$A_n = (-n^{-1},1+n^{-1})$ if n is odd, and $A_n = (1-n^{-1},2+n^{-1})$ if n is even.
Calculate $\overline{\lim}_nA_n$ and $\underline{\lim}_nA_n$ (aka $\limsup$ and $\liminf$ notation wise).
I'm given that $\overline{\lim}_{n\to\infty}A_n=\bigcap_{n\ge 1}\bigcup_{k\ge n}A_k$, and $\underline{\lim}_{n\to\infty}A_n=\bigcup_{n\ge 1}\bigcap_{k\ge n}A_k$, and I have worked out:
$$
\begin{aligned}
\overline{lim}_nA_n=&((-1,2)\cup(-.33,1.33)\cup(-.2,1.2)\cup\cdots\cup(.5,2.5)\cup(.75,2.25)\cup(.83,2.16)\cup\cdots)\\
&\cap((-.33,1.33)\cup(-.2,1.2)\cup\cdots\cup(.75,2.25)\cup(.83,2.16)\cup\cdots))\cap\cdots\\
=&((-1,2)\cup(.5,2.5))\cap((-.33,1.33)\cup(.75,2.25))\cap\cdots\\
=&(-1,2.5)\cap(-.33,2.25)\cap(-.2,2.16)\cap\cdots\\
=&(0,2)
\end{aligned}
$$
$$
\begin{aligned}
\underline{lim}_nA_n=&((-1,2)\cap(-.33,1.33)\cap(-.2,1.2)\cap\cdots\cap(.5,2.5)\cap(.75,2.25)\cap(.83,2.16)\cap\cdots)\\
&\cup((-.33,1.33)\cap(-.2,1.2)\cap\cdots\cap(.75,2.25)\cap(.83,2.16)\cap\cdots)\cup\cdots\\
=&((0,1)\cap(1,2))\cup((0,1)\cap(1,2))\cup\cdots\\
=&\emptyset\cup\emptyset\cup\cdots\\
=&\emptyset
\end{aligned}
$$
Could someone let me know whether what I've done is correct?
My logic is that $\overline{lim}_nA_n=(0,2)$ is the set of numbers which belong to infinitely many $A_n$, and that $\underline{lim}_nA_n=\emptyset$ is the set of real numbers which belong to all $A_n$ for sufficiently large $n$, and since the limits of the odd and even $A_n$ are $(0,1)$ and $(1,2)$, which have $\emptyset$ as their intersection, there are no numbers that can be in both all the odd and even $A_n$'s.
Best Answer
Tackling the question by trying to write out the intervals involved isn't a particularly good approach as I think you've seen. Getting familiar with these kinds of unions and intersections will help you develop mathematically though!
Start with the intervals themselves to understand what we're looking at. When $k=2n+1$ for $n\in \mathbb N$ we see that $A_k$ is always slightly larger than the interval $[0,1]$ Note that this a closed interval, even though $A_k$ is always open, because $0\in A_k$ (you should make sure you understand this; consider the limit of $-1/k$ as $k\rightarrow \infty$). So any intersection of the $A_k$ when $k$ is odd will contain $[0,1]$.
Now let $k=2n$ for $n\in \mathbb N$. By the same reasoning we see that $[1,2] \in A_k$ for all even $k$ and so any intersection of $A_{2n}$ must contain $[0,2]$.
Now we're ready to work out the $\limsup$ and $\liminf$: we have $\limsup_n A_n = \overline{\lim}_n A_n = \bigcap_{n\geq 1} \bigcup_{k \geq n} A_k $. The union of any $A_k$ must contain $[0,1] \cup [1,2]$ because there will always be an even and an odd $k$ in the union $\bigcup_{k \geq n} A_k$ (if there weren't, and you could prove it, you could tell us if countable-infinity were even or odd!). Thus the intersection of all of these unions will contain $[0,1]\cup[1,2] = [0,2]$. Finally we check that no other point can sneak in there: if it did it has the form $-1/n$ or $2+1/n$ as they are the only points that can "stick out" from $[0,2]$. However, any you pick will eventually be excluded when $k\geq n$, so we're good: $\limsup _n A_n = [0,2]$.
For the $\liminf_n A_n = \bigcup_{n \geq 1} \bigcap_{k \geq n} A_k$ we note that $A_k \cap A_{k+1} = \left(1-(k+1)^{-1}, 1+k^{-1} \right)$. Since we take a countable intersection over $k \geq n$, this is the same as considering $\lim_{k \rightarrow \infty} \left(1-(k+1)^{-1}, 1+k^{-1} \right) = \{1\}$ Taking unions of $\{1\}$ still only gives us $\{1\}$ however. This time it's easier to see that there can be no other points in the $\liminf$ because the intersection reduces down to a single point.
If you're doing any kind of measure theory, this interchange between limits and unions/intersections will come up a lot, and very often one way is much easier to use than the other.