Inner Limit in Hausdorff Space by Open Neighborhoods

Notation
Notation 1: Let $\left(\mathcal{X},\mathcal{T}\right)$ be a topological space (not necessarily normed) and $x\in\mathcal{X}$. The set of open neighborhoods of $x$ will be hereinafter denoted as:

$$ \displaystyle \mho\left(x\right):=\left\{V\in\mathcal{T}\ ,\ x\in V\right\} $$

Notation 2: The following notation will be used for the classes of cofinite and cofinal subsets of $\N$:

$$ \mathcal{N}_\infty:= \left\{N\subset \N| \mathbb{N}\setminus N \text{ is finite}\right\} $$

$$ \mathcal{N}_\infty^\#:= \{N\subset \N| N \text{ is infinite}\} $$

Characterization using open neighborhoods
Proposition 1: Let $\left\{C_n\right\}_{n\in\mathbb{N}}$ be a sequence of sets in a Hausdorff topological space $\left(\mathcal{X},\mathcal{T}\right)$. Then the inner limit (a.k.a Limit Inferior) of $\left\{C_n\right\}_{n\in\mathbb{N}}$ is:

$$ \displaystyle \liminf_n C_n = \left\{x|\forall V\in\mho(x),\ \exists N\in \mathcal{N}_\infty, \forall n\in N: C_n\cap V\neq \emptyset\right\} $$

or equivalenty: $$ \displaystyle \liminf_n C_n = \left\{x|\forall V\in\mho(x),\ \exists N_0\in \mathbb{N},\forall n\geq N_0: C_n\cap V\neq \emptyset\right\} $$

Proof: (1). If $x\in\liminf_n C_n$ then we can find a sequence $\left\{x_k\right\}_{k\in\mathbb{N}}$ such that $x_k\to x$ while $x_k\in C_{n_k}$ and $\left\{n_k\right\}_{k\in\mathbb{N}}\subseteq\mathbb{N}$ is a strictly increasing sequence of indices. For any $V\in\mho\left(x\right)$ there is a $N_0\in\mathbb{N}$ such that for all $i\geq N_0$ it is: $x_i\in V$; but also $x_i\in C_{n_i}$. Thus $C_{n_i}\cap V\neq \emptyset$. Therefore $x$ is in the right-hand side set of the equation.

(2). For the reverse direction assume that $x$ belongs to the right-hand side set of the given equation. Then, there is a strictly increasing sequence $\left\{n_k\right\}_{k\in\mathbb{N}}$. Then, for every $V\in\mho\left(x\right)$ we can find a $x_k\in C_{n_k}\cap V$. Hence, $x_k\to x$ ( in the topology $\mathcal{T}$ ).

Application on Normed Spaces
Instead of arbitrary open sets - if $\mathcal{X}$ is a normed space - we may use open balls, i.e. sets of the form: $$ \mathcal{B}\left(\varepsilon\right)=\{x\in\mathcal{X}:\ \|x\|<\varepsilon\} $$ we denote the unit ball of $\mathcal{X}$ by $\mathcal{B}:=\mathcal{B}(1)$. Then $\mathcal{B}\left(\varepsilon\right)=\varepsilon\mathcal{B}$. This leads us to the following corollary:

Corollary 2: Let $\left\{C_n\right\}_{n\in\mathbb{N}}$ be a sequence of sets in a normed space $\left(\mathcal{X},\|\cdot\|\right)$. Then, $$ \displaystyle \liminf_n C_n = \left\{x|\forall \varepsilon>0,\ \exists N\in \mathcal{N}_\infty, \forall n\in N: x\in C_n+\varepsilon\mathcal{B}\right\} $$

Another characterization of the inner limit is given by the following proposition:

Topological Properties
Proposition 3: Let $(\mathcal{X},\mathcal{T})$ 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.

This proposition manifests a very important topological property of the inner limit of any sequence of sets:

Corollary 4: Let $(\mathcal{X},\mathcal{T})$ 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 proposition 3. Arbitrary intersection of closed sets is closed.

Note: 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.

Note that these results can be extended to nets of sets in topological spaces. For example the last proposition becomes:

Proposition 5: Let $\left(\mathcal{X},\mathcal{T}\right)$ be a Hausdorff topological space and $\langle C_\lambda \rangle_{\lambda\in\Lambda}$ be a net of sets in $\mathcal{X}$ - where $\Lambda$ is a partially ordered set. Then, $$ \displaystyle\liminf_n C_n = \displaystyle\bigcap\left\{ \text{cl}\displaystyle\bigcup_{i\in \Sigma} C_i,\ \Sigma \text{ is a cofinal subset of } \Lambda\right\} $$