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Deeper than primes

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doronshadmi said:
No, the whole notion of countable\uncountable sets is a mambo jambo wrong reasoning.
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I thought that you liked and accepted Cantor's diagonal proof of unaccountability of real sets. Remember? You enslaved the diagonal proof and raped it repeatedly into the <0,1> argument.

Let me help you . . .

Definition: Any set is countable.

If so, continuity doesn't exist and there are no lines and curves. If there are no curves, there is no curvature. If there is no curvature, there are no circles. So please stop presenting various circles as some arguments of yours -- unless you want to further participate as a subject in the hitherto unknown medical field called "Analytic Psychopathy."
 
doronshadmi said:
The Man, you still do not know the meaning of: "The set of all circles with different curvatures".
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Evidently that is just you as your assertion before was…

doronshadmi said:
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.
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So does your “set of all circles with different curvatures” include “all circles with different curvatures” or just the ones that intersect your line at two points?


doronshadmi said:
1) Each member of this set has a unique curvature, so if one of the curvatures is used as a tangent, it is not used as one of the centered points, and we get two places along the line, which are not covered by the intersecting points of the circle with the unique curvature that is already used as a tangent.
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Please indentify these “two places along the line, which are not covered by the intersecting points of the circle with the unique curvature that is already used as a tangent”. How many circles “with the unique curvature” do you have? Have you run out for some reason? You do understand that those “two points” would be intersected and thus covered by some other circle, don’t you? Again please learn some basic geometry.

doronshadmi said:
2) The set of all circles with different curvatures (where pi=circumference/diameter holds) does not have the smallest or the largest circles, so anyway the associated points with the set of all circles with different curvatures, are discontinuous w.r.t the center point (total curvature, where pi=circumference/diameter does not hold) and w.r.t the line (total non-curvature, where pi=circumference/diameter does not hold), or in other words, the set of all these intersecting points, does not completely cover an infinitely long straight line.
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However the set of all concentric circles “with different curvatures” that are centered on some point off your line does have a smallest circle that intersects your line as well as a subset of circles “with different curvatures” that do not intersect your line at all. Your line is completely covered by those circles that do intersect your line. Again your assertion was that your line was not completely covered by points yet you could show no point on your line that was not a, well, point. So now you have moved to trying to show that your line can’t be covered by “intersecting points” of “The set of all circles with different curvatures”. Yet again you can identify no point on your line that can not be intersected by a member of some set of concentric circles of different curvatures.


Oh and again…

The Man said:
So by all means please explain to us the difference between changing and unchanging with “no past (before) and no future (after)”?
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Originally Posted by jsfisher
All unit circles have exactly the same curvature, that being 1.
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doronshadmi said:
Nonsense again:
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Your love affair with 1+1=3 not even death could part . . . .
[qimg]http://upload.wikimedia.org/wikipedia/commons/thumb/f/f7/ApollianGasketNested_2-20.svg/360px-ApollianGasketNested_2-20.svg.png[/qimg]
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Your inability to understand the term "unit circle" at least fetched a graphic representation of a countably infinite set. Feel free to demonstrate your "systematic mapping" of its power set and bless the board with some more circles.
 
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epix said:
stop presenting various circles as some arguments of yours -- unless you want to further participate as a subject in the hitherto unknown medical field called "Analytic Psychopathy."
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They say it takes one to know one.
 
ehcks said:
The change in slope per 1 degree rotation is the same, regardless of the radius?
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That's right. The angular change of 1 degree equals 1 degree change in slope for any radius. The rate of change of slope is not the same as the curvature of a circle given by 1/radius. The former measure is used with curves, such as the sine curve, where there is no center point that would connect line m with a given point on the curve resulting in curvature 1/m, like 1/r in the case of the circle.
 
The Man said:
So does your “set of all circles with different curvatures” include “all circles with different curvatures” or just the ones that intersect your line at two points?
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The set of all circles with different curvatures includes those different circles, whether they intersect some infinitely long straight line, or not.

If intersecting, then any intersection is associated to some circle, which its curvature is different than the rest of the circles that are included as unique members (by their different curvatures) of the set of all circles with different curvatures.

Since each circle of the set of all circles with different curvatures is included once and only once as a member of this set, then any unique circle, which does not intersect the considered infinitely long straight line or it is used as a tangent circle with the considered line, its unique curvature is already used, and as a result it does not have a pair of intersecting points, which are associated with it, and this pair does not cover the infinitely long straight line.

As a result the set of all points along an infinitely long striated line, is incomplete, even if one of the unique circles is used as a tangent circle (in this case the other point of the potential pair of the tangent circle, is not on the infinitely long striated line).
 
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epix said:
I thought that you liked and accepted Cantor's diagonal proof of unaccountability of real sets. Remember? You enslaved the diagonal proof and raped it repeatedly into the <0,1> argument.
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Wrong again epix, Cantor's construction method of all P(S) members, is independent of the logical conclusions of his theorem.
 
epix said:
I thought that you liked and accepted Cantor's diagonal proof of unaccountability of real sets. Remember? You enslaved the diagonal proof and raped it repeatedly into the <0,1> argument.
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epix you still do not understand the independence of Cantor's systematic constriction method of all P(S) members, and the logical conclusions of Cantor's theorem, as explicitly shown in
http://www.internationalskeptics.com/forums/showpost.php?p=6986611&postcount=14585

epix said:
Let me help you . . .

Definition: Any set is countable.

If so, continuity doesn't exist and there are no lines and curves. If there are no curves, there is no curvature. If there is no curvature, there are no circles. So please stop presenting various circles as some arguments of yours -- unless you want to further participate as a subject in the hitherto unknown medical field called "Analytic Psychopathy."
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Let me help you, a line segment is not the collection of all the points along it, simply because the smallest pair along the considered line segment, is not satisfied.

In other words the a existence of a line or a circle are independent of the existence of collection of all the points along them.
 
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doronshadmi said:
Wrong again epix, Cantor's construction method of all P(S) members, is independent of the logical conclusions of his theorem.
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LOL. I thought that any conclusion depends on the examination of the material presented. According to you, Cantor examined x|x in R: 0≤x≤1, went to the bathroom and came up with a proof by looking inside the toilet bowl.
 
epix said:
LOL. I thought that any conclusion depends on the examination of the material presented. According to you, Cantor examined x|x in R: 0≤x≤1, went to the bathroom and came up with a proof by looking inside the toilet bowl.
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You are right, he did not look also beyond the toilet bowl.

A researcher really does a valuable research only if he\she is both an external observer AND an internal participator of the researched subject.
 
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jsfisher said:
How about the set {} and its power set {{}}? What would be a bijection between the members of those two sets?
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Please provide the members of {} to be mapped with.

Since they do not exist we get the member of {{}} which is mapped with nothing:

{} ↔


But if the minimal term must be an existing set then the empty set is mapped with the member of the power set of the empty set, where the power set of the empty set is {{}}.

In this case we get:

{} ↔ {}
 
doronshadmi said:
epix you still do not understand the independence of Cantor's systematic constriction method of all P(S) members, and the logical conclusions of Cantor's theorem, as explicitly shown in
http://www.internationalskeptics.com/forums/showpost.php?p=6986611&postcount=14585


Let me help you, a line segment is not the collection of all the points along it, simply because the smallest pair along the considered line segment, is not satisfied.

In other words the a existence of a line or a circle are independent of the existence of collection of all the points along them.
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Just unbelievable. :rolleyes:

The existence of objects with dimension d>0 depends on the continuum, which in turn is a result of any 'a' divided by any 'b' apart from zero equals 'c.' As a new point C appears between A and D, another point B appears between A and C and point B' between C and D with the number of new points appearing exponentially. This "dynamic" disables the process of counting defined by the 1-on-1 correspondence also known as bijection, as Cantor proved:
http://wiki.answers.com/Q/How_do_you_prove_that_the_continuum_is_uncountable

In order to prove that any set is countable you need to disprove Cantor's diagonal proof, which is a piece of cake for you -- any random thought of yours wearing the mask of logic will obviously do.
 
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epix said:
Just unbelievable.
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You are right, no belief of any kind is needed, in order to understand the following facts:

1) The members an infinite set are in a 1-to-1 and onto (bijection) with the members of any arbitrary infinite subset of the considered set ( you are still missing http://www.internationalskeptics.com/forums/showpost.php?p=6986611&postcount=14585 ).

2) Any additional point between the extreme endpoints of a given line segment, is resulted by more lines with end points, etc. ad infinitum, where the points are actually different than each other exactly because given any scale level, there is an uncovered line between any closer pair of points (the closest pair of points does not exist exactly because there is always an uncovered line between them) along the finitely long line.
 
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doronshadmi said:
You are right, no belief of any kind is needed, in order to understand the following facts:

1) The members an infinite set are in a 1-to-1 and onto (bijection) with the members of any arbitrary infinite subset of the considered set ( you are still missing http://www.internationalskeptics.com/forums/showpost.php?p=6986611&postcount=14585 ).
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Your references reside well outside the realm of reason.

Just focus on disproving Cantor's diagonal argument, otherwise your idea of any set being countable wouldn't free itself from the constrains of pure irrational belief.
 
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