Deeper than primes

[jsfisher]

The axiom of an infinite collection:

C is an infinite collection if for all x in C, x can't be picked twice if all x in C are picked.

That is not an axiom by any stretch of the imagination. Please stop trying to say it is.

It tries to be a definition, but it fails. It looks to be circular, but you'd need to define what you mean by the verb, to pick, first, before we can be certain.

As it stands, though, the set of integers would qualify as finite for the same reason I gave before.

 

[doronshadmi]

That is not an axiom by any stretch of the imagination. Please stop trying to say it is.

It tries to be a definition, but it fails. It looks to be circular, but you'd need to define what you mean by the verb, to pick, first, before we can be certain.

As it stands, though, the set of integers would qualify as finite for the same reason I gave before.

Edit:

It is under construction.

The axiom of infinite set:

For all x in C, if all x in C are picked AND no x can be picked twice, then C is infinite.

There is no "first" and no "arbitrary" in this version.

"Picked" is "chosen by distinction".

 

[jsfisher]

You are right if "Order has no significance" is the same as "Order does not exist".

"Order has no significance" is your pet phrase, not ours. Unordered sets are, well, unordered. No ordering exists for such a set. Make of that what you will.

In that case there is a non-empty collection (of distinct members, in this case) but non of its members can be picked and used, because any attempt to pick something from this collection must be the first pick.

You really need to define what you mean by this picking thing. However, selecting an arbitrary member of a set (is that what you mean by "picking") does not require the set have any ordering. You selecting a member, no big deal. The set remains unchanged.

Furthermore, how can one define a collection of distinct members by not being able to get them both particularly (locally) and globally (non-locally)?

"Distinct members" -- there's another of your pet phrases. Sets have no such requirement. {A, A, B, A, B, B, B} is a perfectly fine set. It happens to be the same set as {B, A}, but so what?

In other words, if order does not exist among sets, then their members are not distinct.

Your premises were invalid, but your conclusion still did not follow from them. If A is an element of some set, S, and B is an element of S, it is possible for A to be the same as B.

Since this is not the case, then "Order has no significance among sets" is the right one.

Nope. Unordered.

 

[jsfisher]

Edit:

It is under construction.

Construct all you like. It is still not an axiom.

The axiom of an infinite collection:

For all x in C, if all x in C are picked AND no x can be picked twice, then C is infinite.

There is no "first" and no "arbitrary" in this version.

"Picked" is "chosen by distinction".

Ah! The much heralded Doronetics word shift, in which the need to define one word or phrase is side-stepped by jumping to a different word or phrase in need of definition.

"Chosen by distinction" means what, exactly?

And are we heading towards your definition involving a sequence of steps in one-to-one correspondence with the positive integers? That will complete the circle.

 

[doronshadmi]

{A, A, B, A, B, B, B} is a perfectly fine set. It happens to be the same set as {B, A}, but so what?

No. You reduced redundancy in order to get {A,B} out of {A, A, B, A, B, B, B}.

If you do not do that then {A, A, B, A, B, B, B} is a multiset and {A,B} is a set.

If A is an element of some set, S, and B is an element of S, it is possible for A to be the same as B.

I am talking about members of the same set.

Do you mean that A=3, B=3 and S={A,B} (where S is not a mutiset)?

 

[doronshadmi]

Construct all you like. It is still not an axiom.

Please define the difference between and axiom and a definition,

Ah! The much heralded Doronetics word shift, in which the need to define one word or phrase is side-stepped by jumping to a different word or phrase in need of definition.

Do you understand what "under construction" means?

"Chosen by distinction" means what, exactly?

Means that one picks members according to their distinct properties.

And are we heading towards your definition involving a sequence of steps in one-to-one correspondence with the positive integers? That will complete the circle.

It is exactly not complete the circle, since no distinct member can be picked twice, if the set is infinite.

 

[jsfisher]

No. You reduced redundancy in order to get {A,B} out of {A, A, B, A, B, B, B}.

No, I did not. The two sets I presented are identical according to set theory, uncontaminated by unnecessary Doronetics concepts.

If you do not do that then {A, A, B, A, B, B, B} is a multiset and {A,B} is a set.

There is a multiset that looks like that, but I was clear in calling it a set, and the two sets, {A, A, B, A, B, B, B} and {A,B}, are indistinguishable (and are therefore the same set).

I am talking about members of the same set.

Do you mean that A=3, B=3 and S={A,B} (where S is not a mutiset)?

No. I mean A \in S and B \in S where S = {3,7,33}, for example. (Substitute the normal set membership symbol for \in.) A and B can both be 7.

 

[epix]

The axiom of an infinite collection:

C is an infinite collection if for all x in C, x is arbitrarily the first object of C no more than once.

Example:

Given collection Z of points, any arbitrary chosen point is the first object of collection Z no more than once.

The magic of C = Z. Lols.

Now let's see a theorem held together by the axiom.

 

[doronshadmi]

No, I did not. The two sets I presented are identical according to set theory, uncontaminated by unnecessary Doronetics concepts.

In that case, for example, |{A,A}| = |{A}| by based on, what you call, set theory.

As for no order, for example, (AB,AB) really has no order exactly because the considered framework is under superposition of identities of 2-Uncertainty x 2-Redundancy Distinction Tree.

According to this example (without a loss of generality) sets are based on Distinction State (A,B) = (B,A) under F(1,1).

Please be aware of the fact that (A,B) is just a particular case of Frame (1,1) under 2-Redundancy x 2-Uncertainty Distinction Tree, as follows:

(AB,AB) (AB,A)  (AB,B)  (AB)    (A,A)   (B,B)   (A,B)   (A)     (B)     ()

A * *   A * *   A * .   A * .   A * *   A . .   A * .   A * .   A . .   A . .
  | |     | |     | |     | |     | |     | |     | |     | |     | |     | |
B *_*   B *_.   B *_*   B *_.   B ._.   B *_*   B ._*   B ._.   B *_.   B ._.

(2,2) = (AB,AB)
(2,1) = (AB,A),(AB,B)
(2,0)=  (AB)
[COLOR="magenta"][B](1,1)[/B][/COLOR] = (A,A),(B,B), [COLOR="magenta"][B](A,B)[/B][/COLOR]
(1,0)=  (A),(B)
(0,0)=  ()

Since the members of {A,B} are based on the particular case of DS (A,B) under F (1,1), and since {A,B} has 0-Uncertainy and 0-Redundancy, then its members are pick-able no matter if {A,B} = {B,A}, and being picked has unconditionally first pick, second pick, etc ...

In other words {A,B} = {B,A} means that "order has no significance" (which is not the same as "order does not exist", where "order does not exist" holds, by this example, at DS (AB,AB)) such that the members of this set are picked as follows:

The first pick can be any member.

The second pick is any member that is not the first picked member.

(By going beyond {A,B} case, the third pick is any member that is not the first or the second picked members.

...

The n pick is any member that is not any of the previously picked members.

...

etc... ad infinitum, such that no member can be picked twice, unless the amount of the members is finite).

In other words:

C is a set.

For all x in C, if all x in C are picked AND no x can be picked twice, then C is infinite.

There is a multiset that looks like that, but I was clear in calling it a set, and the two sets, {A, A, B, A, B, B, B} and {A,B}, are indistinguishable (and are therefore the same set).

You are indeed clear about nonessential decelerations, in this case, which is probably contaminated by unnecessary jsfisheretics concepts.

No. I mean A \in S and B \in S where S = {3,7,33}, for example. (Substitute the normal set membership symbol for \in.) A and B can both be 7.

In that case A or B are not used as members in terms of {A,B}, but they are used as different variables of the same S member.

 

[epix]

Construct all you like. It is still not an axiom.

Doron really came up with a peculiar definition and not an axiom, which is a statement mostly establishing a relationship. He would have to present a theorem and prove it with the statement that he regards as an axiom. Since he has shown that his understanding of giving an example differes from what one can normally expect, any inquiry into the meaning of his wild constructions is likely to fail, if he gives an "example" as a form of an explanation. Then there is only a short step to the notorious "strict box."

 

[Squid]

I'm going to advise people not to respond to this thread. doronshadmi has a history of being totally incomprehensible and his threads always go for dozens of pages without any progress being made.

And passing page 410... wow, good luck with this!

 

[doronshadmi]

Doron really came up with a peculiar definition and not an axiom, which is a statement mostly establishing a relationship. He would have to present a theorem and prove it with the statement that he regards as an axiom. Since he has shown that his understanding of giving an example differes from what one can normally expect, any inquiry into the meaning of his wild constructions is likely to fail, if he gives an "example" as a form of an explanation. Then there is only a short step to the notorious "strict box."

C is a set.

The axiom of infinite set:

For all x in C, if all x in C are picked AND no x can be picked twice, then C is infinite.

The axiom of completeness:

For all x in C, if all x in C are picked AND no x can be picked twice, then C is incomplete.

Theorem: If C is infinite, then C is incomplete.

Proof: derived directly from the axioms.

 

[doronshadmi]

No. I mean A \in S and B \in S where S = {3,7,33}, for example. (Substitute the normal set membership symbol for \in.) A and B can both be 7.

Alternatively, you are using the ZF Axiom of extensionality, which sates:

Two sets are equal (are the same set) if they have the same elements.

( http://en.wikipedia.org/wiki/Zermelo–Fraenkel_set_theory )

 

[jsfisher]

Two sets are equal (are the same set) if they have the same elements.

( http://en.wikipedia.org/wiki/Zermelo–Fraenkel_set_theory )

Yes, exactly. Two sets are equal if they have the same elements. Is this a revelation to you? And since the set {1, 1, 1, 2, 2, 2} has the exact same members as {2, 1}, they are the same set.

 

[The Man]

The n pick is any member that is not any of the previously picked members.

Thus by your own requirement of what constitutes your current (n_C) “pick” you can never pick any member previously picked (an n_P pick). So your purported “infinite” and “completeness” are not aspects of any given set but simply of the limitation you have placed on what you defined as the members you can currently pick from (never those that have been picked already). By your own requirement no member of, say, the set {A,B,C} can be both pick n_C and pick n_P as simply being pick n_P excludes it from consideration as pick n_C. So set {A,B,C} must be “infinite” and “complete” by your requirement (as already noted by jsfisher). Be careful when you’re “Under construction” that you don’t just blindly bulldoze yourself into claiming clearly finite sets are infinite.


Oh and making your definition of “complete” the same definition (by essentially simply replacing the word “infinite” with “complete”) as your definition of “infinite” makes your definition of “complete” superfluous and a contrivance to assert what we already understand. That you only want to consider “infinite” sets as “complete”. The simple statement that ‘only “infinite” sets are “complete”” in your notions states clearly and explicitly what you mean without the pretence of thinking you’re defining a different aspect with the same definition you use for “infinite”.

 

[epix]

C is a set.

The axiom of infinite set:

For all x in C, if all x in C are picked AND no x can be picked twice, then C is infinite.

The axiom of completeness:

For all x in C, if all x in C are picked AND no x can be picked twice, then C is incomplete.

Theorem: If C is infinite, then C is incomplete.

Proof: derived directly from the axioms.

You have two statements going where the property for_all_x in C is identical, but there are two different "axioms." What does prevent Doronetics to conditionally define set C as

For all x in C, if all x in C are picked AND no x can be picked twice, then C is an incomplete infinite set.


Your "theorem" wouldn't materialize, would it?

The real axiom of completeness states that every non-empty subset S of R that has an upper bound in R has a least upper bound, or supremum, in R.

And I say that it is not true -- at least one of non-empty subsets... doesn't have supremum.

Can I prove it?

No, there are no axioms available for the task, and so the original statement is not theorem but axiom.

The axioms usually start with "there exists...," but in the set theory they can be constructed different way, so knock yourself off.

 

[epix]

C is a set.

The axiom of infinite set:

For all x in C, if all x in C are picked AND no x can be picked twice, then C is infinite.

Example.

C = {1, 2, 3, 4}

Ordered selection under the condition: {2, 3, 4}, {3, 4}, {4}, { }.

In which way the statement "C is a set" precludes C being any set and therefore a finite set?

The axiom of completeness:

For all x in C, if all x in C are picked AND no x can be picked twice, then C is incomplete.

When you open a jigsaw puzzle box and there is nothing in there, you call it an incomplete box?

 

[doronshadmi]

Thus by your own requirement of what constitutes your current (n_C) “pick” you can never pick any member previously picked (an n_P pick).

There is no requirement to not pick again the first picked object, if all the objects are picked.

For example, it can be done in the case of a non-empty finite set.

But it can't be done in the case of a non-empty infinite set.

This is exactly what the axiom of infinite set asserts:

For all x in C, if all x in C are picked AND no x can be picked twice, then C is infinite.

The Man, you simply replied only to a picked part of what I wrote, by ignoring the next part, here it is again:

The n pick is any member that is not any of the previously picked members.

...

and you ignored:

etc... ad infinitum, such that no member can be picked twice, unless the amount of the members is finite).

Next time please read the whole post before you reply.

 

[doronshadmi]

Yes, exactly. Two sets are equal if they have the same elements. Is this a revelation to you? And since the set {1, 1, 1, 2, 2, 2} has the exact same members as {2, 1}, they are the same set.

Is this a revelation to you that {1, 1, 1, 2, 2, 2} and {2, 1} actually have the same elements in terms of type?

Being the same in terms of type is based on the ability to distinguish between them, where redundancy and uncertainty have 0 value.

So according to your reasoning "the same" is actually limited to 0-Uncertainy x 0-Rerundancy case.

 

[doronshadmi]

The real axiom of completeness states that every non-empty subset S of R that has an upper bound in R has a least upper bound, or supremum, in R.

And I say that it is not true -- at least one of non-empty subsets... doesn't have supremum.

Can I prove it?

No, there are no axioms available for the task, and so the original statement is not theorem but axiom.

I made a typo mistake in the name of the second axiom, here is the right one:

The axiom of incompleteness:

For all x in C, if all x in C are picked AND no x can be picked twice, then C is incomplete.

What does prevent Doronetics to conditionally define set C as

For all x in C, if all x in C are picked AND no x can be picked twice, then C is an incomplete infinite set.

No problem, it can done as follows:

C is a set.

Axiom:

For all x in C, if all x in C are picked AND no x can be picked twice, then C is infinite AND incomplete.

Theorem: If C is infinite, then C is incomplete.

Proof: derived directly from the axiom.