Let $A_n = (−1 + \frac{1}{n}, 2 − \frac{1}{n})$ if $n$ is odd and $[0, n]$ if $n$ is even. Find $\liminf A_n$ and $\limsup A_n$.
This is a question on a past paper for a measure theory module I'm taking and I'm not quite sure if my answer is correct. I have $\limsup A_n = [-1, 2]$ and $\liminf A_n=[0,0]$. I'm not quite sure if either of these are correct as the $[0,n]$ when $n$ is even is throwing me. Since we just have $n$ does that mean that $\limsup A_n$ could be $[-1, ∞ ]$? Any help would be appreciated!
We know that $x \in \limsup_{n\to\infty} A_n$ if and only if $x \in A_n$ for infinitely many $n \in \Bbb{N}$. We claim that $$\limsup_{n\to\infty }A_n = (-1,+\infty).$$
Indeed, let $x \in (-1,+\infty)$. If $x \in [0,+\infty)$ then $x \in A_n$ for every even $n\in\Bbb{N}$ such that $n \ge x$. If $x \in (-1,0)$ then pick $n_0\in\Bbb{N}$ such that $-1+\frac1{n_0} <x$ and hence for every odd $n \ge n_0$ we have $x \in A_n$. We conclude $x \in \limsup_{n\to\infty} A_n$. Conversely, if $x \le -1$, then $x$ is not contained in any $A_n$ and hence $x \notin \limsup_{n\to\infty} A_n$.
Similarly, $x \in \liminf_{n\to\infty} A_n$ if and only if $x \in A_n$ for all except finitely many $n \in \Bbb{N}$. We claim that $$\liminf_{n\to\infty }A_n = [0,2).$$
Indeed, let $x \in [0,2)$. Clearly $x \in A_n$ for every even $n \ge 2$. Pick $n_0 \in \Bbb{N}$ such that $x < 2-\frac1{n_0}$. Then for every odd $n \ge n_0$ we have $x \in A_n$. We conclude that for every $n \ge \max\{2,n_0\}$ holds $x \in A_n$ and hence $x \in \liminf_{n\to\infty} A_n$. Conversely, if $x < 0$ then $x\notin A_n$ for every even $n \in \Bbb{N}$, and if $x \ge 2$ then $x\notin A_n$ for every odd $n \in \Bbb{N}$ and thus $x \notin \liminf_{n\to\infty} A_n$.