# General formula for finding limsup A_n,liminf A_n where A_n is a sequence of sets

#### kalish

##### Member
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 $n\rightarrow \infty$.

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!

#### girdav

##### Member
In general, when we consider the $\limsup$ and $\liminf$ of an arbitrary sequence of sets (not necessarily intervals), we have the definition $\limsup_n A_n:=\bigcap_{n\geqslant 1}\bigcup_{k\geqslant n}A_k$ and $\liminf_n A_n:=\bigcup_{n\geqslant 1}\bigcap_{k\geqslant n}A_k$. That is, $x\in\limsup_n A_n$ if the set $\{n,x\in A_n\}$ is infinite, while $x\in\liminf_nA_n$ if $\{n,x\in A_n\}$ contains all but finitely many positive integers.