User talk:Sven.st

This looks like a proof around a deleted neighborhood.

If, naively, I took the function expression Gi(Ai)=(Bi). I would read this as a proof of proportionality. y=kx

Why could I not rearrange to Gi=(Bi)/(Ai)

In addition if in n≥3

we multiply by unity in the form of 6/6 along the set of n

and arrive at 18/6, 24/6, 30/6, 36/6 and so on

Grouping these in bundles of three, so 18/6 becomes 3 groups of 6 gons, 24/6 becomes 4 groups of 6 gons, 30/6 becomes 5 groups of 6 gons and so on. The numerator labels the circle, each distinct.

By bounding out 1 and 2 we removed parity by bounding out the origin. We return to parity by multiplying by unity in 6/6. Factored as 2*3.

We can then translate and rotate in a very specific ways. Along the surface in translation and and proportionately by multiples of the radius of curvature.

Since n is positive and ascending, I read the resulting spheres as concave always. The spheres second half is mapped to the rear surface of it's front half. Each surface is modeled as a radius of curvature. Any n makes a distinct sphere. It translates around a corkscrew spiral which visits every surface in order from n=3.


 * I am a layman too so take everything I say with a grain of salt and consider asking on the talk page of the article, if you still have questions.


 * First I want to make sure we are talking about the same thing: You seem to be referring to my edit on the Banach–Tarski paradox.
 * So $$A_i$$ and $$B_i$$ are subsets of a set $$X$$ and $$g_i$$ is an element of some group $$G$$.
 * I am not very experienced with group theory so I prefer to just think about the common special case mentioned in the article, where $$X$$ is an $$n$$-dimensional Euclidean space and $$G$$ is the group of all isometries of $$X$$(so $$A_i, B_i$$ are sets of points and $$g_i$$ is a transformation composed of translation, rotation and/or reflection).
 * In my understanding the formula $$g_i(A_i)=B_i$$ does therefore not mean that $$g_i$$ gets multiplied by $$A_i$$, but that the action of $$g_i$$ is applied to $$A_i$$.


 * If you look at the equation this way you will notice that you can not rearrange it by dividing both sides by $$A_i$$ just like you can not rearrange $$\sin(x)=y$$ to $$\sin=\frac{y}{x}$$. In addition to that "dividing" by $$A_i$$ is not even defined, because $$A_i$$ is not a number, but a set of points.


 * I don't think it is right to say, that $$g_i(A_i)=B_i$$ means that $$A_i$$s and $$B_i$$s are proportional (i.e. they can be transformed into each other by multiplication with a constant factor), but that they are congruent(i.e. they can be transformed into each other by a length-preserving transformation).


 * I did not understand the second part of your question completely, maybe you can phrase it in a different way and clarify a few points? $$n$$ is the dimension of the Euclidian space $$X$$? Then n is surely positive, but why should it be "ascending"? What do you want to multiply by 6/6? All possible dimensions $$n$$? I am not sure why the Banach-Tarski-Paradox should be related to deleted neighbourhoods, what exactly looks similar to which proof involving deleted neighbourhoods?


 * PS: According to Signatures you should add a signature to your posts to talk pages by appending four tilde characters  to it. Also if you use a heading then it is easier to see to which discussion a post belongs.


 * Sven.st (talk) 18:10, 22 August 2017 (UTC)