set sequence and limits I have the next question.
$\forall n\in\mathbb{N}$, let $A_{n} \subseteq \mathbb{R}$ be given by $A_{n}=\{ x\in\mathbb{R} | -\frac{1}{n}<x<n\}$.
Find $\limsup A_{n}$ and $\liminf A_{n}$ by finding $\bigcap A_{n}$ and $\bigcup A_{n}$.
I think that we need to prove that $\bigcap A_{n} = (-1,\infty)$ and $\bigcup A_{n} = [0,1)$ but I'm not sure how to do so.
 A: You are $\color{red}{\text{not}}$ right. $\color{red}\bigcap A_{n} = (-1,\infty)$ and $\color{ red} \bigcup A_{n} = [0,1)$ are not true.
The true statements are 
$$\color{green}\bigcup A_{n} = (-1,\infty)\tag a$$
and
$$\color{green} \bigcap A_{n} = [0,1)\tag b$$
This how to prove that.
For $(a)$.
For any point of $(-1,\infty)$ there exists an $n$ so that the point in question belongs to $\left(-\frac1n, n \right)$.
For $(b)$.
$[0,1)$ is the set all of whose points belong to all the sets of the form $\left(-\frac1n, n \right)$.
A: You've got your guesses mixed up.
In any case, let's turn back to the definitions.
For any indexing $I$ we have that
$$\bigcap_{i\in I}X_i=\{x\,|\,\forall i \in I,\,x\in X_i\}$$
while
$$\bigcup_{i\in I}X_i=\{x\,|\,\exists i \in I,\,x\in X_i\}.$$
Therefore:


*

*In order to show that $x\in\bigcap_{i\in I}X_i$, you must show that $x$ lies in all of the $X_i$.

*In order to show that $x\in\bigcup_{i\in I}X_i$, you must show that $x$ lies in at least one of the $X_i$.

