what is the name of this set? and how i prove this problem. $$\mathcal{C}^{(i)}=\left\{ {(x,y)\in{\mathbb{R}^{2}}:x^{2}+y^{2}=i^2} \right\}$$
$$\mathcal{A}^{(i)}=\left\{ {(x,y)\in{\mathbb{R}^{2}}:CH\left( {\mathcal{C^{(i)}}} \right)} \right\}$$
$\mathcal{B}$ can be any bounded set, i.e.,
$$\mathcal{B}=\left\{ {(x,y)\in{\mathbb{R}^{2}}:CH\left( {[1,1],[1,-1],[-1,-1],[-1,1]} \right)} \right\}$$
where CH means the convex hull, and $\mathcal{A}^{(i)}$ is the expanding region of circle
How can I prove that $\lim\limits_{i \to \infty}\mathcal{A}^{(i)}$ becomes a superset of $\mathcal{B}$ at some iteration $1 \le i < \infty$ although this is intuitionally trivial.
Does $\lim\limits_{i \to \infty}\mathcal{A}^{(i)}$ become $\mathbb{R}^{2}$?
What is the name of such set of $\lim\limits_{i \to \infty}\mathcal{A}^{(i)}$?
Is this unbounded set? or bounded set?
Since $\lim\limits_{i \to \infty}\mathcal{A}^{(i)}$ is unbounded set, $\lim\limits_{i \to \infty}\mathcal{A}^{(i)}$ becomes a superset of $\mathcal{B}$. Is this sentence can be a proof?
 A: For the first one:


*

*Show that $\mathcal{A}^{(i)}$ contains every point of norm $i$ or less; 

*Show that every point in $\mathcal{B}$ has norm less than some maximum.


The limit is $\mathbb{R}^2$, in the sense that the $\mathcal{A}^{(n)}$ are strictly increasing as $n$ increases through $\mathbb{N}$, and every point in $\mathbb{R}^2$ is contained in some $\mathcal{A}^{(n)}$ 
A: Notice that your set $\mathcal{A}^{(n)}$ is simply a disk around the origin of radius $n$, i.e. $\mathcal{A}^{(n)} = \overline{B}((0,0),n)$.
Now it is clear that $$\liminf_{n\to\infty} \mathcal{A}^{(n)} = \limsup_{n\to\infty} \mathcal{A}^{(n)} = \mathbb{R}^2$$
where 
$$\limsup_{n\to\infty} \mathcal{A}^{(n)} = \{v \in \mathbb{R}^2 : v \in \mathcal{A}^{(n)} \text{ for infinitely many }n\in\mathbb{N}\}$$
$$\liminf_{n\to\infty} \mathcal{A}^{(n)} = \{v \in \mathbb{R}^2 : v \in \mathcal{A}^{(n)} \text{ for all except finitely many }n\in\mathbb{N}\}$$
Therefore we can say $\lim_{n\to\infty } \mathcal{A}^{(n)} = \mathbb{R}^2$. In particular, $\lim_{n\to\infty } \mathcal{A}^{(n)} \supseteq \mathcal{B}$.
