Timeline for Planar sets closed under intersection of circles

Current License: CC BY-SA 3.0

12 events

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May 18, 2012 at 7:55	comment	added	Denis Serre		@Gjergji. I remark a flaw in my definition of the property: the vertices of the square (or its diagonals as well) are on a circle $C$ which can be written as a $\Gamma_A$ for four differents triplets $A$. Thus $C=\Gamma_A\cap\Gamma_B$ for some $A\ne B$. It seems that we should have $C\subset X$ !!! I modify my definition accordingly.
May 18, 2012 at 7:28	vote	accept	Denis Serre
May 18, 2012 at 7:28	comment	added	Denis Serre		@Gjergji. Im OK with your answer now. The picture with six vertices is interesting. It raises questions.
May 16, 2012 at 17:39	comment	added	Gjergji Zaimi		@Denis, I added a few more details. You're right, a rectangle and it's center is only circularly stable if it is a square.
May 16, 2012 at 17:37	history	edited	Gjergji Zaimi	CC BY-SA 3.0	added 871 characters in body
May 16, 2012 at 6:15	comment	added	Denis Serre		@Gjergji. Isn't it a problem in the second case ? Let us denote $a,b,c,d$ the points on the rectangle ($ab$, ..., $da$ being edges), then $O$ its center. The intersection of $\Gamma_{abO}$ and $bc=\Gamma_{bc\infty}$ has an other point, not in $A$.Can we really have $\|A\|=6$ ? Even a $5$-element circularly stable $A$ is not clear to me.
May 16, 2012 at 5:59	vote	accept	Denis Serre
May 16, 2012 at 6:11
May 15, 2012 at 15:58	history	edited	Gjergji Zaimi	CC BY-SA 3.0	added 308 characters in body
May 15, 2012 at 15:27	comment	added	Gjergji Zaimi		@Misha, thank you, I changed projective to Mobius :)
May 15, 2012 at 15:24	history	edited	Gjergji Zaimi	CC BY-SA 3.0	edited body
May 15, 2012 at 15:22	comment	added	Misha		@Gjerhji: I think, meant complex-projective (or Moebius) transformation (since real ones do not map circles to circles).
May 15, 2012 at 15:09	history	answered	Gjergji Zaimi	CC BY-SA 3.0