$\limsup $ and $\liminf$ of a sequence of subsets relative to a topology From Wikipedia

if $\{A_n\}$ is a sequence of subsets of a topological space $X$,
  then:
$\limsup A_n$, which is also called the outer limit, consists of those
  elements which are limits of points in $A_n$ taken from (countably)
  infinitely many n. That is, $x \in \limsup A_n$ if and only if there
  exists a sequence of points $\{x_k\}$ and a subsequence $\{A_{n_k}\}$ of
  $\{A_n\}$ such that $x_k \in A_{n_k}$ and $x_k \rightarrow x$ as $k
 \rightarrow \infty$.
$\liminf A_n$, which is also called the inner limit, consists of those
  elements which are limits of points in $A_n$ for all but finitely many
  n (i.e., cofinitely many n). That is, $x \in \liminf A_n$  if and only
  if there exists a sequence of points $\{x_k\}$ such that $x_k \in A_k$
  and $x_k \rightarrow x$ as $k \rightarrow \infty$.

According to the above definitions (or what you think is right), my questions are:


*

*Is $\liminf_{n} A_n \subseteq \limsup_{n} A_n$?

*Is $(\liminf_{n} A_n)^c = \limsup_{n} A_n^c$?

*Is $\liminf_{n} A_n =
    \bigcup_{n=1}^\infty\overline{\bigcap_{m=n}^\infty A_m}$? This is
based on the comment by Pantelis Sopasakis following my previous
question.

*Is $\limsup_{n} A_n =
\bigcap_{n=1}^\infty\overline{\bigcup_{m=n}^\infty A_m}$, or
$\limsup_{n} A_n =
\bigcap_{n=1}^\infty\operatorname{interior}(\bigcup_{m=n}^\infty
A_m)$, or ...?


Thanks and regards!

I must admit that I did not know these definitions, either.


*

*Yes, because if $x \in \liminf A_n$ you have a sequence $\{x_k\}$ with $x_k \in A_k$ and $x_k \rightarrow x$ and you can choose your subsequence $\{A_{n_k}\}$ to be your whole sequence $\{A_n\}$.

*For $X = \mathbb{R}$ take $A_n = \{0\}$ for all $n$. Then you have $(\liminf A_n)^c = \{0\}^c = \mathbb{R}\setminus{\{0\}}$, but $\limsup A_n^c = \mathbb{R}$. So in general you do not have equality.

*Of course you have "$\supseteq$". But if you take $A_n := [-1,1- \frac{1}{n}) \subseteq \mathbb{R}$ you have the sequence $\{x_k\}$ defined by $x_k := 1 - \frac{2}{k}$ which converges to $1$ which is not in $\overline{\bigcap_{n = m}^{\infty} A_n} = [-1,1-1/m]$ for any $m \in \mathbb{N}$. But it works of course, if $\{A_k\}$ is decreasing.

*In your first assumption you have "$\subseteq$". A sequence $\{x_k\}$ with $x_k \in A_{n_k}$ with $\{A_{n_k}\}$ some subsequence of $\{A_n\}$ lies eventually in $\bigcup_{n = m}^\infty A_n$ for all $m$ by the definition of subsequence. Now for metric spaces you have "$\supset$", too, as you can start by choosing $x_1 \in A_{n_1}$ with $d(x_1,x) < 1$ and proceed by induction choosing $x_{n_k} \in A_{n_k}$ with $n_k > n_{k-1}$ and $d(x_{n_k},x) < \frac{1}{k}$. I doubt that this holds in general (at least this way of proving it does not).
