Let ${A_n}$ be one set sequence, why lower limitation is $\varliminf_{n\rightarrow \infty}A_n=\bigcup_{n=1}\bigcap_{k=n}A_n$, How to understand it?

Question

I'm confused about the definition of upper limit and lower limit of a set sequence. Could I think the lower limitation of one set sequence as "The largest intersection while $n$ goes to infinity" and the upper limitation as "The smallest union while $n$ goes to infinity"? If there is a graph to describe then best!

Answer 1

For real numbers $(x_n)$ we have : $$\liminf_{n\to +\infty} x_n = \lim_{n\to +\infty} \inf_{k\geq n} x_k = \sup_{n\in \mathbb N} \inf_{k\geq n} x_k$$ where the last equality holds because the sequence $(\inf_{k\geq n} x_k)_{n\in\mathbb N}$ is increasing.

Sets are ordered by inclusion, the infinimum of a family of sets is the intersection, while the supremum is the union. Therefore, for sets $(A_n)$, we have : $$\liminf_{n\to +\infty} A_n = \bigcup_{n\in\mathbb N}\bigcap_{k\geq n} A_k$$

Edit : The explanation above motivates the definition (why it is natural to define it that way). Then, to understand what it actually does on a family on set, we do the following : \begin{align} x\in \liminf_{n\to +\infty } A_n &\Longleftrightarrow x\in\bigcup_{n\in\mathbb N}\bigcap_{k\geq n}A_k\\ &\Longleftrightarrow \exists n\in\mathbb N, x\in \bigcap_{k\geq n}A_k \\ &\Longleftrightarrow \exists n\in\mathbb N, \forall k\geq n, x\in A_k \end{align} In other words, $\liminf_{n\to +\infty} A_n$ is the set of elements $x$ which to all $A_k$ with $k$ large enough (ie who belong to all $A_k$ but a finite number).