Show that lim inf Bn and lim sup Bn equals to a null set Suppose that ${B_n: n \geq 1}$ is a sequence of disjoint set.

Show that 

$$\begin{align}\limsup_{n\rightarrow \infty}B_n &amp;= \emptyset \text{ and}\\
\liminf_{n \rightarrow \infty}B_n&amp;= \emptyset \end{align}$$

Where $$\begin{align}\limsup_{n\rightarrow \infty}B_n &amp;= \{x \mid  x\in B_n \text{ for infinitely many } n\} \text{ and}\\
\liminf_{n \rightarrow \infty}B_n&amp;= \{x \mid x \notin B_n \text{ for at most finitely many } n\}. \end{align}$$

Can someone help me to solve this?
Since it's a disjoint set, means there is no intersection in the sequence.  So the union of intersection of Bn will be empty set and that's why lim inf  Bn is and empty set too?
Please help me, thanks..
As a given element $x$ cannot be in more than one $B_n$, in particular it can't be in infinitely many, giving both results as the lim sup requires $B_n$ to be in infinitely many, and the lim inf is even stronger requiring it to be in all but finitely many.
Sorry, that was expressed in a confusing way. Start again!

As you know already, the default ordering for sequences of sets is inclusion. So in your case $B_{n+1}\subset B_n$. 

The problem is the definition of lim sup and lim inf for sequences of sets. 

lim sup is defined as the set of all points which belong to infinitely many $B_n$, and lim inf is defined as the set of all points which belong to all but finitely many $B_n$. 

Since $k$ belongs to just $k$ of the $B_n$, these two sets are both empty.