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Apr 24, 2023 at 16:00 history edited Glorfindel CC BY-SA 4.0
broken link fixed, cf. https://meta.mathoverflow.net/q/5301/70594
Mar 3, 2016 at 10:25 vote accept SBF
S Dec 12, 2015 at 13:23 history suggested Ali Taghavi
I add a tag
Dec 12, 2015 at 12:52 review Suggested edits
S Dec 12, 2015 at 13:23
Jan 1, 2014 at 21:06 answer added 7891user timeline score: 4
Jan 1, 2014 at 17:32 comment added SBF @D.Kelleher: sorry, I misread your previous comment - now I understand what you meant, thanks for clarifying. The idempotence of $\operatorname{sco}$ holds at least for analytic sets (see my answer below).
Jan 1, 2014 at 17:30 answer added SBF timeline score: 2
Jan 1, 2014 at 16:25 comment added D. Kelleher @Ilya Ahh, I'll clarify. The Choquet theorem doesn't use idempotence of $\operatorname{sco}$. What it says is that Convex sets are exactly the Range of $\operatorname{sco}$. My suggestion was: if we knew that $\operatorname{sco}$ is idempotent, the Choquet theorem implies that convex sets are all of the fixed points of $\operatorname{sco}$. But since I don't know for a fact this is true, I didn't feel right calling it an "answer."
Jan 1, 2014 at 14:35 comment added SBF @D.Kelleher: I see, could you clarify where the assumption of the idempotence of $\operatorname{sco}$ is used in Choquet's theorem? I don't see it is mentioned in the formulation of the result.
Dec 28, 2013 at 15:06 comment added D. Kelleher @Ilya The Choquet theorem is on the wiki page (though it should be in "Lecture notes on" too), and essentially says that a convex set is $\operatorname{sco} P$ for some $P$. I would think that it should be idempotent pretty generally, but i don't know off hand, and generalizing the proof from convex combinations runs into measure issues.
Dec 28, 2013 at 13:03 comment added SBF @D.Kelleher: can you suggest any sufficient conditions for the idempotence of $\operatorname{sco}$? Also, the Choquet's theorem you are talking about can be found in "Lecture notes on" I guess, isn't it?
Dec 28, 2013 at 4:03 answer added Gerald Edgar timeline score: 3
Dec 27, 2013 at 22:09 comment added D. Kelleher The study of this sort of convex hull is the domain of Choquet theory. In fact assuming $\operatorname{sco}$ is idempotent, the Choquet theorem implies that convex sets are exactly the fixed points of $\operatorname{sco}$. The wiki page has some good references en.wikipedia.org/wiki/Choquet_theory
Dec 27, 2013 at 18:51 comment added SBF @MichaelGreinecker: by a convex combination I mean the integral over a "weighting" measure $\nu$ which gives a full outer measure to $P$ (at least I'm not aware of any other definition to be used here). Can you say that $\operatorname{sco}$ is idempotent?
Dec 27, 2013 at 17:40 comment added Michael Greinecker Why is this "exactly the set of all convex combinations"? And wouldn't that answer the question since taking convex hulls is an idempotent operation?
Dec 27, 2013 at 15:58 history edited SBF CC BY-SA 3.0
added 210 characters in body
Dec 27, 2013 at 13:48 history asked SBF CC BY-SA 3.0