Finding a sequence of sets whose intersection is a null set 
Find a sequence of sets $I_n=\{r:r \in \mathbb{Q}, a_n\le r \le b_n\} $ in $\mathbb{Q}$, where $a_n, b_n \in\mathbb{Q}$ such that $$I_{n+1} \subset I_n\forall n\in\mathbb{N}$$ $\lim_{n \to \infty}(b_n - a_n)= 0 $ but $$\cap_{n=1}^{\infty} I_n= \emptyset$$ 

Can anyone explain the question and how to go about its solution?
 A: consider a irrational number $i \in \mathbb{R}$...we know that $\mathbb{Q}$ is dense in $\mathbb{R}$....consider a incersing sequence of rationals converging to $i$ as {$a_n$} and consider a decreasing sequence {$b_n$} converging to $i$ and consider $J_n$= {$r \in \mathbb{R}| a_n \leq r \leq b_n$} then $J_n \cap \mathbb{Q} = I_n$ ...then by cantor intersection property $\bigcap_{n} J_n$ is singleton and contain $i$ only...so   $\bigcap_{n} I_n = \phi$
A: Pick your favorite irrational number $\alpha$. Choose sequences $\{a_n\}$ and $\{b_n\}$ of rational numbers such that $\{a_n\}$ is increasing, $a_n<\alpha$, $\lim_{n\to\infty}a_n=\alpha$, $\{b_n\}$ is decreasing, $b_n>\alpha$ and $\lim_{n\to\infty}a_n=\alpha$.
For instance, if
$$
\alpha=\pi=3.141592\dots
$$
take $a_1=3$, $a_2=3.1$, $a_3=3.14$, $a_4=3.141$, $a_5=3.1415$, $a_6=3.14159$, and so on, 
and $b_1=4$, $b_2=3.2$, $b_3=3.15$, $b_4=3.142$, $b_5=3.1416$, $b_6=3.14160$, ...
A: For each $n \in \mathbb{N}$, take $a_n$ a rational number in the interval $ (\pi - \frac{1}{n}, \pi)$, and $b_n$ a rational number in $(\pi, \pi + \frac{1}{n})$. Now,
$$
\bigcap_{n=1}^{\infty} (a_n,b_n) = \lbrace \pi \rbrace.
$$
However, with your definition,
$$
I_n = (a_n,b_n) \cap\mathbb{Q},
$$
then
$$
\bigcap_{n=1}^{\infty} I_n = \bigcap_{n=1}^\infty (a_n,b_n) \cap \mathbb{Q} = \lbrace \pi \rbrace \cap \mathbb{Q} = \emptyset.
$$
Remind that the set of rational numbers is not complete, but the set of real numbers is complete.
