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?
countable intersection of closed convex bounded subsets reflexive banach space is non empty.
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banach-spaces
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0How does one mark some answers as correct? – 2012-06-10
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0http://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – 2012-06-10
2 Answers
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Hint: Do you know the Eberlein-Shmulyan theorem?
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0Would 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. – 2012-06-10
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0I 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? – 2012-06-10
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0Also, I have seen many versions of the theorem on the internet. Which one is applicable here? – 2012-06-10
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1If $x_n \in C_n$, Eberlein-Shmulyan says some subsequence has a weak limit point $x$. Show that $x$ is in all the $C_n$. – 2012-06-10
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the proof can be seen in Introduction to Banach Space Theory by Robert E. Megginson.
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1Please try to describe as much here as possible in order to make the answer self-contained. – 2013-04-30