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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$\begingroup$ How does one mark some answers as correct? $\endgroup$– nadaCommented Jun 10, 2012 at 18:43
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$\begingroup$ meta.stackexchange.com/questions/5234/… $\endgroup$– utdiscantCommented Jun 10, 2012 at 19:00
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2 Answers
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Hint: Do you know the Eberlein-Shmulyan theorem?
answered Jun 10, 2012 at 18:38
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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. $\endgroup$– nadaCommented Jun 10, 2012 at 18:45
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$\begingroup$ 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? $\endgroup$– nadaCommented Jun 10, 2012 at 20:15
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$\begingroup$ Also, I have seen many versions of the theorem on the internet. Which one is applicable here? $\endgroup$– nadaCommented Jun 10, 2012 at 20:19
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1$\begingroup$ If $x_n \in C_n$, Eberlein-Shmulyan says some subsequence has a weak limit point $x$. Show that $x$ is in all the $C_n$. $\endgroup$ Commented Jun 10, 2012 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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1$\begingroup$ Please try to describe as much here as possible in order to make the answer self-contained. $\endgroup$– robjohn ♦Commented Apr 30, 2013 at 13:29