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Deeper than primes

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Not at all.

If the powerset of set A is {{},A} it means that |A|=1

You are the first to introduce any "set A". Why do you insist on irrelevant asides?

It means that also |{A}|=1

No, it doesn't. The cardinality of {A} is 1 because that's how cardinality works. It has nothing to do with what "set A" happens to be. That was just an irrelevant aside from you.

...and in the case P({A}) = {{},{A}}

The would follow directly from the meaning of "power set". So, you have ventured from the irrelevant to the bloody obvious. Please continue.

By using Cantor's construction method
...being the method that Cantor never provided...
we get a 1-to-1 correspondence between 1,2 and {},{A}, which is not different than the 1-to-1 correspondence between 1,2 and {},A

Were you expecting a "1-to-1 correspondence between 1,2 and {},{A}" to be different from the "1-to-1 correspondence between 1,2 and {},{A}"? Seriously?

...such that:

1 ↔ {} ↔ {}
|
2 ↔ A ↔ {A}

and we get the same 1-to-1 correspondence for both cases.

The original set is {A} and its power set is {{},{A}}. Your trivial mapping, above, for "both cases" does not provide a mapping between the members of the two sets at issue. Since you didn't provide a mapping between members of {A} and {{},{A}}, it is unlikely to be a bijective mapping.

Again, Doron, you have failed. You tried but failed to show any bijection between the members of {A} and its power set.
 
LOL. Considering a finite case: if |S| = 3, then |P(S)| = 23. According to you, 3 = 23.

EDIT:

You are still missing my argument about S and P(S).

According to it the cardinality of natural numbers with any collection of different objects, exists between non-bijection to bijection.

When we are dealing with finite collections of different objects, if we understand each collection in terms of fixed cardinality, one of the results is that two finite collections have different amount of objects, exactly as can be seen in the particular case of {1,2,3} and {{},{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3}}.

By using the set of natural numbers together with Cantor's constriction method, we are no longer closed under the different amounts of {1,2,3} and {{},{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3}}, as follows:

.
.
.
|
|
|-----------------------
|1 ↔ {1} |
|2 ↔ {2} | Provides 1 ↔ {}
|3 ↔ {3} |
|-----------------------
|
|
|-----------------------
|1 ↔ { } |
|2 ↔ {2} | Provides 2 ↔ {1}
|3 ↔ {3} |
|-----------------------
|
|
|-----------------------
|1 ↔ {1} |
|2 ↔ { } | Provides 3 ↔ {2}
|3 ↔ {3} |
|-----------------------
|
|
|-----------------------
|1 ↔ {1} |
|2 ↔ {2} | Provides 4 ↔ {3}
|3 ↔ { } |
|-----------------------
|
|
|-----------------------
|1 ↔ { } |
|2 ↔ {3} | Provides 5 ↔ {1,2}
|3 ↔ {1,3} |
|-----------------------
|
|
|-----------------------
|1 ↔ { } |
|2 ↔ {2} | Provides 6 ↔ {1,3}
|3 ↔ {1,2} |
|-----------------------
|
|
|-----------------------
|1 ↔ {1} |
|2 ↔ {3} | Provides 7 ↔ {2,3}
|3 ↔ {1,2} |
|-----------------------
|
|
|-----------------------
|1 ↔ { } |
|2 ↔ {1} | Provides 8 ↔ {1,2,3}
|3 ↔ {2} |
|-----------------------
|
|
.
.
.

We actually deal with a DNA-like code between 3 and 8 objects, and by using it are able to define any mapping degree between |S| and |P(S)| for our purpose.

Such flexibility is impossible by the rigid |S| < |P(S)| particular case, which is wrongly taken as some universal principle of the possible relations between S and P(S).


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


Furthermore, by using infinite sets we discover that the whole idea of mapping between sets is changed, because we are also able to define a bijection between S and its proper subset, as can be seen bet between the natural numbers and, for example, their proper subset of even (or odd) numbers.

We also are able to define a bijection between rational numbers and their proper subset of natural numbers.

This ability does not stop also in the case of real numbers, where natural numbers are their proper subset, and the possible bijection between real numbers (which are equivalent by their amount to the powerset of natural numbers) is shown by using Cantor's construction method, as follows:

Cantor's construction method constructs explicit |{{},...,{1,2,3,...}}l {{},...,{1,2,3,...}} members as follows:

1) Some defined explicit {{},...,{1,2,3,...}} member is the result of a 1-to-1 correspondence between {1,2,3,...} members and the same amount of members taken from {{},...,{1,2,3,...}}.

2) This explicit constructed {{},...,{1,2,3,...}} member ( notated as D. Please read more about D construction in http://en.wikipedia.org/wiki/Cantor's_theorem ) includes a {1,2,3,...} member only if this {1,2,3,...} member does not exist as one of the members of the {{},...,{1,2,3,...}} member that is mapped with it.

By using this construction method with amount |{{},...,{1,2,3,...}}l , one enables to define a bijection between {1,2,3,...} members
and {{},...,{1,2,3,...}} members, where D is actually a placeholder for |{{},...,{1,2,3,...}}l {{},...,{1,2,3,...}} members.

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

In both cases, whether the considered set is finite or not, by using Cantor's construction method, one enables to define any wished degree of mapping between two sets, and use the resulted mapping for some useful purpose.
 
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EDIT:

You are the first to introduce any "set A". Why do you insist on irrelevant asides?
Not exactly, in http://www.internationalskeptics.com/forums/showpost.php?p=6922616&postcount=14407 you wrote:
jsfisher said:
Let's take for example S = {A}. It's power set P(S) = {{},A}.
but I show that also if the powerset is {{},A} then:

1) |A| must be 1.

2) By using Cantor's construction method, we are able to define a 1-to-1 correspondence between the members of {{},{A}} (or {{},A}) and the appropriate amount of natural numbers, and in this case, any 2 different natural numbers can be used in order to demonstrate the 1-to-1 correspondence.

jsfisher, your rigid reasoning still missing http://www.internationalskeptics.com/forums/showpost.php?p=6963549&postcount=14542 .
 
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By using Cantor's construction method
...which exists only in your imagination...
we are able to define a 1-to-1 correspondence between the members of {{},{A}} (or {{},A}) and the appropriate amount of natural numbers

...which was both (a) trivial to do and (b) unrelated to the required task.

The required task is now and has always been: Show a bijection between the members of {A} and its power set.


Again you fail to accomplish that which you claimed you could do.
 
QUOTE=jsfisher;6963761...which was both (a) trivial to do and (b) unrelated to the required task.[/QUOTE
...which is both (a) general to do an (b) exactly related to the required task, which is the existence of collections of different objects, where the properties of the sets have no significance, by generalization.
 
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QUOTE=jsfisher;6963761...which was both (a) trivial to do and (b) unrelated to the required task.[/QUOTE
...which is both (a) general to do an (b) exactly related to the required task, which is the existence of collections of different objects, where the properties of the sets have no significance, by generalization.


Formatting errors aside, you still haven't provided a bijection between the members of {A} and it power set.

You again fail.
 
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“objects” that are all specifically natural numbers. Are you claiming that some member(s) of your “set of natural numbers” is/(are) not (a) natural number(s)?
Since you do not understand present continuous in terms of "ever increasing", let us express it as "permanently changing", where the permanent (invariant) aspect is like pi among the collection of circles, and the change (variant) aspect is like the curvature among the collection of circles.

The present continuous state, as described above, is the fundamental property of any infinite collection of different objects.

For more details, please look at http://www.internationalskeptics.com/forums/showpost.php?p=6963549&postcount=14542 .
 
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If the powerset of set A is {{},A} it means that |A|=1
Yes, it means that if A = {$}, and $ is put in 1:1 correspondence with the set of natural numbers, you get

1 <---> $

and the cardinality of set A is therefore |A| = 1
By using Cantor's construction method we get a 1-to-1 correspondence between 1,2 and {},{A}, which is not different than the 1-to-1 correspondence between 1,2 and {},A, such that (...)

1 ↔ {}{}
|
2A{A}
Where does that 2 come from? You are matching A with P(A), and A returned only one natural number and that's 1.
 
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Yep, both those links provide evidence of how wrong Doron is. Repeated failed attempts to show a bijection between the members of a set and its power set.
You are unable (yet) to understand the generalization of sets as collections of different objects (where the property of any given set is insignificant), and how by this generalization, any given set has a 1-to-1 correspondence with some arbitrary objects of natural numbers (where, by generalization, this 1-to-1 correspondence is some particular case of many degrees of mapping between collections of different objects).

So let us do it in baby steps.

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

By baby steps, there is an interesting difference between mapping among finite sets and infinite sets, such that there is no 1-to-1 correspondence between a non-empty finite set and its proper subset, but there is a 1-to-1 correspondence between an infinite set and its proper subset.

Among infinite sets we are able to define a bijection between S and its proper subset, as can be seen between the natural numbers and, for example, their proper subset of even (or odd) numbers.

We are also able to define a bijection between rational numbers and their proper subset of natural numbers.

This ability does not stop also in the case of real numbers, where natural numbers are their proper subset, and the bijection between real numbers (which are equivalent by their amount to the powerset of natural numbers) is shown by using Cantor's construction method, as follows:

Cantor's construction method constructs explicit |{{},...,{1,2,3,...}}l {{},...,{1,2,3,...}} members as follows:

1) Some defined explicit {{},...,{1,2,3,...}} member is the result of a 1-to-1 correspondence between {1,2,3,...} members and the same amount of members taken from {{},...,{1,2,3,...}}.

2) This explicit constructed {{},...,{1,2,3,...}} member ( notated as D. Please read more about D construction in http://en.wikipedia.org/wiki/Cantor's_theorem ) includes a {1,2,3,...} member only if this {1,2,3,...} member does not exist as one of the members of the {{},...,{1,2,3,...}} member that is mapped with it.

By using this construction method with amount |{{},...,{1,2,3,...}}l , one enables to define a bijection between {1,2,3,...} members
and {{},...,{1,2,3,...}} members, where D is actually a placeholder for |{{},...,{1,2,3,...}}l {{},...,{1,2,3,...}} members.
 
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As for collections of different objects, let us look at Nicholas of Cusa's ( http://www.integralscience.org/cusa.html )model of infinity.

By using a collection of circles, he wished to show that there is a continuous relation between a circle and a stright line, as shown in the following diagram:

5519866444_4c55f67312.jpg


I added a point to this diagram, which is the central point of the considered circles.

It has to be stressed that a form is considered as a circle only if the invariant pi is defined as the result of the ratio between its circumference and its diameter.

Pi is not found at point or straight line forms, so they can't be considered as objects of the collection of circles, but they are used as the inaccessible limits of this collection, such that infinitely many circles with increased curvature degrees can't be a point AND infinitely many circles with decreased curvature degrees can't be a line.

In other words, the collection of circles is a permanently changing model , where the permanent aspect is pi and the changing aspect is the circle's curvature degrees.

If the collection of circles is finite, we are able to define a complete collection of different circle's curvature degrees. But such completeness is not found in the case of infinite collection because the amount of different circle's curvature degrees is not satisfied.

Following the same principle about the collection natural, rational and irrational numbers, their amount is not satisfied if they have infinitely many objects.
 
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You are unable (yet) to understand the generalization of sets as collections of different objects (where the property of any given set is insignificant)

It is only you, Doron, that seems to have trouble with this basic properties of sets. You belabor the trivial.

and how by this generalization, any given set has a 1-to-1 correspondence with some arbitrary objects of natural numbers (where, by generalization, this 1-to-1 correspondence is some particular case of many degrees of mapping between collections of different objects).

More trivia. No one disputes there is a bijection between any set of n objects and the first n natural numbers. That is a trivial observation. The fact you dwell on it as significant speaks to your most limited understanding of Mathematics.

However, you made a couple of claims, different from above, that are flatly untrue. You claimed that for any set, S,

(1) a bijection exists between S and its power set, P(S).
(2) a bijection exists between P(S) and the natural numbers.

S does not mean the original set S plus other things to make it the same size as P(S). Natural numbers means all of them, not just a few of them.

Your claims are wrong. You have failed. You also claimed some construction method is to be found in Cantor's Theorem. You failed to establish that, too.

In things mathematical, Doron, you do little else than fail.
 
This explicit constructed {{},...,{1,2,3,...}} member ( notated as D. Please read more about D construction in http://en.wikipedia.org/wiki/Cantor's_theorem ) includes a {1,2,3,...} member only if this {1,2,3,...} member does not exist as one of the members of the {{},...,{1,2,3,...}} member that is mapped with it.

Here it is, Doron, the entire proof from your wikipedia reference. Please highlight the constuction method for the member "notated as D".

Two sets are equinumerous (have the same cardinality) if and only if there is a one-to-one correspondence between them. To establish Cantor's theorem it is enough to show that given any set A no function f from A into P(A), the power set of A, can be surjective, i.e. to show the existence of at least one subset of A that is not an element of the image of A under f. Such a subset, B ∈ P(A), is given by the following construction:
B = { x ∈ A : x ∉ f(x) }​
This means, by definition, that for all x in A, x ∈ B if and only if x ∉ f(x). For all x the sets B and f(x) cannot be the same because B was constructed from elements of A whose images (under f) did not include themselves. More specifically, consider any x ∈ A, then either x ∈ f(x) or x ∉ f(x). In the former case, f(x) cannot equal B because x ∈ f(x) by assumption and x ∉ B by the construction of B. In the latter case, f(x) cannot equal B because x ∉ f(x) by assumption and x ∈ B by the construction of B.

Thus there is no x such that f(x) = B; in other words, B is not in the image of f. Because B is in the power set of A, the power set of A has a greater cardinality than A itself.

That's the whole proof, Doron. There is no set D to be found. Your reading comprehension skills have failed you again. You have failed again.
 
That's the whole proof, Doron. There is no set D to be found. Your reading comprehension skills have failed you again. You have failed again.

You simply ignore this:

Using this idea, let us build a special set of natural numbers. This set will provide the contradiction we seek. Let D be the set of all non-selfish natural numbers. By definition, the power set P(N) contains all sets of natural numbers, and so it contains this set D as an element. Therefore, D must be paired off with some natural number, say d. However, this causes a problem. If d is in D, then d is selfish because it is in the corresponding set. If d is selfish, then d cannot be a member of D, since D was defined to contain only non-selfish numbers. But if d is not a member of D, then d is non-selfish and must be contained in D, again by the definition of D.

This is a contradiction because the natural number d must be either in D or not in D, but neither of the two cases is possible. Therefore, there is no natural number which can be paired with D, and we have contradicted our original supposition, that there is a bijection between N and P(N).

Note that the set D may be empty. This would mean that every natural number x maps to a set of natural numbers that contains x. Then, every number maps to a nonempty set and no number maps to the empty set. But the empty set is a member of P(N), so the mapping still does not cover P(N).

There is no bijection between N and P(N), only if (each D is taken separately) AND (it is not mapped with some N member).

By using this dead end "proof", one misses the ability to demonstrate the bijection between P(N) and its proper subset N.

By the way, by your provided part taken from http://en.wikipedia.org/wiki/Cantor's_theorem , D is called B.

In other words, you have no case.
 
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