I am learning real analysis, I encountered the definition \begin{equation} \overline{\lim_{n \to \infty }}A_n = \bigcup_{n=1} \bigcap_{m=n} A_m \end{equation} I can understand this definition but I want to know the motivation about it, although I know that the limit superior set contains the points that is contained in infinity $A_m$s but I feel I want to know the deeper motivation.
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Related: https://math.stackexchange.com/questions/107931/lim-sup-and-lim-inf-of-sequence-of-sets – Hans Lundmark Mar 11 '18 at 10:23
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We define $\limsup_{n}A_{n}=\displaystyle\bigcap_{n=1}\bigcup_{m\geq n}A_{m}$ and $\liminf_{n}A_{n}=\displaystyle\bigcup_{n=1}\bigcap_{m\geq n}A_{m}$, one has $\liminf_{n}A_{n}\subseteq\limsup_{n}A_{n}$, this is similar to the sequence version: $\liminf_{n}x_{n}\leq\limsup_{n}x_{n}$.
If $\limsup_{n}A_{n}=\liminf_{n}A_{n}$, then we let $\lim_{n}A_{n}=\limsup_{n}A_{n}=\liminf_{n}A_{n}$.
Moreover, if we consider the usual pointwise ordering of functions, we have $\liminf_{n}\chi_{A_{n}}\leq\limsup_{n}\chi_{A_{n}}$. And $\liminf_{n}\chi_{A_{n}}=\limsup_{n}\chi_{A_{n}}=\lim_{n}\chi_{A_{n}}$ if and only if $\lim_{n}A_{n}=\limsup_{n}A_{n}=\liminf_{n}A_{n}$, this is a standard exercise in many measure theory books.
user284331
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Notice for a (bounded) sequence of numbers:
$$\limsup_{n \to \infty} x_n = \lim_{n \to \infty} \sup_{k \geq n}{x_k} = \inf \{\sup\{x_k \mid k \geq n\} \mid n \geq 0\}$$
An infinum takes the greatest lowerbound, so it makes things smaller. Therefore, it intuitively makes sense to let it correspond with an intersection. On the other hand, a supremum makes things larger, so we let this correspond with a union. Therefore, it makes sense to define:
$$\limsup_{n \to \infty} A_n = \bigcap_{n}^\infty \bigcup_{k=n}^ \infty A_n$$
and analogue for $\liminf$