Limsup and liminf of a set Let E $\subset$ $\mathbb R$ and t $\in \mathbb R$
Define $E + t = \{ x + t \mid x \in E\}$
Pick a sequence $t_n$ > 0 and $t_n \to $ 0 strictly decrasing and define:
$E_n$ = E + $t_n$
Show that if E = (-$\infty$,0) then lim $E_n$ exists and is equal to (-$\infty$,0]
My question is: How may I show the existence of the limit by computing the liminf and limsup?

*

*I applied the translation, in order to work on the set (-$\infty$, $t_n$)


*Just by drawing a graph is easy to show that the limit is (-$\infty$,0] and the zero is also included since picking a x < $t_n$ and $t_n$ goes to 0, so x will be $\le$ zero.


*If I would like to show it by computing lim inf and lim sup, my procedure was:
Liminf $\rightarrow$ If I pick x $\in$ (-$\infty$,0] it will be in there in all but finitely many times but also for every n, so i basically showed even the limsup.
To be sure about the limsup, if I'd pick a x $\in$ (0,$\infty$) it will be inside this interval only for a finite number of n, so the limsup would be empty and even the liminf, so I would find a contradiction.
Is the whole reasoning correct or there is anything missing?
Another question is on how I could find similar exercise to this, about lim, liminf and limsup of intervals and translation of this kind.
Thank you!
 A: If $(A_n)_n$ is a sequence of sets then it has a limit iff $\limsup A_n=\liminf A_n$.
I general we have: $$\liminf A_n\subseteq\limsup A_n\tag0$$ so for proving that the sequence has a limit it is enough to show that $\limsup A_n\subseteq\liminf A_n$.
In this answer I provide a more general setup that might be useful for similar exercises.

Proving that $\lim A_n$ exists with $S=\lim A_n$ actually boils down to proving that:$$\limsup A_n\subseteq S\subseteq\liminf A_n\tag1$$
Combining $(1)$ and $(0)$ we find immediately that:$$\limsup A_n= S=\liminf A_n\tag2$$
Further in some situations it might be handsome to make use of:$$\limsup A_n\subseteq S\iff S^c\subseteq\liminf A_n^c\tag3$$which is justified by the generally true statement $(\limsup A_n)^c=\liminf A_n^c$.
So it is enough to prove that:

*

*For every $s\in S$ some $n_0$ exists such that $x\in A_n$ for every $n\geq n_0$.


*For every $s\notin S$ some $n_0$ exists such that $x\in A_n^c$ for every $n\geq n_0$.
Applying that in the described situation it appears to be enough to prove that:

*

*For every $x\leq0$ some $n_0$ exists such that $x\in E+t_n=(-\infty,t_n)$ for every $n\geq n_0$.

*For every $x>0$ some $n_0$ exists such that $x\in (E+t_n)^c=[t_n,\infty)$ for every $n\geq n_0$.

Both statements are evidently true. In the first bullet we can just take $n_0=1$ and in the second it is used that $t_n\to0$ strictly decreasing.
A: So suppose that $E=(-\infty, 0)$. As $\{t_n\}$ is positive decreasing, for any $N \in \mathbb N$, you have $E_N=(-\infty,t_N)$ and
$$\bigcup_{n \ge N} E_n = E_N$$ and therefore
$$\lim\limits_{n \to \infty} E_n = \limsup E_n =\bigcap_{N \in \mathbb N} \bigcup_{n \ge N} E_N = (-\infty ,0]$$
If you use $\liminf$, then just notice that for any $N \in \mathbb N$
$$\bigcap_{n \ge N} E_n =(-\infty,0]$$ and
$$\lim\limits_{n \to \infty} E_n = \liminf E_n =\bigcup_{N \in \mathbb N} \bigcap_{n \ge N} E_N = (-\infty ,0]$$
