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

• Oct 14th 2013, 09:02 AM
abscissa
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] $

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 $n\geq1$,

$limsup_{n\rightarrow \infty}a_n = 3$, $liminf_{n\rightarrow \infty}a_n = 1$, $limsup_{n\rightarrow \infty}A_n = (0,3)$, $liminf_{n\rightarrow \infty}A_n = (0,1)$. " alt="

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!
• Oct 14th 2013, 10:08 AM
SlipEternal
Re: General formula for finding $limsupA_n, liminfA_n$, where $A_n$ is a sequence of
Let $\mathcal{A}_s^+ = \left\{A_n \mid a_n \ge \limsup_{n\to \infty} a_n \right\}$, $\mathcal{A}_s^- = \left\{A_n \mid a_n < \limsup_{n \to \infty} a_n \right\}$, $\mathcal{A}_i^+ = \left\{A_n \mid a_n > \liminf_{n \to \infty} a_n \right\}$, $\mathcal{A}_i^- = \left\{A_n \mid a_n \le \liminf_{n \to \infty} a_n \right\}$.

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

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

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

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

Then $\limsup_{n\to \infty} A_n = A_s^+ \cup A_s^-$ and $\liminf_{n \to \infty} A_n = A_i^+ \cup A_i^-$
• Oct 14th 2013, 10:19 AM
abscissa
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.