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

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 $\limsup$.
