Deeper than primes

[jsfisher]

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.

There are many sets for which there are bijections between their members and those of specific subsets. Is this a general truth for any set and any of its proper subsets? No.


Now, about that bijection you claim exists between members of {A} and {{}, {A}}....

 

[epix]

You are wrong epix.

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.

Read again:

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?

When do you think you finish systematically bijecting the subsets with a single element and start bijecting the subsets with two elements? I had the opportunity to see the unbelievable: you actually started mapping {a}, {b}, {c}, . . . thinking that one nice day you will continue "systematic mapping" with {a,b}, {a,c}, {a,d}, . . . and then {a,b,c}, {a,b,d}, {a,b,e}, . . . Even Bernie can figure out what will probably take a billion light years for you to comprehend . . .

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

. . . and another trillion light years to comprehend what is written bellow.
http://en.wikipedia.org/wiki/Uncountable_set

There are many equivalent characterizations of uncountability. A set X is uncountable if and only if any of the following conditions holds:

1. There is no injective function from X to the set of natural numbers.

2. X is nonempty and every ω-sequence of elements of X fails to include at least one element of X. That is, there is no surjective function from the natural numbers to X.
.
.
.

There is no bijection without injection AND surjection.

Just keep systematic mapping of {a}, {b}, {c}, . . . and I hope that just before Milky Way merges with Andromeda, God finally succeeds to stuff your head with the concept of a function called "surjection."

 

[epix]

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.

Of course there is a bijection.

1 <--> {2, 4, 6, . . .}

So is

2 <--> {4, 6, 8, . . .}
3 <--> {6, 8, 10, . . .}
4 <--> {8, 10, 12, . . .}
5 <--> {10, 12, 14 . . .}
.
.
.

When you reach the infinity by travelling alongside the even road you can start mapping the subsets made of odd numbers

infinity <--> {1, 3, 5, . . .}
infinity+1 <--> {3, 5, 7, . . .}
.
.
.

When you reach another infinity that ends the odds, you can start mapping the primes and all its infinite subsets.

2*infinity <--> {2, 3, 5, . . .}
2*infinity+1 <--> {3, 5, 7, . . .}
2*infinity+2 <--> {5, 7, 11 . . .}
.
.
.

Traveling the road of phantasmagoric multidimensional infinities, you will stop for a breakfast. A waitress named Surjection brings the pancakes to you. But the real Surjection causes two functions -- erection and ejaculation -- to join the math.

 

[The Man]

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.

Again that is just your self imposed limitation and once again concentric circles where the center point in not a point on your meets that limitation. Also the first of those circles that intersects your line will do so only at one point (a tangent). The rest will get your pairs.

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).[/qoute]

See above.

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.

complex plane? So now you have added a -j (-n^1/2) axis?. Again see above.

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.

Again see above and once again please learn some basic geometry.

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

Exactly what? 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 at all, if this set is infinite, and the considered subject deals only with infinite sets, in case you have missed it.

Guess what? “The set of all circles" includes only circles, in case you have missed it.

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.

Again see above. Concentric circles where the center point is not a point on your line meets your self-imposed limitations. The first of those circles that intersects your line will do so only at one point (a tangent). The rest will get your pairs. Agian as mentioned before even with your concentric circles centered on a point of your line, as designated by you, they are superficially centered on a point of your, well, line. Again please learn some basic geometry.

 

[doronshadmi]

...the first of those circles that intersects your line will do so only at one point (a tangent). The rest will get your pairs.

It does not matter.

1) This first circle is omitted from the rest of the circles that get the pairs, so there is a pair of points that does not cover the line, simply becuse there are no two circles with the same curvature in the set of all circles with different curvatures.

2) Your arbitrary first circle is not the smallest and not the largest circle of the collection of all circles with different curvatures, so nothing was changed by your "tangent trick".

In other words, you still have to learn Set theory and Geometry, before you deal with:

http://www.internationalskeptics.com/forums/showpost.php?p=6993998&postcount=14607

http://www.internationalskeptics.com/forums/showpost.php?p=6998970&postcount=14629

or

http://www.internationalskeptics.com/forums/showpost.php?p=6999117&postcount=14631

 

[epix]

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.

Why is it so that CURVATURE is associated through pi with DOES NOT EXIST the same way as the opposite NON-CURVATURE? Have you been reading Loss Leader's seminal work and found it to your liking?

Why do you relate curvature to the ratio between circumference and diameter at all when the magnitude of a curvature is given by the ratio 1/radius? In other words, the curvature is a reciprocal of circle's radius.

Now I truly understand the meaning of "fire at will."

 

[doronshadmi]

Is this a general truth for any set and any of its proper subsets? No.

You are wrong again, the answer is Yes.

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

Now, about that bijection you claim exists between members of {A} and {{}, {A}}....

In this case you are using a finite case, so Cantor's systematic construction method of {{},{A}} members is done as follows:

A ↔ {A} provides {}
A ↔ {} provides {A}

So, as can be seen, no {{},{A}} was omitted, and by using this fact the following mapping holds:

{} ↔ A
{A} ↔ B

Furthermore, also the two next cases hold:

Only one mapping:
{} ↔ A
{A} ↔

No mapping:
{} ↔
{A} ↔


In other words,there is no universal law about mapping, because we are free to define any wished degree of it for our purpose.

Any attempt to force universality on mapping is based on limited understanding of this fine subject.

Let's take for example Hilbert's Hotel ( http://en.wikipedia.org/wiki/Hilbert's_paradox_of_the_Grand_Hote ):

We actually can empty any wished number of rooms from an infinite set of rooms (where an empty room is equivalent to no-mapping), so the whole idea of mapping with some given set is defined according to our wish, and the wished degree is from no-mapping to bijection.

 

[sympathic]

It does not matter.



In other words, you still have to learn Set theory and Geometry, before you deal with:

Says he who learned neither, and is not willing to learn either.

 

[doronshadmi]

Why is it so that CURVATURE is associated through pi with DOES NOT EXIST the same way as the opposite NON-CURVATURE?

Any attempt to reach a point or a line form, starting from some arbitrary circle, is done by discontinuity between an arbitrary circle and a point, or by discontinuity between an arbitrary circle and a line.

Why do you relate curvature to the ratio between circumference and diameter at all when the magnitude of a curvature is given by the ratio 1/radius? In other words, the curvature is a reciprocal of circle's radius.

Use 2pi instead of pi, it does not matter.

 

[epix]

Now, about that bijection you claim exists between members of {A} and {{}, {A}}....

As it turned out, that reminder has proven itself a very powerful diagnostic tool. Don't miss the moment when Doron pulls rabbit B from the { }-hat.

 

[doronshadmi]

As it turned out, that reminder has proven itself a very powerful diagnostic tool. Don't miss the moment when Doron pulls rabbit B from the { }-hat.

Time is not involved here, so there is no such moment.

Actually because of focusing on time, you are missing what is written in http://www.internationalskeptics.com/forums/showpost.php?p=7006557&postcount=14647 .

 

[epix]

Originally Posted by epix
Why do you relate curvature to the ratio between circumference and diameter at all when the magnitude of a curvature is given by the ratio 1/radius? In other words, the curvature is a reciprocal of circle's radius.

Use 2pi instead of pi, it does not matter.

But 2pi doesn't equal 1/r.

First things first: Do you know what year it is?

 

[doronshadmi]

But 2pi doesn't equal 1/r.

Yes it is, if the formula is circumference/radius (instead of circumference/diameter).

 

[epix]

Yes it is, if the formula is circumference/radius (instead of circumference/diameter).

Doron, give it some rest. Just go for a walk and try not to think about math.

 

[doronshadmi]

Doron, give it some rest. Just go for a walk and try not to think about math.

I am resting at the source of my mind, where no thoughts exist ( http://www.youtube.com/watch?v=wGLLDEYFAzs ).

 

[jsfisher]

Is this a general truth for any set and any of its proper subsets? No.

You are wrong again, the answer is Yes.

You really want to go with that answer? Let's take, for example, the set of natural numbers as "the set" and the null set as one of "its proper subsets".

There is no bijection between the members of N and {}.

Now, about that bijection you claim exists between members of {A} and {{}, {A}}....

In this case you are using a finite case, so Cantor's systematic construction
method of {{},{A}}

Whoa!!! Wait a second. This interesting. Given a set, S, and it's power set, P(S), you propose to construct P(S)? That is very adventurous of you, Doron.

...members is done as follows:

A ↔ {A} provides {}

This mapping is not a bijection between S and P(S) nor is it a surjection. So, whatever it is you are doing here, it does not correspond to anything in the reference proof. Just how do you get {} from A ↔ {A}?

A ↔ {} provides {A}

Ditto.

So, as can be seen, no {{},{A}} was omitted

So, you started with P(S) and ended up with P(S)? That is remarkable. We are all very impressed. You did include a few extra steps, though.

...and by using this fact the following mapping holds:

{} ↔ A
{A} ↔ B

Why is this diversion important? The goal was a bijection between the members of {A} and {{},{A}}. This is not such a bijection.

Again, you fail in your attempt to demonstrate your claim.

 

[doronshadmi]

You really want to go with that answer? Let's take, for example, the set of natural numbers as "the set" and the null set as one of "its proper subsets".

There is no bijection between the members of N and {}.



Whoa!!! Wait a second. This interesting. Given a set, S, and it's power set, P(S), you propose to construct P(S)? That is very adventurous of you, Doron.


This mapping is not a bijection between S and P(S) nor is it a surjection. So, whatever it is you are doing here, it does not correspond to anything in the reference proof. Just how do you get {} from A ↔ {A}?

Ditto.



So, you started with P(S) and ended up with P(S)? That is remarkable. We are all very impressed. You did include a few extra steps, though.



Why is this diversion important? The goal was a bijection between the members of {A} and {{},{A}}. This is not such a bijection.

Again, you fail in your attempt to demonstrate your claim.

Well, your last reply demonstrates again that since you a running in circles within a closed box, you simply unable to understand http://www.internationalskeptics.com/forums/showpost.php?p=7006557&postcount=14647 .

So enjoy your run, I am not a participator of it.

 

[jsfisher]

Well, your last reply demonstrates again that since you a running in circles within a closed box, you simply unable to understand http://www.internationalskeptics.com/forums/showpost.php?p=7006557&postcount=14647 .

So enjoy your run, I am not a participator of it.

Excellent. You completely avoided any points raised. Well done!!

 

[doronshadmi]

Excellent. You completely avoided any points raised. Well done!!

Wrong again, jsfisher, your points are closed within your box, therefore they are not relevant.

Just for example, {} is not a member of N.

Furthermore, we are not talking about finite amount of elements that are mapped with infinitely many elements, so again your reply is a load of nonsense.

 

[jsfisher]

Wrong again, jsfisher, your points are closed within your box, therefore they are not relevant.

Whatever.

We are all still in awe, though, because you have perfected a method to construct a power set given the power set as a starting point. Well done!