Give a family of sets $\{F_a\}$ with each $F_a\subseteq (0,1)$ and $F_a \cap F_b \neq \emptyset$ but $\bigcap_aF_a = \emptyset$. Give a family of sets $\{F_a\}$ with each $F_a\subseteq (0,1)$ and $F_a \cap F_b \neq \emptyset$ but $\bigcap_aF_a = \emptyset$.
What is an example of such a family of sets?
 A: A standard example in analysis is to let your $F_n = (0,1/n), n \in \mathbb{N}$. This ensures that any two $F_n$ intersect, but the intersection of all of the $F_n$ is $\emptyset$. 
Specifically, for $n < m$, $F_n \cap F_m = F_m$. However, for any $a > 0$, $F_n$ does not contain $a$ when $1/n < a$.
A: You can go much easier on this. Pick any three numbers $a_1,a_2$ and $a_3$ in $(0,1)$ and let
$$F = \{\{a_1, a_2\}, \{a_1, a_3\}, \{a_2, a_3\}\}$$
The family is finite and indexed by $\{1,2,3\}$, $F_1\cap F_2 = \{a_1\}, F_1\cap F_3 = \{a_2\}, F_2\cap F_3 = \{a_3\}$ but
$$\bigcap_{i=1}^3 F_i = \emptyset$$
This example is minimal in the sense that if the index set is any smaller ($2$ sets), it doesn't work anymore and if the sets are any smaller ($1$ element per set), it doesn't work either.

For a general construction prescribed a finite index set $I \subset \mathbb N$ with $|I| \ge 3$ you can simply pick an indexed set $A$ and define
$$A(S) := \{a_s \mid s\in S\}\\
F_i := A(I \setminus \{i\})$$
Then $F_i\cap F_j = A(I\setminus\{i,j\}) \ne \emptyset$ since $|I| > 2$ and
$$\bigcap_{i\in I} F_i = A\left(I \setminus \bigcup_{i\in I} \{i\}\right) = \emptyset$$
A: Let me try.
Take the numbers $K_2 \subset \mathbb{N}$ divisible by $2$, and in general $K_n$ divisible by $n > 2$. 
For any $n,m$ the intersection of $K_n, K_m$ is non-empty containing the multiples of $n\times m$.
But the intersection $\cap_a K_a = \emptyset$
Now take the set $S_2$ of the form $\frac{1}{k_2}, k_2 \in K_2$, and in general $S_n$ of the form $\frac{1}{k_n}, n > 2$, that is, the inverse of the previous sets. 
They satisfy $\cap_a S_a = \emptyset$ and $S_a \subseteq (0,1)$.
