Inner Limit of Sequence of Sets in Normed Space

Theorem
Let $\left({\mathcal X,\left\|{\cdot}\right\|}\right)$ be a normed space and $\left \langle{C_n}\right \rangle_{n \in \N}$ be a sequence of sets in $\mathcal X$.

The inner limit of $\left \langle{C_n}\right \rangle_{n \in \N}$ is:
 * $\displaystyle \liminf_n \ C_n = \left\{{x \in X : \lim_n \ d \left({x, C_n}\right) = 0}\right\}$

Proof
$(1): \quad$ We need to show that:
 * $\displaystyle \limsup_n \ d \left({x, C_n}\right) = 0$

Let us assume that:
 * $\displaystyle \limsup_n \ d \left({x, C_n}\right) > 0$

that is, that there exists an increasing sequence of indices $\left \langle{n_k}\right \rangle_{k \in \N}$ so that $d \left({x,C_{n_k}}\right) \to_k a > 0$.

This suggests that there is a $\epsilon_0 > 0$ such that:
 * $\forall k \in \N: d \left({x, C_{n_k}}\right) > \epsilon_0$

However, according to Inner Limit in Hausdorff Space by Set Closures:


 * $\displaystyle x \in \operatorname{cl} \left({\bigcup_{k \in \N} C_{n_k}}\right)$

while:
 * $\displaystyle d \left({x, \operatorname{cl} \left({\bigcup_{k \in \N} C_{n_k}}\right)}\right) \ge \epsilon_0$

which is a contradiction.

Hence:
 * $\displaystyle \limsup_n \ d \left({x, C_n}\right) = 0$

That is:
 * $\displaystyle \lim_n \ d \left({x, C_n}\right) = 0$

and thus we have proven that $x$ is in the right-hand side set.

$(2): \quad$ Assume that $x$ in the right-hand side of the given equation.

This is:
 * $\displaystyle \lim_n \ d \left({x, C_n}\right) = 0$

For any $\epsilon > 0$, we can find $n_0 \in \N$ such that:
 * $\forall n \ge n_0: d \left({x, C_n}\right) \leq \dfrac \epsilon 2$

By definition:
 * $d \left({x, C_n}\right) = \inf \left\{{\left\|{x - y}\right\|: y \in C_n}\right\}$

Thus we can find a $y_n \in C_n$ such that:


 * $\left\|{y_n - x}\right\| < d \left({x, C_n}\right) + \dfrac \epsilon 2 = \epsilon$

That is:


 * $\exists y_n \in C_n: \left\|{y_n - x}\right\| < \epsilon$

Therefore:
 * $x \in C_n + \epsilon \mathcal B$

where $\mathcal{B}$ is the open unit ball of $\mathcal{X}$, i.e. $\mathcal{B}:=\left\{ x\ |\ \left\|x\right\|<1 \right\}$.

Note also that $\epsilon \mathcal B=\left\{ x\ |\ \left\|x\right\|<\epsilon \right\}$.

From this observation, it follows that:


 * $\displaystyle x \in \liminf_n C_n$

according to the Inner Limit in Hausdorff Space by Open Neighborhoods.