First we introduce the following notation:
$$
\mathcal{N}_\infty:= \{N\subset \mathbb{N}| \mathbb{N} \text{\ }N \text{ is finite}\}
$$
and
$$
\mathcal{N}_\infty^\#:= \{N\subset \mathbb{N}| N \text{ is infinite}\}
$$
In most textbooks of real analysis, the limit inferior is defined in one of the following two ways:
$$
\liminf_n C_n = \bigcup_{n=1}^\infty \bigcap_{m=n}^\infty C_m
$$
or
$$
\liminf_n C_n = \left\{x \in \mathcal{X} | x\in C_k \text{ ultimately for all } k \right\}
$$
We need to show that:
$$
\liminf_n C_n = \bigcap_{N\in \mathcal{N}_\infty^\#} \overline{\bigcup_{n\in N}C_n}
$$
where the overline denotes the set closure in the respective topology. For the limit superior we need to show that:
$$
\limsup_n C_n = \bigcap_{N\in \mathcal{N}_\infty} \overline{\bigcup_{n\in N}C_n}
$$
These properties appear in [p.110, 1] as exercises.
[1] R.T. Rockafellar and R. J-B. Wets, "Variational Analysis", Grundlehren der mathematischen Wissenschaften, vol. 317, 1998.
Best Answer
Thanks to Martin Sleziak (for pointing at the book of G. Beer [p.145, Prop. 5.2.2. in 1]) the proof is as follows:
Proposition 1. Let $(\mathcal{X},\mathcal{T})$ be a Hausdorff topological space. Then: $$ \liminf_n C_n = \bigcup_{N\in\mathcal{N}_\infty^\#}\overline{\bigcap_{v\in N}C_v} $$
Proof.
(1). Let $x\in\liminf_n C_n$ and let $\Sigma\in\mathcal{N}_\infty^\#$. Let $W$ be a neighborhood of $x$. There is a $N_0\in\mathbb{N}$ sucht that for all $n\geq N_0$ such that $n\in\Sigma$: $$ W\cap C_n \neq \emptyset $$ Thus, $$ x\in\overline{\bigcup_{n\in\Sigma}C_n} $$ (2). Assume that $x\notin \liminf_n C_n$. Then, there is an open neighborhood of $x$, let $W\ni x$, such that $\Sigma_0:=\{n\in\mathbb{N}| W\cap C_n = \emptyset\}$. Therefore, $x\notin \overline{\bigcup_{n\in\Sigma_0}C_n}$. This completes the proof. $\square$
Note 1: Characterization of the closure of a set: Let $C\subset \mathcal{X}$ and $\bar{C}$ denote its closure which is defined as:
$$ \bar{C}=\bigcap\{F\supset C| F^c\in \mathcal{T}\} $$ Then:
$$ x\in\bar{C} \Leftrightarrow \forall V\in\mathcal{T},\ V\ni x:\ V\cap C \neq \emptyset $$
Note 2: This result is stated in [1] for nets of sets in $\mathcal{X}$, $\{C_n\}_{n\in\Lambda}$ where $\Lambda$ is a partially ordered set. Then the class $\mathcal{N}_\infty^\#$ is replaced by the family of cofinal sets of $\Lambda$. Set set $\Sigma$ is called a cofinal subset of $\Lambda$ if for all $\lambda\in\Lambda,\ \exists\sigma\in\Sigma:\ \sigma\geq\lambda$.
Corollary 2. Let $\{C_n\}_{n\in\mathbb{N}}$ be a sequence of subsets of $\mathcal{X}$. Then: $$ \bigcap_{n\in\mathbb{N}}C_n \subseteq \overline{\bigcap_{n\in\mathbb{N}}C_n} \subseteq \liminf_n C_n $$
Proof. It follows from Proposition 1 taking $N=\mathbb{N}\in\mathcal{N}_\infty^\#$. $\square$
[1] G. Beer, "Topologies on Closed and Closed Convex Sets", Kluwer Academic Publishers, ISBN: 0-7923-2531-1.