Take the 2-minute tour ×

Here is part of the question in my HW. Let$\ \{C_n\}\subset X $ be a bounded nested decreasing sequence of closed and convex sets. I am asked to show that $\bigcap C_n \not= \emptyset$ iff X is reflexive. My intuitive tell me that Kakutani theorem can possibly be used to find a convergent subsequence in each $C_n$. However,I can't show that these sequences converge to a same point. Alternatively, The hint says that the sets $C_n$ are weakly closed by using the Separating Hyperplane theorem and its corollaries. If I can show the weakly closed part , then by weakly compactness there might be an accumulation point in the intersection.

Could anyone tell me how to show the sequence is weakly closed as suggested in the hint and the nonempty intersection part? Thanks in advance !

share|improve this question

1 Answer 1

up vote 1 down vote accepted

A closed an convex set $C$ in a Banach space is also weakly closed: Suppose $x \not\in C$, then by Hahn-Banach (the seperating hyperplane theorem, I suppose) there is a functional $x^* \in X^*$ and a $\alpha \in \mathbb R$ such that $\sup_{y\in C}\Re x^*(y) \le \alpha$, $\Re x^*(x) > \alpha$. Now $\{\Re x^* > \alpha\}$ is a weakly neighbourhood of $x$ disjoint from $C$, so $X\setminus C$ is weakly open, hence $C$ is weakly closed.

For the non-empty intersection part, we use the weak compactness: As the $C_n$ are decreasing and not empty, finitely many of them have non-empty intersection, as they are compact, $\bigcap_n C_n \ne \emptyset$.

share|improve this answer
Thank you ! But why by weakly compactness, it suffices to show any finite subsequence has nonempty intersection? did you try to use the fact that X is reflexive somewhere ? –  teshile Oct 23 '12 at 12:13
This method works for strong compactness, but why does it work for weak compactness ? –  teshile Oct 23 '12 at 12:35
I used the reflexivity in using weak compactness of the $C_n$: A space $X$ is reflexive iff all its closed convex subsets are weakly compact. For the finite intersection part: That's a property of compactness as a topological property. We use it for the weak topology here, it follow from the fact, that if $\bigcap C_n = \emptyset$, $\{X \setminus C_n \mid n \ge 2\}$ where an weakly open cover of $C_1$ without a finite subcover. –  martini Oct 23 '12 at 12:50
@martiniThanks again ! is there a proof (maybe a link...) for A space X is reflexive iff all its closed boundede subsets are weakly compact? –  teshile Oct 23 '12 at 13:03
The direction needed here is easy: If $X$ is reflexive, $(B_X, \text{weak})$ is homeomorphic to $(B_{X^{**}}, \text{weak$^*$})$ and the latter is compact by Banach-Alaoglu. Now a closed bounded convex set is a weakly closed (by the above) subset of some $n\cdot B_X$, hence compact. –  martini Oct 23 '12 at 13:15

Your Answer


