Kuratowski $\limsup$ and $\liminf$ of a sequence of subsets of a topological space? From Wikipedia:

Let $(X, d)$ be a metric space. For any point $x ∈ X$ and any
  non-empty compact subset $A ⊆ X$, let $$
     d(x, A) = \inf \{ d(x, a) \mid a \in A \}. $$ For any sequence of such subsets $A_n ⊆ X, n ∈ \mathbb{N}$, the Kuratowski limit inferior
  (or lower closed limit) of $A_n$ as $n → ∞$ is 
  $$\begin{align*}
     \mathop{\mathrm{Li}}_{n \to \infty} A_{n} &= \left\{ x \in X \;\left|\; \limsup_{n \to \infty} d(x, A_{n}) = 0 \right. \right\}\\
         &= \left\{ x \in X \;\left|\; \begin{matrix} \mbox{for all open neighbourhoods } U \mbox{ of } x, \\ U \cap A_{n} \neq \emptyset
 \mbox{ for large enough } n \end{matrix} \right. \right\};
\end{align*} $$ 
  and the
  Kuratowski limit superior (or upper closed limit) of $A_n$ as $n → ∞$
  is $$\begin{align*}
     \mathop{\mathrm{Ls}}_{n \to \infty} A_{n} &= \left\{ x \in X \;\left|\; \liminf_{n \to \infty} d(x, A_{n}) = 0 \right. \right\}\\
 &= \left\{ x \in X \;\left|\; \begin{matrix} \mbox{for all open neighbourhoods } U \mbox{ of } x, \\ U \cap A_{n} \neq \emptyset
 \mbox{ for infinitely many } n \end{matrix} \right. \right\}.\end{align*} $$

The second part of each definition is purely topological and has nothing to do with metric. So I try to represent them using operations on subsets, such as union, intersection, closure, interior, .... 
Questions:


*

*Is it correct that:


*

*"For all open neighbourhoods  $U$ of $x$, $ U \cap A_{n} \neq
    \emptyset$" is equivalent to $x \in \mathrm{closure}(A_n)$?

*So $$ \left\{ x \in X \;\left|\; \begin{matrix} \mbox{for all open
    neighbourhoods } U \mbox{ of } x, \\ U \cap A_{n} \neq \emptyset 
    \mbox{ for large enough } n \end{matrix} \right. \right\} =
    \cup_{i=1}^\infty \cap_{n=i}^\infty \mathrm{closure}(A_n), $$ 
and $$ \left\{ x \in X \;\left|\;
    \begin{matrix} \mbox{for all open neighbourhoods } U \mbox{ of } x,
    \\ U \cap A_{n} \neq \emptyset  \mbox{ for infinitely many } n
    \end{matrix} \right. \right\} = \cap_{i=1}^\infty \cup_{n=i}^\infty \mathrm{closure}(A_n)? $$


*I don't think they are the same as those provided by Andre,
since here there is no taking closure after taking union to make the results to be closed. But this is contrary to the linked Wikipedia article:

$\mathop{\mathrm{Li}}_{n \to \infty} A_{n}$ and $\mathop{\mathrm{Ls}}_{n \to \infty} A_{n}$ are always closed sets in the metric topology on $(X, d)$

Which one is right,  Wikipedia or I?

*How can it be shown that the first one in part 1 is $\left\{ x \in X \left|
\limsup_{n \to \infty} d(x, A_{n}) = 0 \right. \right\}$, and the
second is $\left\{ x \in X \left| \liminf_{n \to \infty} d(x, A_{n})
= 0 \right. \right\}$?


Thanks and regards!
 A: (1) is false, even in the context of metric spaces (in fact, even for the real line $\mathbb{R}$).  For $n \in \mathbb{N}$ let $A_n = [2^{-(n+1)} , 2^{-n} ]$.  It is easy to show that $$\mathrm{Li}_{n \rightarrow \infty} A_n = \{ 0 \} = \mathrm{Ls}_{n \rightarrow \infty} A_n.$$
It is also quite easy to establish that $$\bigcup_{i=1}^\infty \bigcap_{n=i}^\infty \overline{A_n} = \bigcup_{i=1}^\infty \bigcap_{n=i}^\infty A_n = \bigcup_{i=1}^\infty \emptyset = \emptyset,$$ and also that $$\bigcap_{i=1}^\infty \bigcup_{n=i}^\infty \overline{A_n} = \bigcap_{i=1}^\infty \bigcup_{n=i}^\infty A_n = \bigcap_{i=1}^\infty ( 0 , 2^{-i} ] = \emptyset.$$
A: (1) "For all open neighbourhoods $U$ of $x$, for $n$ large enough (resp. infinitely many $n$), $U \cap A_n \neq \emptyset$" 
is not equivalent to 
(2) "For $n$ large enough (resp. infinitely many $n$), for all open neighbourhoods $U$ of $x$, $U \cap A_n \neq \emptyset$"
(1) is the one present in the definition of the Kuratowski limit, and (2) is the one present in your definition using closure(An)
When you switch the order of quantifiers from a $\forall x \exists y P(x,y)$ into a $\exists y \forall x P(x,y)$, the numbers of things your choice of $y$ can depend on gets smaller, so you almost never get equivalent statements. 
However, there is an easy implication between the two : $\exists y \forall x P(x,y) \implies \forall x \exists y P(x,y)$.
So $\cup_{i\ge 1} \cap_{n\ge i} A_n \subset \liminf A_n$ and $\cap_{i\ge 1} \cup_{n\ge i} A_n \subset \limsup A_n$ 
$\limsup A_n$ and $\liminf A_n$ are always closed : $x \notin \limsup A_n$ means that there is an open neighboorhood $U$ of $x$ such that $P(U)$ for some property $P$, and so for every $y \in U$ it is again the case that there is a neighboorhood of $y$, namely $U$, such that $P(U)$.
So in fact, we have $\cup_{i\ge 1} \cap_{n\ge i} A_n \subset \overline{\cup_{i\ge 1} \cap_{n\ge i} A_n} \subset \liminf A_n$ and $\cap_{i\ge 1} \cup_{n\ge i} A_n \subset \overline{\cap_{i\ge 1} \cup_{n\ge i} A_n} \subset \cap_{i\ge 1} \overline{\cup_{n\ge i} A_n} = \limsup A_n$.
There should be examples where all the inequalities are strict.
Also, I've not thought about cases where we don't necessarily pick compact sets for $A_n$.

Since we are dealing with positive sequences, and since the open balls form a basis for the topology,
$$\limsup (d(x,A_n)) = 0 \Leftrightarrow \forall \varepsilon > 0 \exists n \forall m \ge n, d(x,A_m) < \varepsilon \\ \Leftrightarrow \forall \varepsilon > 0 \exists n \forall m \ge n, B(x,\varepsilon) \cap A_m \neq \emptyset \\ \Leftrightarrow \forall U  \exists n \forall m \ge n, U \cap A_m \neq \emptyset  $$
and
$$\liminf (d(x,A_n)) = 0 \Leftrightarrow \forall \varepsilon > 0 \forall n \exists m \ge n, d(x,A_m) < \varepsilon \\ \Leftrightarrow \forall \varepsilon > 0 \forall n \exists m \ge n, B(x,\varepsilon) \cap A_m \neq \emptyset \\ \Leftrightarrow \forall U \forall n \exists m \ge n, U \cap A_m \neq \emptyset  \\ \Leftrightarrow  \forall n \forall U, U \cap (\cup_{m\ge n} A_m) \neq \emptyset  $$
