We define the limit inferior and limit superior of a sequence $(A_n:n\in \mathbb N)$ of subsets of a set $X$ by setting
(1) $$\begin{split}\lim \inf _{n\rightarrow\infty} A_n&=\bigcup_{n\in\mathbb N} \bigcap _{k\ge n} A_k\\ &= \{x\in X:x\in A_n \text{ for all but finitely many } n\in\mathbb N\}\end{split}$$
(2) $$\begin{split}\lim \sup_{n\rightarrow \infty} A_n &= \bigcap_{n\in\mathbb N}\bigcup _{k\ge n} A_k\\ &=\{x\in X:x\in A_n \text{ for infinitely many } n\in \mathbb N\}\end{split}$$
Now for my own expansion of their definition:
$$\begin{split}\lim\inf_{n\rightarrow\infty} A_n &= \left(\bigcap_{k=1,2,3,...} A_k \right)\cup\left(\bigcap_{k=2,3...} A_k \right)\cup...\end{split}$$
I do not see how the definition of liminf as "$x \in A_n$ for all but finitely many $n$" could work. If it is in finitely many $A_k$, then the number will not appear in the intersection of any $\cap A_k$ and thus not appear in the overall union. What gives?
However, I do see how limsup definition works, if
$$\begin{split}\lim\sup_{n\rightarrow\infty} A_n &= \left(\bigcup_{k=1,2,3,...} A_k \right)\cap\left(\bigcup_{k=2,3...} A_k \right)\cap...\end{split}$$
Thus there must be no end point to when the number stops appearing to appear in the limsup.
In English the phrase "all but" has two different meanings. However, in mathematics, "$x \in A_n$ for all but finitely many $n$" means "$x \in A_n$ is false for finitely many $n$." See also this answer.
In particular, if $N$ is larger than the finitely many $n$ for which "$x \in A_n$" is false, then $x \in \bigcap_{k = N, N+1, \ldots} A_k$.