This is from Elements of Integration and Lebesgue Measure by Bartle.
Exercise 2.H. If $(A_n)$ is a sequence of subsets of $X$ … Give an example of a sequence $(A_n)$ such that $$\lim\inf A_n=\emptyset\,\,\lim\sup A_n=X $$
The definitions given are
$$\lim\sup A_n=\bigcap_{m=1}^\infty\left[\bigcup_{n=m}^\infty A_n\right]$$
and
$$\lim\inf A_n=\bigcup_{m=1}^\infty\left[\bigcap_{n=m}^\infty A_n\right]$$
I had a few ideas as follows:
Idea 1:
Let $X={\Bbb N}$ then $A_n=\{n, 2n, 3n, …\}$. However, this gives $\cap_{n=m}^\infty A_n=\emptyset$ so $\lim\sup A_n=\emptyset$.
Idea 2:
Again let $X={\Bbb N}$ then $A_n=\{n,n+1,…\}$. This fails for the same reason as idea 1, also it is a monotone decreasing sequence so by exercise 2.G. it would not have worked anyway.
Idea 3:
Again let $X={\Bbb N}$ then $A_n=\{1,2,..,n\}$. This fails since for a monotone increasing sequence $\lim\sup A_n=\lim\inf A_n$. (This result is exercise 2F in the book!)
Idea 4:
Let $X=\{x\in{\Bbb Q}:x\in(0,1)\}$ then $A_n=\{1/n,2/n,..,(n-1)/n\}$. I think $\cup_{n=m}^\infty A_n=X$ here because of duplicated fractions, and $\cap_{n=m}^\infty A_n=\emptyset$, because the intersection over coprime numerators is empty. However I'm "reaching" a bit here.
Any better ideas or extensions of these notions?
Best Answer
HINT: Let $\emptyset\neq B\neq X$. Let $A_{2n}=B$ and $A_{2n+1}=X\setminus B$.