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Let S be a subset of a linear space. Let S1 be the union of all line segments that join pairs of points in S. Now what happens if we repeat this process and construct S2, S3,....(Thus for example S2 is the union of line segments in S1)?

My guess is that the end-result should be the convex hull of the set S, but I am not able to prove/disprove this.

Thank you,


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Unless I'm missing something, it is. Outline of argument: your set is clearly convex and contains S; conversely, a convex set containing S must contain Sn for each n (induction). –  Yemon Choi Mar 2 '11 at 0:52
Ohhh yes indeed!! I guess that settles it....One question though... Does this constructed chain S1, S2, S3,... have to be infinite to become the convex hull? In R^2 it seems to me that S2 of any set becomes the convex hull. –  smilingbuddha Mar 2 '11 at 0:58
@Sanjeev, by Carathéodory's theorem $S_n$ is the convex hull in $\mathbb R^n$. –  Gjergji Zaimi Mar 2 '11 at 1:17
@Zaimi : You do not answer the question of finiteness, if S is finite it looks true (intuitively) that the S1..Sn is finite. If S is infinite I think it is false : Let S be a set of enumerable points on the circumference with one point of accumulation. –  Jérôme JEAN-CHARLES Mar 2 '11 at 1:51
Am I missing something? Caratheodory's theorem implies that any point of the convex hull is inside an n-simplex with vertices in $S$. The cardinality of $S$ doesn't seem to have anything to do with this... –  Gjergji Zaimi Mar 2 '11 at 2:35

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