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If $X$ is a reflexive Banach space and $(C_n), n \in \mathbb{N}$ is a sequence of closed convex bounded sets with $C_{n+1}$ contained in $C_n$ for all $n \in \mathbb{N}$. How does one show that the countable intersection of $C_n$ for $n \in \mathbb{N}$ is not the empty set?

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Hint: Do you know the Eberlein-Shmulyan theorem?

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  • $\begingroup$ Would it be possible to please give me a few more details? I am not too aware of how to use it to prove the statement. $\endgroup$ – nada Jun 10 '12 at 18:45
  • $\begingroup$ I need to prove that X being reflexive is weak sequentially compact(by the Eberlein Shmulyan theorem). But how does that follow from the statement given in the question? $\endgroup$ – nada Jun 10 '12 at 20:15
  • $\begingroup$ Also, I have seen many versions of the theorem on the internet. Which one is applicable here? $\endgroup$ – nada Jun 10 '12 at 20:19
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    $\begingroup$ If $x_n \in C_n$, Eberlein-Shmulyan says some subsequence has a weak limit point $x$. Show that $x$ is in all the $C_n$. $\endgroup$ – Robert Israel Jun 10 '12 at 22:00
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the proof can be seen in Introduction to Banach Space Theory by Robert E. Megginson.

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    $\begingroup$ Please try to describe as much here as possible in order to make the answer self-contained. $\endgroup$ – robjohn Apr 30 '13 at 13:29

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