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Aug 26, 2016 at 13:28 vote accept Amin
Aug 26, 2016 at 12:47 answer added Tony Prochazka timeline score: 4
Aug 25, 2016 at 22:03 comment added Amin If $x \not\in (x_n)_{n=0}^{\infty}$ then in particular $x \neq 0$, and this implies that for some $k$, the $k$-th coordinate of $x$ is not zero, hence we can choose $E(x) := \{k\}$ (I think). So I think your example is correct. I'm just curious to know whether an explicit representation could be worked out. The proofs of Corson-Lindenstrauss or Mazur property do not immediately provide the explicit representation, at least my attempts have been futile.
Aug 25, 2016 at 16:23 comment added Ben W It looks like the first condition (every weakly compact convex set is the intersection of cells) follows from the Mazur Intersection Property for $\ell_2$. (However it is apparently a deep result, and not true for arbitrary Banach spaces, although it is true for reflexive spaces with Frechet-differentiable norm.) I'm still thinking about the second condition.
Aug 24, 2016 at 23:18 comment added Amin This is an interesting example.
Aug 24, 2016 at 20:47 comment added Ben W I'm guessing that every weakly compact convex set is the intersection of cells. If so, then it is clear that for each finite set $E$ of $\mathbb{N}_0:=\mathbb{N}\cup\{0\}$, the set $(x_n)_{n\in E}\cup\overline{co}(x_n)_{n\notin E}$ the intersection of finite unions of cells, where $\overline{co}$ denotes "closed convex hull" and $(x_n)_{n=0}^\infty$ is formed from $x_0=0$ and $x_n=e_n$ for $n\in\mathbb{N}$. Now, is it true that for any $x\notin(x_n)_{n=0}^\infty$ we can find a finite subset $E(x)$ of $\mathbb{N}_0$ such that $x\notin\overline{co}(x_n)_{n\notin E(x)}$? Again I'm guessing yes.
Aug 22, 2016 at 21:52 history asked Amin CC BY-SA 3.0