Let $$A_n = \mathbb{B}\left(\left(\frac{(-1)^n}{n},0\right),1\right) \ , \ \forall n \in \mathbb{R}^2$$
I don't know How can I calculate $\limsup A_n$ as $n \rightarrow \infty$ and $\liminf A_n$ as $n \rightarrow \infty$ of this set $A_n$?
I have that $$A_1 = \mathbb{B}((-1,0),1)$$ $$A_2 = \mathbb{B}\left(\left(\frac{1}{2},0\right),1\right)$$ $$A_3 = \mathbb{B}\left(\left(-\frac{1}{3},0\right),1\right)$$ etc..
For a sequence of subsets $A_n$ of a set $X$, the $\limsup A_n= \bigcap_{N=1}^\infty \left( \bigcup_{n\ge N} A_n \right)$ and $\liminf A_n = \bigcup_{N=1}^\infty \left(\bigcap_{n \ge N} A_n\right)$.
But How can I calculate this in $\mathbb{R}^2$?
Could someone help me to solve and understand this, pls...
Thanks for your time and help.
You are probably aware of the following characterizations:
$$x \in \limsup_n A_n \iff x \in A_n \text{ for infinitely many } n \in \mathbb{N}$$ $$x \in \liminf_n A_n \iff x \in A_n \text{ for all except finitely many } n \in \mathbb{N}$$
I claim that $\liminf_n A_n = B((0,0),1)$ and $\limsup_n A_n = \overline{B}((0,0),1) \setminus \{(0, \pm 1)\}$.
Indeed, for any $(x,y )\in B((0,0),1)$ we have that $$d\left((x,y), \left(\frac{(-1)^n}{n},0\right)\right)^2 = \left(x - \frac{(-1)^n}{n}\right)^2 + y^2 \xrightarrow{n\to\infty} x^2 + y^2 < 1$$
so there exists $n_0 \in \mathbb{N}$ such that $$n \ge n_0 \implies d\left((x,y), \left(\frac{(-1)^n}{n},0\right)\right)^2 < 1 \implies (x,y) \in A_n$$
This shows that $B((0,0),1) \subseteq \liminf_n A_n, \limsup_n A_n$.
Now take any point $(x,y) \in \partial B((0,0),1)$ which is not $(0, \pm1)$. We will show that $(x,y) \in \limsup_n A_n$.
Indeed, if $x > 0$, for even $n \in \mathbb{N}$ we have:
$$d\left((x,y), \left(\frac{(-1)^n}{n},0\right)\right)^2 = \left(x - \frac{(-1)^n}{n}\right)^2 + y^2 = \underbrace{\left(x - \frac1{n}\right)^2}{} + y^2 < x^2 + y^2 = 1$$
so $(x, y) \in \limsup_n A_n$.
For odd $n \in \mathbb{N}$ however, we have:
$$d\left((x,y), \left(\frac{(-1)^n}{n},0\right)\right)^2 = \left(x - \frac{(-1)^n}{n}\right)^2 + y^2 = \underbrace{\left(x - \frac1{n}\right)^2}_{>1} + y^2 > 1$$
so $(x,y) \notin \liminf_n A_n$. Analogously we would prove it for $x < 0$.
So, $$\partial B((0,0),1) \setminus \{(0, \pm 1)\} \subseteq \limsup_n A_n$$
For any $(x,y) \notin \overline{B}((0,0),1)$, similarly we have:
$$d\left((x,y), \left(\frac{(-1)^n}{n},0\right)\right)^2 = \left(x - \frac{(-1)^n}{n}\right)^2 + y^2 \xrightarrow{n\to\infty} x^2 + y^2 >1$$
so there exists $n_1 \in \mathbb{N}$ such that $$n \ge n_1 \implies d\left((x,y), \left(\frac{(-1)^n}{n},0\right)\right)^2 > 1 \implies (x,y) \notin A_n$$
For $(x,y) = (0, \pm 1)$ we have:
$$d\left((x,y), \left(\frac{(-1)^n}{n},0\right)\right)^2 = \left( \frac{(-1)^n}{n}\right)^2 + y^2 \xrightarrow{n\to\infty} \frac1{n^2}+1 >1$$
so $(x,y) \notin \liminf_n A_n, \limsup_n A_n$.
This shows that $\liminf_n A_n, \limsup_n A_n \subseteq \overline{B}((0,0),1) \setminus \{(0, \pm 1)\}$. We also previously showed that elements of $\partial B((0,0),1)$ are not in $\liminf_n A_n$ so $\liminf_n A_n \subseteq B((0,0),1)$.
Hence, $\liminf_n A_n = B((0,0),1)$ and $\limsup_n A_n = \overline{B}((0,0),1) \setminus \{(0, \pm 1)\}$.