Determine liminf$A_{n}$ and limsup$A_{n}$ for the $A_{n}$ of the following sets

Question

$$A_{n} = \begin{cases} (\frac{1}{n}-2 , 1) \ \ \ \ \ \text{if} \ n \ \text{is odd} \\ (0,3 + \frac{1}{n}) \ \ \text{if} \ n \ \text{is even} \end{cases}$$

This question is on a past measure theory final and I think I have the right answer but would just like a second opinion.

I have:

• limsup($\frac{1}{n}-2,1)$= ($-1,1$)
• liminf($\frac{1}{n}-2,1)$=($-2,1$)
• limsup($0,3 + \frac{1}{n})$=($0,3.5$)
• liminf($0,3 + \frac{1}{n})$=($0,3$)

I was also wondering if the way I formatted my solutions would be acceptable on an exam? Thank you in advance.

Answer 1

I am afraid you are on the wrong track. The question seems to be about the whole sequence $A_n$, and you have separately evaluated two subsequences of it (the odd and the even members). Whatever the result, it must have a single $\lim\sup_{n\in\mathbb N}A_n$ and a single $\lim\inf_{n\in\mathbb N}A_n$.

To solve the problem: recall that $\lim\sup_{n\in\mathbb N}A_n$ is the set of all elements contained in infinitely many $A_n$'s. This will turn out to be $(-2,3]$. Similarly, $\lim\inf_{n\in\mathbb N}A_n$ is the set of those elements contained in all but finitely many of $A_n$'s, and it is easy to see that this set is $(0,1)$.

Answer 2

We have $\limsup_{n\to\infty} 3+1/n = 3$, $\liminf_{n\to\infty} 1/n- 2 = -2$, $\liminf_{n\to\infty} 0 =0$, $\limsup_{n\to\infty} 1=1$. So in summary

• $\limsup_{n\to\infty} A_n = [0,3]$
• $\liminf_{n\to\infty} A_n = [-2,1]$