Find $\liminf _{n \to \infty}A_n $ and $\limsup_{n \to \infty}A_n $ Define
$$A_n= \begin{cases}
 \left ( \frac{1}{2}-\frac{1}{2n},1+\frac{1}{2n}\right )& \text{ if n is odd }  \\
 \left ( \frac{1}{2n},\frac{3}{4}-\frac{1}{2n} \right )& \text{ if n is even }  
\end{cases}$$
Find $\liminf _{n \to \infty}A_n $ and $\limsup_{n \to \infty}A_n $
$\limsup_{n \to \infty}A_n= \cap_{n=1}^{\infty}(\cup_{n\le N})$
$\liminf _{n \to \infty}A_n=\cup_{n=1}^{\infty}(\cap_{n\le N})$
$A_1=\left ( 0,1+ \frac{1}{2}\right )\\
A_2=\left ( \frac{1}{4}, \frac{1}{2}\right )\\
A_3=\left (\frac{1}{3},1+ \frac{1}{6}\right )$
Can you help someone here i'm confusing
 A: A good way to think is like this:-
1.Lim sup of a sequence of sets $A_{n}$ is the set of all $x$ such that $x\in A_{n}$ for infinitely many $n$.
2.Lim inf of $A_{n}$ is the set of all $x$ such that $x\in A_{n}$ for all but finitely many $n$.
you see that all $x\in [\frac{1}{2},1]$ . It lies in $A_{n}$ for all odd $n$. and all $x\in (0,\frac{3}{4})$ lies in $A_{n}$ for infinitely many even integers $n$.
So $\lim\sup A_{n}=[\frac{1}{2},1]\cup (0,\frac{3}{4})=(0,1]$.
Now only $x$ such that $x\in[\frac{1}{2},\frac{3}{4})$ lies in $A_{n}$ for all but finitely many $n$.
So $\lim\inf A_{n}=[\frac{1}{2},\frac{3}{4})$.
A: I find it helpful to think about what the following two sequences of sets are doing.
For the $\liminf$, I think about the sequence
\begin{align}
\bigcap_{j = 1}^{\infty} A_j\\
\bigcap_{j = 2}^{\infty} A_j\\
\bigcap_{j = 3}^{\infty} A_j\\
\vdots
\end{align}
The $\liminf$ is the union of all of these, and since they're clearly getting larger I don't have to worry about the first however many. I can just focus on the ultimate behavior.
The left-hand side of $A_n$ is oscillating, half the time approaching $(0$ and half the time approaching $[\frac{1}{2}$. The right-hand side of $A_n$ is also oscillating, half the time approaching $\frac{3}{4})$ and half the time approaching $1]$.
So turning back to the $\liminf$, what's the ultimate behavior of the increasing sequence above? It's clearly ultimately going to contain $[\frac{1}{2}, \frac{3}{4})$, so that's the $\liminf$.
For the $\limsup$ I can consider the sequence
\begin{align}
\bigcup_{j = 1}^{\infty} A_j\\
\bigcup_{j = 2}^{\infty} A_j\\
\bigcup_{j = 3}^{\infty} A_j\\
\vdots
\end{align}
The $\limsup$ is the intersection of all of these, and since they're clearly getting smaller I don't have to worry about the first however many. I can just focus on the ultimate behavior.
Based on the analysis above, we take the other parts of the oscillation, because we're dealing with unions this time. The interval that's in at least one of the $A_i$ as $i$ becomes large is $(0, 1]$, so that's the $\limsup$.
