Given a sequence $a_n$, what can be said about $\limsup A_n$ and $\liminf A_n$ if $A_n=(-\infty, a_n)$?
I considered the following scenarios (1) $a_n$ is montonic, (2) $a_n$ is not monotonic.
$(1)$. It can be (a) increasing or (b) decreasing.
$(1a)$ If it is convergent to $\alpha$, then $(-\infty, a_{n+1})\subset(-\infty, a_n)$ then $\limsup A_n = \liminf A_n =\bigcup_n A_n = (-\infty, \alpha).$
If it is divergent, then $\limsup A_n = \liminf A_n = \mathbb{R}$.
$(1b)$ If it is convergent to $\alpha$, then $(-\infty, a_{n})\subset(-\infty, a_{n+1})$ then $\limsup A_n = \liminf A_n =\bigcap_n A_n = (-\infty, \alpha).$
If it is divergent, then $\limsup A_n = \liminf A_n = \emptyset$.
$(2)$ But what if $a_n$ si not monotonic? I fail to find a pattern here.
For example a convergent sequence could be $a_n=\{\frac{1}{(-1)^n n}, n\in \mathbb{N }\}$. Then the $A_n$ sets will be $\{ (-\infty, -1), (-\infty, \frac{1}{2}), (-\infty, -\frac{1}{3}), (-\infty, \frac{1}{4}), \dots \}$.
In this case it seems $\liminf A_n = \{(-\infty, -1), (-\infty, -\frac{1}{3}), (-\infty, -\frac{1}{5}, \cdots), (-\infty, -\frac{1}{2n-1}) \}$. But it doesn't reveal any general pattern that will apply to all monotonic convergent sequences.
In the general case,$$\limsup\nolimits_n(-\infty,a_n)=\left(-\infty,\limsup\nolimits_na_n\right)\text{ or }\limsup\nolimits_n(-\infty,a_n)=\left(-\infty,\limsup\nolimits_na_n\right].$$In fact, if $a>\limsup_na_n$, then, if $N$ is large enough, $a_n<a$. So, $a\in(-\infty,a_n)$ only finitely many times. Therefore, $a\notin\limsup_n(-\infty,a_n)$. Amd, if $a<\limsup_na_n$, then there are infinitely many $n$'s for which $a_n>a$. So, $a\in\limsup_n(-\infty,a_n)$.
By a similar argument,$$\liminf\nolimits_n(-\infty,a_n)=\left(-\infty,\liminf\nolimits_na_n\right)\text{ or }\liminf\nolimits_n(-\infty,a_n)=\left(-\infty,\liminf\nolimits_na_n\right].$$
Of course, both cases may occur. For instance, if $a_n=\frac1n$, then$$\limsup\nolimits_n(-\infty,a_n)=\liminf\nolimits_n(-\infty,a_n)=(-\infty,0],$$whereas if $a_n=-\frac1n$,$$\limsup\nolimits_n(-\infty,a_n)=\liminf\nolimits_n(-\infty,a_n)=(-\infty,0).$$