Explain confusion about the identity $\lim \inf x_n = \lim \inf A_n$, with $A_n = \{x_k | k \ge n\}$ I am confused about one thing during the lecture. 
Let $x_n = n$ and  $A_n = \{x_k | k \ge n\} = \{n, n+1, n+2, ...\}$.
Then, $\inf A_n = n $, and $\sup A_n = \infty$. 
My lecturer also said that $\lim\inf x_n = \lim\inf A _n=\lim n$. 
My thinking is that $\{x_n\}_{n=1}^{\infty}=\{1, 2, 3, .....\}$. Shouldn't $\inf x_n = 1$?? Then, $\lim \inf x_n =1$, which is not equal to $\lim n$.
Could you tell me if I am wrong? 
 A: Because we define $\liminf\limits_{n\rightarrow\infty}x_{n}=\lim\limits_{n\rightarrow\infty}\left(\inf\limits_{k\geq n}x_{k}\right)=\lim\limits_{n\rightarrow\infty}\left(\inf A_n\right)$.
It is not defined as $\lim\limits_{n\rightarrow\infty}\inf\{x_{k}:k=1,2,...\}$. Note that $\inf\{x_{k}: k=1,2,...\}$ is an extended real number independent of $n$, hence $\lim\limits_{n\rightarrow\infty}\inf\{x_{k}: k=1,2,...\}$ is simply $\inf\{x_{k}: k=1,2,...\}$.
@Did has noted a good comment.
A: Indeed strange since we usually define$$\liminf_{n\to\infty} A_n = \bigcup_{n=1}^\infty \left(\bigcap_{k=n}^\infty A_n\right)$$
which is clearly a set.
It can be shown that 
$$\liminf_{n\to\infty} A_n = \{x \in \mathbb{R} : x \in A_n \text{ for all except finitely many } n \in \mathbb{N}\}$$
so in this case we have $$\liminf_{n\to\infty} A_n = \emptyset$$

Another interpretation is $$\lim_{n\to\infty}( \inf A_n) = \lim_{n\to\infty} n = \lim_{n\to\infty} x_n = \liminf_{n\to\infty} x_n = +\infty$$
since $\inf A_n = n = x_n$, $\forall n \in \mathbb{N}$.
