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Deeper than primes

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You are right, it is between {} and itself.

Of course I am right. You presented something completely unrelated to the task at hand. Are you looking to try again, or will you finally admit your claim was bogus?

In case you forgot (or wanted to dodge and weave away from), the claim was that a bijection exists between the members of any set S and its power set.
 


Nope, didn't miss it. That link is to a post where you link to a post where you tried and failed. Nothing difficult there to understand. You were attempting to present a particular bijection. You got to the end, proudly wrote "Ta da!!", and then sat back smugly. The only problem, of course, was that your result was completely wrong and nothing to be proud of.

So, will you now retract your bogus claim?
 
The set of all circles with different curvatures includes those different circles, whether they intersect some infinitely long straight line, or not.

So the first circle that intersects your line is not “omitted” as you claimed before.


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.

Glad we could clear that up.



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.

Don’t forget there is a whole subset of “circles with different curvatures” that do not intersect your line when those concentric circles are not centered on your line.


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.

How many “circles with different curvatures” do you have? Have you run out for some reason?

You seem to be thinking as if you had only a finite number of circles to use. So tell us Doron how many circles “with different curvatures” are in your “set of all circles with different curvatures”?

Again as noted above: there is a whole subset of “circles with different curvatures” that do not intersect your line when those concentric circles are not centered on your line. That some of those circles do not intersect your line in no way restricts the rest of those concentric circles from intersecting every point on your line. So you still need to identify what point(s) on your line can not be intersected by some member of that set of concentric circles that is not centered on your line.

You’ve gone a long way around once again only to end up right where you were before, unable to identify any point on your line that is not covered by a point and now even just not intersected by a concentric circle not centered on your 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).

Doron the fact that one can define a set that includes some, none or even all but one point of your line in no way makes your line incomplete. In fact we can define any number of sets that include all points along your line, the simplest one being, well, your line.

Again this is just your typical self contradictory nonsense: Please identify what member(s) of “the set of all points along an infinitely long striated line” “is not on the infinitely long striated line”.
 
Don’t forget there is a whole subset of “circles with different curvatures” that do not intersect your line when those Don’t forget there is a whole subset of “circles with different curvatures” that do not intersect your line when those concentric circles are not centered on your line. are not centered on your line.

Again you demonstrate your misunderstanding of the expression "the set of all circles with different curvatures".

This set is exactly those concentric circles, and any given circle with a unique curvature is already one of those concentric circles, so your argument does not hold water.
 
Not at all. That's the post where you tried but failed to present a bijection between the elements of {} and {{}}.

Why do you continue to highlight your failures, Doron?
Wrong, that is a post that your weak reasoning can't comprehend.

Why do you continue to use this weak reasoning?
 
Again you demonstrate your misunderstanding of the expression "the set of all circles with different curvatures".

This set is exactly those concentric circles, and any given circle with a unique curvature is already one of those concentric circles, so your argument does not hold water.

Again you demonstrate your (apparently deliberate) misunderstanding of the fact that all points along your line are still just points along your line. That can all be intersected by any number of sets of geometrical objects, including concentric circles. Also your, again apparently deliberate, misunderstanding of the fact that simply defining a set that doesn’t include some, all, or just one point along your line does not make that set nor the set of all points along your line incomplete.



ETA: So even though you quoted quoted it twice it twice you’re still deliberately forgetting that it is a subset of your “set of all circles with different curvatures" when those concentric circles are not centered on your line?
 
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Wrong, that is a post that your weak reasoning can't comprehend.

Why do you continue to use this weak reasoning?


Doron, stop trying to change the subject. You made a claim. You were called out on it. You have tried repeatedly to support the claim, but every single time you tried to support it with an example you have failed.

You failed every single time.

No amount of your "Oh, but you don't understand this other, equally bogus claim" reverses reality: You failed on every single attempt.

Let's just deal with one of your failures at a time. Once you retract the nonsense claim about bijections between the members of any set and its power set, then we can move on to another of your failures. But let's not jump around.
 
Again you demonstrate your (apparently deliberate) misunderstanding of the fact that all points along your line are still just points along your line. That can all be intersected by any number of sets of geometrical objects, including concentric circles. Also your, again apparently deliberate, misunderstanding of the fact that simply defining a set that doesn’t include some, all, or just one point along your line does not make that set nor the set of all points along your line incomplete.
Wrong The Man,

The set of all circles with different curvatures is associated with all possible points along the infinitely long straight line, and yet there is discontinuity between these associated points and the center point, and the infinitely long straight line is permanently beyond the set of these associated points, because of a very simple reason:

pi is not a property of the center point (total curvature) or the infinitely long straight line (total non-curvature).

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

Since the Circle's game can't be comprehended by you, I have changed it to the Line segment game, as follows:

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.
 
You made a claim.
But you, jsfisher, can't comprehend it.


Again, you project your own failings onto others. You claimed there was a bijection between the members of any set and its power set. The claim is simple enough to understand, yet you don't. Is there a specific part that is causing you the most trouble?

Are you yet willing to retract it, since it is a bogus claim? It is also key to another of your failings, that for any set S, |S| = |P(S)|, but let's continue on just one failure at a time. Jumping around only confuses you more, Doron.
 
By using his weak reasoning, jsfisher simply fails to get the following post:

http://www.internationalskeptics.com/forums/showpost.php?p=7019615&postcount=14694

He simply can't comprehend that this post is about Set's fundamental foundations.


Still highlighting your failure, eh? Oh, wait, you aren't trying to compound it, now, are you? Are you perhaps now claiming you did provide a bijection between the members of {} and {{}}??

Your failures only compound, Doron.

Simple question, Doron, that even you should comprehend: Do you still claim there is a bijection between the elements of any set and its power set?
 
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