Alternative definition of limsup and liminf

Question

I am reading a Chinese textbook on Real analysis and it introduces the definition of limsup and liminf as the following:

$$\varliminf A_n\equiv\liminf_{n} A_n\equiv\{x; \text{there exists finitely(*) many n, such that } x\notin A_n\}$$

$$\varlimsup A_n\equiv\limsup_{n} A_n\equiv\{x; \text{there exists infinitly many n, such that } x\in A_n\}.$$

Then it gives an example:

Let $A_n=[\frac{1}{n},3+(-1)^n],n=1,2,3,...$, we have that $\liminf_{n} A_n=(0,2], \limsup_n A_n=(0,4]$.

The notion that for any series of sets $A_1,A_2,...A_n,...$ we have: $$\liminf_n A_n=\bigcup^\infty_{m=1}\bigcap^\infty_{i=m} A_i$$ $$\limsup_n A_n=\bigcap^\infty_{m=1}\bigcup^\infty_{i=m} A_i$$ follows the example and is introduced as the property of liminf and limsup. It is really hard for me to understand how the example relates to the definition. It seems easy to see if $x\in(0,4]$ it satisfies the definition of limsup. But what if $x\in(0,2]$, how do I connect it to the definition of liminf given in the text?

(*) The original text is written in Chinese, so I am not sure if the word finitely is translated correctly. By direct translation, the text would be "there is limited many n", but I have never heard such term in English.

Answer 1

It should be$$\liminf_nA_n=\{x\mid x\in A_n\text{ for all but finitely many $n$'s}\}$$or$$\liminf_nA_n=\{x\mid x\notin A_n\text{ only for finitely many $n$'s}\}.$$

You have$$A_n=\begin{cases}\left[\frac1n,2\right]&\text{ if $n$ is odd}\\\left[\frac1n,4\right]&\text{ if $n$ is even.}\end{cases}$$So, if $x>2$ or $x\leqslant0$, then $x\notin\liminf_nA_n$, since there are infinely many $n$'s such that $x\notin A_n$. But if $x\in(0,2]$, then $x$ belongs to every $A_n$; in particular, $x\in\liminf_nA_n$.

Answer 2

The sets $A_n$ in the example are either of the form $[\frac{1}{n}, 2]$ (odd $n$) or $[\frac{1}{n}, 4]$ (even $n$). So, it is clear that, if $x\gt 2$, $x$ won't belong to those sets for odd $n$, and obviously if $x\le 0$ it won't belong to any of those sets at all.

All $x\in[1,2]$ will belong to all of those sets. As for $x\in(0,1)$, there will be finitely many sets which won't contain $x$ (precisely those where $\frac{1}{n}\gt x$), but from some $n$ onwards, those sets will all contain $x$. In other words, $x$ will not be contained in only finitely many of those sets, as per your definition.