10. [6] Find the number of nonempty sets $\mathcal{F}$ of subsets of the set $\{1, \ldots, 2014\}$ such that:
(a) For any subsets $S_{1}, S_{2} \in \mathcal{F}, S_{1} \cap S_{2} \in \mathcal{F}$.
(b) If $S \in \mathcal{F}, T \subseteq\{1, \ldots, 2014\}$, and $S \subseteq T$, then $T \in \mathcal{F}$.
$\{1, \ldots, 2014\}$. It can be checked that the sets $\mathcal{F}_{S}$ satisfy the conditions 1 and 2 . We claim that the $\mathcal{F}_{S}$ are the only sets of subsets of $\{1, \ldots, 2014\}$ satisfying the conditions 1 and 2. (Thus, the answer is the number of subsets $S$ of $\{1, \ldots, 2014\}$, which is $2^{2014}$.)
Suppose that $\mathcal{F}$ satisfies the conditions 1 and 2 , and let $S$ be the intersection of all the sets of $\mathcal{F}$. We claim that $\mathcal{F}=\mathcal{F}_{S}$. First, by definition of $S$, all elements $T \in \mathcal{F}$ are supersets of $S$, so $\mathcal{F} \subseteq \mathcal{F}_{S}$. On the other hand, by iterating condition 1 , it follows that $S$ is an element of $\mathcal{F}$, so by condition 2 any set $T$ with $S \subseteq T \subseteq\{1, \ldots, 2014\}$ is an element of $\mathcal{F}$. So $\mathcal{F} \supseteq \mathcal{F}_{S}$. Thus $\mathcal{F}=\mathcal{F}_{S}$.