Deeper than primes

[doronshadmi]

You continue with your willful gibberish. Cantor's Theorem is |S| < |P(S)|, nothing more.



With this gibberish are you trying to say that the set B, is defined in the reference proof, must be an element of P(S)?



...except that there are two problem with this. First, your list of mappings is not a bijection, as required in the reference proof, so it cannot be used in constructing B. Second, you begin with the claim, a ↔ {a}, but then change it was something else, a ↔ {}.

Double Doron fail.

Thank you for that post, jsfisher.

It demonstrates the exact meaning of "The inability to get things beyond one's box" syndrome.

you begin with the claim, a ↔ {a}, but then change it was something else, a ↔ {}.

It is not a claim, jsfisher.

It is a systematic construction method of all P(S) members.

 

[jsfisher]

you begin with the claim, a ↔ {a}, but then change it was something else, a ↔ {}.

It is not a claim, jsfisher.

It is a systematic construction method of all P(S) members.

Really? So when you wrote this:

Such B is {} if:

a ↔ {a}
b ↔ {b}
c ↔ {c}
d ↔ {d}
...

where a ↔ {}

Please tell us what you meant by "a ↔ {a}" and how it isn't contradicted by "a ↔ {}".

Also, please tell us how B is constructed from this since the only "construction method" under consideration is from the reference proof and it requires a mapping with properties your mapping (either version) does not have.

Your double Doron fail remains.

 

[The Man]

The Man, once again your poor reasoning is demonstrated.

In this case you are unable to understand that, whether different circles are organized around a single point or not, these are the same circles, where their organization method is not significant.

It isn’t an “organization method”, Doron it is just a matter of where the center of the circle is located. Try concentric circles but with that center not on your line.

Furthermore, the set of all different circles (which is a one and only one set, no matter how its members are organized) is incomplete exactly because it does not have the smallest or the biggest circle.

Indeed your poor reasoning has to be upgraded in order to deal with real Geometry or real Set theory.

So according to you a circle with its center point on your line is no different than a circle where the center point is not on that line? While both circles may have the same radius and even be coplanar there is one specific difference between them. If you can’t figure it out, I’ll give you a hint: it has something to do with the, well, center points.

 

[The Man]

I am talking exactly about the set of all circles with different curvatures.

That all have the same center point, a point you are continually and intentionally ignoring as well as the fact that said center point is specifically one that is , by your own designation, on your line.

In this case, each circle's curvature exists exactly once as property of a member of this set.

So what? Again your chosen limitations don’t limit anyone but just you.

By using the members of this set along an infinitely long straight line, their intersections with this line do not completely cover that line, no matter how the circles are organized along this line.

Try concentric circles where the center point is not on your line.

The reason for that remains the same (at any organized way) which is:

Your deliberate and intentional ignorance.

The set of all circles with different curvatures does not have the smallest circle or the biggest circle, as shown in:
[qimg]http://farm6.static.flickr.com/5134/5533739885_1b5a702131_b.jpg[/qimg]

“The set of all circles”? So “all” is no longer a Doronic no-no? Guess what? “The set of all circles with different curvatures does not have the smallest circle or the biggest circle” simply because such circles can’t be defined within that set.

and any attempt to be (totally curved (being a point)) AND (totally straight (being a line)) is possible only by discontinuity, and resulted by an uncovered line.

Well you've certainly got plenty of discontinuity, particularly in that assertion. I guess that’s why just as a circle isn’t a point a line isn’t a point either. Would you like to try using, I don’t know, say, points, as points?

Furthermore, let the center point be any arbitrary real number along the infinitely long straight line, it does not change the facts described above.

Again try non-concentric circles or concentric circle where the center point is not on your line. Deliberate ignorance will not help you Doron, even though you apparently seem to think it does.

Oh, and again…

So by all means please explain to us the difference between changing and unchanging with “no past (before) and no future (after)”?

 

[epix]

It is not a claim, jsfisher.

It is a systematic construction method of all P(S) members.

Your systematic mapping between the members of infinite S and P(S) is the peasant proof of P(S) being uncountable.

a <--> {}
b <--> {a}
c <--> {b}
d <--> {c}
e <--> {d}
.
.
.

The systematic construction doesn't allow a multi-element member of P(S) to enter the mapping. Since the cardinality of S is aleph0 and the cardinality of single-element members of P(S) is aleph0 as well, coz there is a bijection, as you nicely demonstrated, then it follows that the cardinality of P(S) must be higher than aleph0. But sets with C>aleph0 are uncountable, coz the cardinality of natural numbers is aleph0. This logic allowed to invent and sustain farming, so we are still here and not extinct due to famine.

I brought this particular set organization to your attention on several occasions, but you either ignored it or no comprende.

There is no N such that subset EVEN NUMBERS precedes subset ODD NUMBERS or vice versa, coz both subsets are infinite and neither of the subsets can appear after the other, coz the membership of both subsets is endless.

Your "systematic construction" of P(S) only constructs subsets of P(S) with cardinality C=1, and the construction of these subsets is heading toward infinity with no chance of the subsets with C>1 ever completing the P(S).

 

[epix]

Your "systematic construction" of P(S) only constructs subsets of P(S) with cardinality C=1, and the construction of these subsets is heading toward infinity with no chance of the subsets with C>1 ever completing the P(S).

The notion of "complete infinity" can materialize only in your central nervous system, epix, and it may contribute to the extinction of human species if you don't leave next door Heather alone. Try "complementing" the next time.

Aah, His Irreversible Wisdom shuffled in.

 

[doronshadmi]

Please tell us what you meant by "a ↔ {a}" and how it isn't contradicted by "a ↔ {}".

In general, any systematic construction method of an infinite set, such that no member of that set is omitted, allows a bijection with any infinite arbitrary proper subset of the systematically constructed set.

I provide a systematic construction method of P(S) members such that no member of P(S) is omitted, and by using this simple fact, there is a biojection between S members (where S is some proper subset of P(S)) and P(S).

So is the case with the systematic construction method of, for example, N and Q members, which allows the bijection between their members and any arbitrary proper sets of them.

No claims, assumptions or contradictions are involved by using systematic construction methods of infinite sets, and this simple fact can't comprehended by a reasoning that is based on claims, assumptions, contradictions etc. .

 

[doronshadmi]

The systematic construction doesn't allow a multi-element member of P(S) to enter the mapping.

There is no such thing like a set with multi-element members, because a member of a given set appears once and only once as a member of the considered set.

According to any given set theory {a,a,b}={a,b}, so your argument does not hold.

If you do not follow this restriction, you actually deal with Multi-set theory, which is defiantly not the discussed subject about mapping between sets (multi-sets are not involved), and the mapping between infinite P(S) set and S set, is done only between sets.

 

[doronshadmi]

So what? Again your chosen limitations don’t limit anyone but just you.

So you are also a person that do not understand that a circle with a certain curvature (where an object is considered as a circle only if pi is its essential property) exists once and only once as a member of the set of all circles with different curvatures.

Any circle not on the line or any circle along the line is already a certain member of the set of all circles with different curvatures (the location of the center point of a given circle is insignificant).

Furthermore, by using a concentric organization of the set of all circles with different curvatures, there is no point on the complex plane (where the real line is a proper subset of the complex plane) which is not associated with some circle with unique curvature.

Yet, it does not change the fact that a point (total curvature, where pi=circumference/diameter does not exist) or an infinitely long straight line (total non-curvature, where pi=circumference/diameter does not exist), are not members of the set of all circles with different curvatures, and any attempt to include them as members of the set of all circles with different curvatures, is resulted by an inherent discontinuity of that extended set, because the smallest circle or the largest circle do not exist.

This discontinuity is considered as an external fact if a point and an infinitely long straight line are not members of the set of all circles with different curvatures.

Actually, there is no difference if the discontinuity is internal or external, because in both cases it is derived from the same reason, which is:

The set of all circles with different curvatures (where an object is considered as a circle only if pi is its essential property) does not have the smallest or the largest circle, and as a result (and because of the discontinuity) the set of all circles with different curvatures can't fully cover an infinitely long straight line.

 

[doronshadmi]

“The set of all circles”? So “all” is no longer a Doronic no-no?

Exactly, and yet this infinite set is incomplete because its smallest or largest circles do not exist.

“The set of all circles”? So “all” is no longer a Doronic no-no? Guess what? “The set of all circles with different curvatures does not have the smallest circle or the biggest circle” simply because such circles can’t be defined within that set.

Guess what? “The set of all circles with different curvatures does not have the smallest circle or the biggest circle” simply because such circles can’t be defined at all, if this set is infinite, and the considered subject deals only with infinite sets, in case you have missed it.

 

[doronshadmi]

Well you've certainly got plenty of discontinuity, particularly in that assertion. I guess that’s why just as a circle isn’t a point a line isn’t a point either. Would you like to try using, I don’t know, say, points, as points?

The ability to understand X depends on one's mind ability to be simultaneously beyond AND at X's domain.

Since you do not understand that fundamental fact, you are not able to get the following diagram:

[qimg]http://farm6.static.flickr.com/5134/5533739885_1b5a702131_b.jpg[/qimg]

which demonstrate the existence of points along an infinitely long straight line, by getting them simultaneously
(beyond (by different circle's curvatures)) AND (at the level of collection of intersecting points along an infinitely long straight line).

For example, an anthropologist really does a valuable research only if he\she is both an external observer AND an internal participator of the researched subject.

 

[epix]

There is no such thing like a with multi-element members, because a member of a given set appears once and only once as a member of the considered set.

According to any given set theory {a,a,b}={a,b}, so your argument does not hold.

If you do not follow this restriction, you actually deal with Multi-set theory, which is defiantly not the discussed subject about mapping between sets (multi-sets are not involved), and the mapping between infinite P(S) set and S set, is done only between sets.

This is an example of some members of P(N): {1,2,6}, {78}, {100}, {2,4,6,...}. These members are subsets of N and are made of a different number of elements. When the number of the elements exceeds just one element, it's normal to refer to these subsets as multi-element subsets as opposed to single-element subsets, such as {100}. "Multi-element members of the power set" is quite different term from your fantasmagoric translation that is highligted in the quote.

 

[doronshadmi]

This is an example of some members of P(N): {1,2,6}, {78}, {100}, {2,4,6,...}. These members are subsets of N and are made of a different number of elements. When the number of the elements exceeds just one element, it's normal to refer to these subsets as multi-element subsets as opposed to single-element subsets, such as {100}. "Multi-element members of the power set" is quite different term from your fantasmagoric translation that is highligted in the quote.

EDIT:

In other words, you still do not get http://www.internationalskeptics.com/forums/showpost.php?p=6996924&postcount=14618 , which systematically provides the bijection of P(S) members (where P(S) is an infinite set) with the members of S, where S is an infinite proper subset of P(S).

 

[epix]

EDIT:

In other words, you still do not get http://www.internationalskeptics.com/forums/showpost.php?p=6996924&postcount=14618 , which systematically provides the bijection of P(S) members (where P(S) is an infinite set) with the members of S, where S is an infinite proper subset of P(S).

It's obvious that you never come to grasp the fact that bijection is a union of two functions: injection AND surjection. It's like there is no kid/bijection born without function male and function female. There is no bijection when there is no surjection and Cantor's proof explicitly shows that there can be no surjection between X and P(X) due to a contradiction. And if there is no surjection there is no bijection. Like if there is no male there can't be a human fetus, apart for the Jesus fetus. Once again, a <--> b by itself doesn't represent any bijection. You need to prove the existence of both functions injection, which is apparent, and surjection, which can be far from apparent.

Your attempt to disprove Cantor's Theorem amounts to proving that human parthenogenesis is a real deal just by tossing the same photo around:

{V.M.} <---> J.

 

[doronshadmi]

It's obvious that you never come to grasp the fact that bijection is a union of two functions: injection AND surjection. [/size]

epix, you still do not grasp the simple fact, which is:

If one shows a systematic construction method of some infinite set, such that no member of the constructed set is omitted, then any arbitrary infinite proper subset of that set is in bijection with this set.

 

[doronshadmi]

Please tell us what you meant by "a ↔ {a}" and how it isn't contradicted by "a ↔ {}".

jsfisher I use a very simple principle:

If one shows a systematic construction method of some infinite set, such that no member of the constructed set is omitted, then any arbitrary infinite proper subset of that set is in bijection with this set, that's all.

In other words, it does not matter in what order the members of some infinite proper subset are mapped with systematically constructed infinite set (where "systematically constructed" means that no one of the members of the constructed set is omitted), there is a bijection between the infinite proper subset and the systematically constructed infinite set.

Some examples of infinite sets:

This simple principle is shown between N and, for example, the set of even numbers, which is a proper subset of N, and it does not matter what order is used among the proper subset members.

This simple principle is shown between Q and, for example, the set of natural numbers, which is a proper subset of Q, and it does not matter what order is used among the proper subset members.

This simple principle is shown between P(S) and, for example, S, which is a proper subset of P(S) ( as show in http://www.internationalskeptics.com/forums/showpost.php?p=6996924&postcount=14618 ), and it does not matter what order is used among the proper subset members.

 

[epix]

epix, you still do not grasp the simple fact, which is:

If one shows a systematic construction method of some infinite set, such that no member of the constructed set is omitted, then any arbitrary infinite proper subset of that set is in bijection with this set.

The power set of of the set of natural numbers is not just "some infinite set." I already asked you to finish the "systematic mapping" of subsets {1}, {2}, (3}, ... and start mapping the next category {1,2}, {1,3}, {1,4}, ... , but you are not done yet with the singletons, are you?

In the case of x in R², that means {a,b}, you can substitute a=numerator and b=denominator and see how Cantor showed that there is a surjection between the naturals and rational numbers.

Cantor spent all members of N bijecting the rationals. If there were some numbers involving numerator, denominator, and one other __ator, Cantor would organize the numbers in a cube (x in R³), and spend all rational numbers on that bijection. What space is required to show surjection for x in R¹ AND R² AND R³ for example?

Your "systematic mapping" of the power set is a fantasy. Actually it is not a fantasy -- it's turbulent mayhem of neural character caused by The Delinquent Pancake of Unusual Desire dilligently spread over time T_0-->always.

 

[doronshadmi]

The power set of of the set of natural numbers is not just "some infinite set." I already asked you to finish the "systematic mapping" of subsets {1}, {2}, (3}, ... and start mapping the next category {1,2}, {1,3}, {1,4}, ... , but you are not done yet with the singletons, are you?

You are wrong epix.

First we are talking about the general P(S);S form, where P(N);N is some particular case of it.

In the general systematic construction of P(S) members {},{a,b,c,...} and all the subsets between {} and {a,b,c,...} are systematically constructed, without omitting even a single P(S) member, and we do not need more than that in order to define a bijection between S members (where S is a proper subset of P(S)) and P(S) members.

All the countable\uncountable mambo jambo does not hold water.

Your "systematic mapping" of the power set is a fantasy.

Again you demonstrate your current inability to get http://www.internationalskeptics.com/forums/showpost.php?p=7003057&postcount=14636 and the linked posts in it.

A natural conclusion of such current inability is understood by you as fantasy.

 

[jsfisher]

jsfisher I use a very simple principle:

If one shows a systematic construction method of some infinite set, such that no member of the constructed set is omitted, then any arbitrary infinite proper subset of that set is in bijection with this set, that's all.

The gibberish continues. Sets are not "in bijection" with sets. If you cannot express yourself without speaking nonsense, don't expect to be an effective communicator.

Be that as it may, you have NOT shown a systematic construction method. Moreover, the phrase "no member of the constructed set is omitted" is tautological. Finally, constructing an infinite set does not establish any sort of bijection.


How about we go back to the basic claim you made, Doron. You have said a bijection can be established between the members of ANY set and its power set. Let's stick to the case where the set is {A}.

Either retract your claim or show a bijection between the members of {A} and {{},{A}}.

 

[doronshadmi]

The gibberish continues. Sets are not "in bijection" with sets. If you cannot express yourself without speaking nonsense, don't expect to be an effective communicator.

The ignorance continues.

If you are unable the understand that there is a bijection between the members of ,for example, set N and the members of some proper subset of it, for example, the set of even numbers, then your replies do not hold water.

Be that as it may, you have NOT shown a systematic construction method.

Wrong jsfisher, by using your weak reasoning you are failing to see the systematic construction method, as show in http://www.internationalskeptics.com/forums/showpost.php?p=7003057&postcount=14636 (and do not forget to follow the links of this post).

Finally, constructing an infinite set does not establish any sort of bijection.

Unless the construction of the infinite set is systematic (no object is omitted), and as a result the members of any arbitrary infinite proper subset of systematically constructed set have, a bijection (1-to-1 and onto mapping ) with the members of infinite systematically constructed set.

How about we go back to the basic claim you made, Doron. You have said a bijection can be established between the members of ANY set and its power set. Let's stick to the case where the set is {A}.

Either retract your claim or show a bijection between the members of {A} and {{},{A}}.

Before we go back, let's first see if you are able to understand that the members of any arbitrary infinite proper subset of systematically constructed set, have a bijection (1-to-1 and onto mapping) with the members of infinite systematically constructed set.