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A collection $\mathcal{F}$ has finite intersection property if for every finite collection of sets from $\mathcal{F}$, namely $\{F_1,...,F_n\}\subset \mathcal{F}$, we have $$\bigcap_{i=0}^{n}F_i\neq\emptyset.$$

Also, we have a theorem: Let $X$ be a topological space. The space $X$ is compact iff for every collection of closed sets $\mathcal{F}$ having a finite intersection property, we have $$\bigcap_{F\in \mathcal{F}}F\neq\emptyset.$$

Is it true, if we consider a closed ball instead, for example, of a normed space $X$? For instance, consider $B(\theta,1)$. From Banach - Alaoglu theorem, it is weakly compact. Suppose that $\mathcal{F}\subset B(\theta, 1)$ is a family of weakly closed subsets which has finite intersection property. Then, can we have $$\bigcap_{F\in \mathcal{F}}\overline{F}^w\neq\emptyset\,?$$

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  • $\begingroup$ Banach-Alaoglu tells you that unit ball of $X^*$ is compact for the weak$^*$ topology. You probably want to consider the unit ball of $X^*$ for the weak$^*$ topology or assume $X$ to be reflexive (the unit ball of $X$ is weakly compact iff $X$ is reflexive). $\endgroup$ – Rhys Steele Jun 4 '18 at 13:04
  • $\begingroup$ You're right, we shall use Kakutani theorem instead $\endgroup$ – zorro47 Jun 4 '18 at 13:11
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Yes. Assume that you don't have $\bigcap\limits_{F\in\mathcal{F}}\bar{F}^w \neq\emptyset$. Let $U = (\bar{F}^w)^c$ be the complement of $\bar{F}^w$ in $B(\theta,1)$ and $\mathcal{U} = \{U=(\bar{F}^w)^c:F\in\mathcal{F}\}$, then $\bigcup\limits_{U\in\mathcal{U}}U$ is a weak open cover for $B(\theta,1)$, which is compact, and thus $\exists \{U_1,\ldots,U_n\}$ a finite weak open subcover. But then $\bigcap\limits_{i=1}^n \bar{F_i}^w=\emptyset$ which contradicts that $\mathcal{F}$ has the FIP.

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