Weak closure of intersection in reflexive Banach space

Question

Let $X$ be a reflexive Banach space. Let $\mathcal{S}$ be a family of finite-dimensional subspaces of $X$. Consider a bounded sequance $(x_Y)_{Y\in \mathcal{S}}\subset X$. Define $$C_Y=\bigcup_{Y'\supseteq Y, Y'\in \mathcal{S}}\{x_{Y'}\}.$$ How to prove that $$\bigcap_{Y\in \mathcal{S}}\overline{C_{Y}}^{\,w}\neq \emptyset\quad?\qquad (1)$$

First, observe that there exists $M>0$ large enough such that $C_Y\subset \overline{B_M}$. I considered a family $\mathcal{F}=\{C_Y\,|\, Y\in \mathcal{S}\}$. Pick $n\in \mathbb{N}$ and take $Y_1, Y_2,..., Y_n \in \mathcal{S}$. Without a loss of generality we may assume that $$Y_1\subseteq Y_2\subseteq... \subseteq Y_n.$$ Then, we have $$C_{Y_n}\subseteq...\subseteq C_{Y_2}\subseteq C_{Y_1}$$ and so $\bigcap_{i=1}^nC_{Y_i}\neq \emptyset$. Now, from finite intersection property $\bigcap_{Y\in \mathcal{S}}C_{Y}\neq \emptyset$. Does it imply (1)? Is really reflexivity needed?

Answer 1

Let me give a counterexample in $X = \ell^2$. We consider the unit sequences $\{e_n\}$, $Y_n = \operatorname{span}(e_n)$, $S = \{Y_1,Y_2,\ldots\}$ and $x_{Y_n} = e_n$.

Then, $C_{Y_n} = \{e_n\}$ and thus $$ \bigcap_{Y \in S} \overline{C_Y}^w = \emptyset. $$

Answer 2

For any $Y\in \mathcal{S}$ we know that $C_Y\subseteq\overline{B_M}$ for sufficiently large $M>0$. From Kakutani theorem we know that $\overline{B_M}$ is weakly compact. Pick $n\in \mathbb{N}$ and let $Y_1,...,Y_n\in \mathcal{S}$. Consider $$\overline{C_{Y_i}}^{\, w}$$ for $i=1,...,n$. Since $\overline{C_{y_i}}^{\,w}$ are weakly closed and $$\bigcap_{i=1}^{n}\overline{C_{Y{i}}}\neq \emptyset,$$ then from finite intersection property we obtain (1).