarding Pablo Picasso's death in the Year of Our Lord 1973.

That's correct, Your Holiness, but let's be remindful of a certain place . . .

Oh, do you mean THAT place?

Yes, Your Holiness. THAT place. See, professor Shadmi's Unique Codes is a mathematical proof of Heaven being as real as our world is, and that's why Picasso's collection of canvas paintings is not complete. The additional artwork done by Picasso in Heaven is "not within the range," as professor Shadmi places the ever-growing complements. Nevertheless, there is the mathematical proof of their existence.

I see. But that's wonderful! Say, is that mathematical proof rigorous enough to initiate the steps leading toward the elevation of professor Shadmi into the sainthood?

Yes, Your Holiness, it is. The proof is as tight as virgin p... p... rofessor Shadmi unfortunately teaches Organic Mathematics at the University of Jerusalem.

I don't see that as an obstacle. He can be canonized after converting to Christianity.

 

[doronshadmi]

Looking beyond the Matrix

Who is this fictional "we" to which you refer?

"we" is the common jargon for formal writing style.

Define "incomplete". So far, your usage is been that incomplete means the set doesn't contain an element it doesn't contain.

Definition A:
"Let X be a placeholder for any finite or infinite cardinality."

Definition B::
"Given any S collection of <0,1> X different objects, such that each object has X bits, S is incomplete if there is <0,1> object with X bits, which is not in the range of the given S collection."

Theorem A:
Given any S collection with X different objects, such that each object is <0,1> form of X bits, S is incomplete.

Proof:
1) Let S be any collection of X different objects, such that each object is <0,1> form of X bits (we get a matrix of X^2 bits).

2) Each different object of P(S) is <0,1> form of X bits (we get a list of 2^X different objects, where each object has X bits).

3) According to (1) and (2) there are <0,1> forms of X bits each, which are not in the range of any given S.

Conclusion: S is incomplete.

Q.E.D

Furthermore, since the collection of all sub-collections does not exist, then any given S is incomplete.

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

A solution of GCH ( http://en.wikipedia.org/wiki/Continuum_hypothesis ):

Theorem:
GCH is false.

Proof:
By theorem A, there are collections of different objects of X <0,1> bits each, where K is some cardinality
of these collections such that X < K < 2^X.

Q.E.D

 

[doronshadmi]

The conclusion obviously doesn't include the collection of objects bounded by Emptiness, which has no predecessor, and Fullness, which has no successor. Or did you forget about your friends from your previous voyage into insanity?

Emptiness is weaker than the existence of collections.

Fullness is stronger than the existence of collections.

 

[sympathic]

"we" is the common jargon for formal writing style.




Definition A:
"Let X be a placeholder for any finite or infinite cardinality."

Definition B::
"Given any S collection of <0,1> X different objects, such that each object has X bits, S is incomplete if there is <0,1> object with X bits, which is not in the range of the given S collection."

Theorem A:
Given any S collection with X different objects, such that each object is <0,1> form of X bits, S is incomplete.

Proof:
1) Let S be any collection of X different objects, such that each object is <0,1> form of X bits (we get a matrix of X^2 bits).

2) Each different object of P(S) is <0,1> form of X bits (we get a list of 2^X different objects, where each object has X bits).

3) According to (1) and (2) there are <0,1> forms of X bits each, which are not in the range of any given S.

Conclusion: S is incomplete.

Q.E.D

Furthermore, since the collection of all sub-collections does not exist, then any given S is incomplete.

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

A solution of GCH ( http://en.wikipedia.org/wiki/Continuum_hypothesis ):

Theorem:
GCH is false.

Proof:
By theorem A, there are collections of different objects of X <0,1> bits each, where K is some cardinality
of these collections such that X < K < 2^X.

Q.E.D

Stick to Direct Perception. Formalism is not your turf.

 

[doronshadmi]

Stick to Direct Perception. Formalism is not your turf.

It is a formal Direct Perception now.

 

[sympathic]

It is a formal Direct Perception now.

To quote you: "not even in your dreams".

 

[doronshadmi]

Some corrections:

Definition A:
"Let X be a placeholder for any finite or infinite cardinality."

Theorem A:
"Given any S collection of <0,1> X different objects, such that each object has X bits, S is incomplete if there is <0,1> object with X bits, which is not in the range of the given S collection."

Proof:
1) Let S be any collection of X different objects, such that each object is <0,1> form of X bits (we get a matrix of X^2 bits).

2) Each different object of P(S) is <0,1> form of X bits (we get a list of 2^X different objects, where each object has X bits).

3) According to (1) and (2) there is at least one <0,1> form of X bits, which is not in the range of a given S.

Conclusion: S is incomplete.

Q.E.D

Furthermore, since the collection of all sub-collections does not exist, then any given S is incomplete.

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

A solution of GCH ( http://en.wikipedia.org/wiki/Continuum_hypothesis ):

Theorem:
GCH is false.

Proof:
By theorem A, there are collections of different objects of X <0,1> bits each, where K is some cardinality
of these collections such that X < K < 2^X.

Q.E.D

 

[doronshadmi]

To quote you: "not even in your dreams".

Please support your claim by a detailed reply to http://www.internationalskeptics.com/forums/showpost.php?p=6904202&postcount=14339 .