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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?

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

2 Answers 2

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

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  • $\begingroup$ 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 $A_{2n+2}$? $\endgroup$ Commented Nov 29, 2014 at 2:03
  • $\begingroup$ @SuzuHirose It means: if the index is even, the set is $B$. If it is odd, the set is $X\setminus B$. $\endgroup$
    – user147263
    Commented Nov 29, 2014 at 2:22
  • $\begingroup$ @Raff thank you. This gives a correct answer. $\endgroup$ Commented Nov 29, 2014 at 2:26
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Let $X= \Bbb N$ and $A_n=[2,n]$ where $n=1,2,\cdots$

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

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