Inner Limit of Sequence of Sets in Normed Space

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

Proof.

(1). We now need to show that $\limsup_n d\left(x,C_n\right)=0$. Let us assume that $\limsup_n d\left(x,C_n\right)>0$, i.e. there exists an increasing sequence of indices $\left\{n_k\right\}_{k\in\N}$ so that $d\left(x,C_{n_k}\right)\to_k a > 0$. This suggests that there is a $\varepsilon_0>0$ such that for all $k\in\N$ one has that $d\left(x,C_{n_k}\right)>\varepsilon_0$. However, according to this proposition, $x\in\text{cl}\bigcup_{k\in\N}C_{n_k}$ while $d(x,\text{cl}\bigcup_{k\in\N}C_{n_k})\geq\varepsilon_0$ which is a contradiction. Hence, $\limsup_n d\left(x,C_n\right)=0$, i.e. $\lim_n d\left(x,C_n\right)=0$ 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 $\lim_n d\left(x,C_n\right)=0$. For any $\varepsilon>0$, we can find $n_0\in\N$ such that $d\left(x,C_{n}\right)\leq \frac{\varepsilon}{2}$ for all $n\geq n_0$. By definition, we have that $d\left(x,C_{n}\right)=\inf\{\left\|x-y\right\|,\ y\in C_{n}\}$, thus we can find a $y_n\in C_{n}$ such that

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

That is:

$$ \exists\ y_n\in C_{n}:\ \left\|y_n-x\right\|<\varepsilon $$

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