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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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Maybe you should start marking some answers as correct on your other questions. An accept rate of 0% does not encourage people to help you. –  utdiscant Jun 10 '12 at 18:42
    
How does one mark some answers as correct? –  nada Jun 10 '12 at 18:43
    
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2 Answers 2

Hint: Do you know the Eberlein-Shmulyan theorem?

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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. –  nada Jun 10 '12 at 18:45
    
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? –  nada Jun 10 '12 at 20:15
    
Also, I have seen many versions of the theorem on the internet. Which one is applicable here? –  nada Jun 10 '12 at 20:19
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If $x_n \in C_n$, Eberlein-Shmulyan says some subsequence has a weak limit point $x$. Show that $x$ is in all the $C_n$. –  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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Please try to describe as much here as possible in order to make the answer self-contained. –  robjohn Apr 30 '13 at 13:29
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