I am struggling with this question from an old measure theory question set, mainly because I'm confused by the notation.
Determine $\liminf A_n$ and $\limsup A_n$ for the sequence $(A_n)_{n∈\mathbb{N}}$ of sets given by: $$ A_n = \begin{cases} (-\frac{1}{n}-2,1) & n \text{ odd}\\ (0,3+\frac{1}{n}) & n \text{ even}. \end{cases} $$
Clearly the odd sequence is converging to $(-2,1)$ and the even series is converging to $(0,3)$, but I don't know what $\limsup$ and $\liminf$ mean on ordered pairs like this.
Edit: Clearly I was over-thinking this - I thought I was dealing with ordered pairs from $\mathbb{R}^2$, when clearly these are open intervals on $\mathbb{R}$. Thank you to the people who set me straight.
If $0<x<1$ the $x$ clearly belongs to all of these sets. Suppose $y$ belongs to $A_n$ for all $n >n_0$ for some integer $n_0$. Then $-2-\frac 1 n <y<1$ for $n >n_0$ which implies $-2 \leq y <1$. Similarly we get $0 <y \leq 3$. Combining these we get $0<y<1$. Hence $\lim \inf A_n =(0,1)$. Can you now verify that a point $x$ belongs to infinitely many of the set $A_n$ iff $x \in [-2,1) \cup (0,3]$. This part is easy so I will let you handle this.