Inner Limit in Hausdorff Space by Set Closures

Theorem
Let $\left(\mathcal{X},\mathcal{T}\ \right)$ be a Hausdorff topological space and $\{C_n\}_{n\in\mathbb{N}}$ be a sequence of sets in $\mathcal{X}$. Then,
 * $\displaystyle\liminf_n C_n = \displaystyle\bigcap_{N\in\mathcal{N}_\infty^\#}\text{cl}\displaystyle\bigcup_{n\in N} C_n$

where $\text{cl}$ stands for the closure of a set.

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,


 * $x\in\overline{\displaystyle\bigcup_{n\in\Sigma}C_n}$

(2). Assume that $x\notin \liminf_n C_n$. Then, there is an open neighborhood of $x$, let $W\in \mho\left(x\right)$, such that the set $\Sigma_0:=\left\{n\in\mathbb{N}| W\cap C_n = \emptyset\right\}$ is cofinal. Therefore, $x\notin \overline{\bigcup_{n\in\Sigma_0}C_n}$. This completes the proof.

Implications of the Theorem
This theorem manifests a very important topological property of the inner limit of any sequence of sets:

Corollary : Let $\left(\mathcal{X},\mathcal{T}\ \right)$ be a Hausdorff topological space and $\left\{C_n\right\}_{n\in\mathbb{N}}$ be a sequence of sets in $\mathcal{X}$. Then the set $\liminf_n C_n$ is closed. Additionally, the inner limit depends only on the closure of the sets of the sequence, i.e. if $\left\{D_n\right\}_{n\in\mathbb{N}}$ is any other sequence in $\mathcal{X}$ such that $\overline{C_n}=\overline{D_n}$ for all $n\in\mathbb{N}$, then $\liminf_n C_n=\liminf_n D_n$.

Proof.

This fact is an immediate result of the above stated theorem since arbitrary intersection of closed sets is closed.

Example 1
In order to better understand the topological nature of the inner limit, consider the simple case of a constant sequence of sets $C_n=C$ for all $n\in\mathbb{N}$. Then the inner limit is $\liminf_n C_n=\bar{C}$ (as well as $\limsup_n C_n=\bar{C}$). If for instance $\mathcal{X}=\mathbb{R}^m$ and $C_i=\mathbb{Q}^m$ for all $i\in\mathbb{N}$, then $\liminf_n C_n = \limsup_n C_n = \mathbb{R}^m$ (not $\mathbb{Q}^m$). The outer limit has exactly the same topological properties, i.e. is always closed and depends only on the closures of the sets that participate in the sequence.