# Thread: General formula for finding $limsupA_n, liminfA_n$, where $A_n$ is a sequence of sets

1. ## General formula for finding $limsupA_n, liminfA_n$, where $A_n$ is a sequence of sets

I would like to know if there is a general formula, and if so, what it is, for finding the $limsup$ and $liminf$ of a sequence of sets $A_n$ as $n\rightarrow \infty$.[/tex]$\displaystyle I know the following examples: **(1)** for$A_n=(0,a_n], (a_1,a_2)=(10,200)$,$a_n=1+1/n$for$n$odd and$a_n=5-1/n$for$n$even, and$n\geq 3$,$limsup_{n\rightarrow \infty}a_n = 5$,$liminf_{n\rightarrow \infty}a_n = 1$,$limsup_{n\rightarrow \infty}A_n = (0,5)$,$liminf_{n\rightarrow \infty}A_n = (0,1]$. **(2)** for$A_n=[0,a_n), (a_1,a_2,a_3,a_4)=(10,100,1000,10000)$,$a_{2n+1}=2-1/(2n+1)$for$n\geq2$and$a_{2n}=4+1/(2n)$for$n\geq4$,$limsup_{n\rightarrow \infty}a_n = 4$,$liminf_{n\rightarrow \infty}a_n = 2$,$limsup_{n\rightarrow \infty}A_n = [0,4]$,$liminf_{n\rightarrow \infty}A_n = [0,2)$. **(3)** for$A_n=(0,a_n], (a_1,a_2)=(50,20)$,$a_{3n}=1+1/(3n), a_{3n+1}=1+1/(3n+1), a_{3n+2}=3-(1/3n+2)$for$n\geq1$,$limsup_{n\rightarrow \infty}a_n = 3$,$liminf_{n\rightarrow \infty}a_n = 1$,$limsup_{n\rightarrow \infty}A_n = (0,3)$,$liminf_{n\rightarrow \infty}A_n = (0,1)$.$

**Is there a general formula describing $limsup_{n\rightarrow \infty}A_n$ and $liminf_{n\rightarrow \infty}A_n$ with the open/closed interval notation, for an arbitrarily defined $\{a_n\}$?**

Thanks for any help!

2. ## Re: General formula for finding $limsupA_n, liminfA_n$, where $A_n$ is a sequence of

Let $\displaystyle \mathcal{A}_s^+ = \left\{A_n \mid a_n \ge \limsup_{n\to \infty} a_n \right\}$, $\displaystyle \mathcal{A}_s^- = \left\{A_n \mid a_n < \limsup_{n \to \infty} a_n \right\}$, $\displaystyle \mathcal{A}_i^+ = \left\{A_n \mid a_n > \liminf_{n \to \infty} a_n \right\}$, $\displaystyle \mathcal{A}_i^- = \left\{A_n \mid a_n \le \liminf_{n \to \infty} a_n \right\}$.

$\displaystyle \mbox{Let } A_s^+ = \begin{cases} \emptyset & \mbox{if } \exists N \in \mathbb{Z}^+, \forall n \ge N, a_n < \limsup_{n\to \infty} a_n \\ \bigcap_{A \in \mathcal{A}_s^+} A & \mbox{otherwise} \end{cases}$.

$\displaystyle \mbox{Let } A_s^- = \begin{cases} \emptyset & \mbox{if } \exists N \in \mathbb{Z}^+, \forall n \ge N, a_n \ge \limsup_{n\to \infty} a_n \\ \bigcup_{A \in \mathcal{A}_s^-} A & \mbox{otherwise} \end{cases}$.

$\displaystyle \mbox{Let } A_i^+ = \begin{cases} \emptyset & \mbox{if } \exists N \in \mathbb{Z}^+, \forall n \ge N, a_n \le \liminf_{n\to \infty} a_n \\ \bigcup_{A \in \mathcal{A}_i^+} A & \mbox{otherwise} \end{cases}$.

$\displaystyle \mbox{Let } A_i^- = \begin{cases} \emptyset & \mbox{if } \exists N \in \mathbb{Z}^+, \forall n \ge N, a_n > \liminf_{n\to \infty} a_n \\ \bigcap_{A \in \mathcal{A}_i^-} A & \mbox{otherwise} \end{cases}$.

Then $\displaystyle \limsup_{n\to \infty} A_n = A_s^+ \cup A_s^-$ and $\displaystyle \liminf_{n \to \infty} A_n = A_i^+ \cup A_i^-$

3. ## Re: General formula for finding $limsupA_n, liminfA_n$, where $A_n$ is a sequence of

Unbelievable that you were able to figure this out so quickly!! Kudos.