# An example of a sequence of subsets such that $\lim\inf A_n=\emptyset$, $\lim\sup A_n=X$

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?

• for idea 1 how the union is $\phi$ Commented Nov 29, 2014 at 2:12
• @learnmore You are right. That should have been $\cap$ Commented Nov 29, 2014 at 2:19

HINT: Let $\emptyset\neq B\neq X$. Let $A_{2n}=B$ and $A_{2n+1}=X\setminus B$.

• I have no idea what you are hinting at. If $A_{2n}=B$ and $A_{2n+1}=X\setminus B$, what is $A_n$ or $A_{2n+2}$? Commented Nov 29, 2014 at 2:03
• @SuzuHirose It means: if the index is even, the set is $B$. If it is odd, the set is $X\setminus B$.
– user147263
Commented Nov 29, 2014 at 2:22
• @Raff thank you. This gives a correct answer. Commented Nov 29, 2014 at 2:26

Let $X= \Bbb N$ and $A_n=[2,n]$ where $n=1,2,\cdots$

• I get $\lim\inf A_n=\{2\}\neq\emptyset$ for that. Commented Nov 29, 2014 at 2:01
• Note that $A_1=[2,1]=\emptyset$
– Paul
Commented Nov 29, 2014 at 2:04
• $\lim\inf A_n=\cup_{m=1}^\infty\cap_{n=m}^\infty[2,n]=\emptyset\cup \{\cup_{m=2}^\infty\{2\}\}=\{2\}$. Commented Nov 29, 2014 at 2:17
• This example has both limits equal to $\{2,3,4,\dots\}$.
– user147263
Commented Nov 29, 2014 at 2:18
• @Raff is right. This is a monotone increasing sequence so $\lim\sup=\lim\inf$. Commented Nov 29, 2014 at 2:21