Inner Limit of Sequence of Sets in Normed Space

Properties of the Inner Limit on Normed Spaces
Proposition 1: Let $(\mathcal{X},\|\cdot\|)$ be a normed space and $\{C_n\}_{n\in\mathbb{N}}$ be a sequence of sets in $\mathcal{X}$. The inner limit (a.k.a Limit Inferior) of a sequence of sets is: $$ \liminf_n C_n = \{ x\in X | \lim_n d(x,C_n)=0 \} $$

Proof.

(1). We now need to show that $\limsup_n d(x,C_n)=0$. Let us assume that $\limsup_n d(x,C_n)>0$, i.e. there exists an increasing sequence of indices $\{n_k\}_{k\in\mathbb{N}}$ so that $d(x,C_{n_k})\to_k a > 0$. This suggests that there is a $\varepsilon_0>0$ such that for all $k\in\mathbb{N}$ one has that $d(x,C_{n_k})>\varepsilon_0$. However, according to this proposition, $x\in\text{cl}\bigcup_{k\in\mathbb{N}}C_{n_k}$ while $d(x,\text{cl}\bigcup_{k\in\mathbb{N}}C_{n_k})\geq\varepsilon_0$ which is a contradiction. Hence, $\limsup_n d(x,C_n)=0$, i.e. $\lim_n d(x,C_n)=0$ and this way we have proven that $x$ is in the right-hand side set.

(2). Assume that $x$ in the right-hand side of the given equation. This is $\lim_n d(x,C_n)=0$. For any $\varepsilon>0$, we can find $n_0\in\mathbb{N}$ such that $d(x,C_{n})\leq \frac{\varepsilon}{2}$ for all $n\geq n_0$. By definition, we have that $d(x,C_{n})=\inf\{\|x-y\|,\ y\in C_{n}\}$, thus we can find a $y_n\in C_{n}$ such that

$$\|y_n-x\|<d(x,C_{n})+\frac{\varepsilon}{2}=\varepsilon$$

That is:

$$ \exists\ y_n\in C_{n}:\ \|y_n-x\|<\varepsilon $$ Therefore, $x\in C_{n} + \varepsilon \mathcal{B}$ from which it follows that $x\in\liminf_n C_n$ (According to this proposition).

Proposition 2: Let $(\mathcal{X},\|\cdot\|)$ be a normed space and $\{C_n\}_{n\in\mathbb{N}}$ be a sequence of sets in $\mathcal{X}$. The inner limit of a sequence of sets is: $$ \limsup_n C_n = \{ x\in X | \liminf_n d(x,C_n)=0 \} $$