$\lim\sup_n A_n$ and $\lim\inf_n A_n$ when $A_n$ is a unit circle with center at $((-1/n)^n,0)$ I was reading the book Probability and Measure Theory by R.B.Ash and stumbled upon the following problem.
Let $\Omega=\mathbb{R}^2, A_n$ the interior of the circle with center at $((-1/n)^n,0)$ and radius $1$. Find $\lim\sup_n A_n$ and $\lim\inf_n A_n$.

So, it kinda seems intuitive to me as both the limit would be a circle at the origin with radius $1$ but I cannot prove it rigorously.I thought about the answer like this- I imagines two circle of unit radius, one at $(-1,0)$ and other at $(-1/2,0)$ and the centers of these two circles coming towards the origin with $1/n$ unit closer in each iteration and when $n$ is very large the centers coincide with the origin.
But this of course does not 'prove' it. So any help would be very much helpful!
 A: The difference between $\liminf A_n$ and $\limsup A_n$ is very simple : $\limsup A_n$ is all those points that belong to infinitely many $A_n$. But $\liminf A_n$ is all those points that belong to all but finitely many $A_n$.
Another way of putting the $\liminf$ : $x$ doesn't belong in $\liminf A_n$ if(and only if) it doesn't belong in infinitely many of the $A_n$.
To do this, of course we must understand the nature of $A_n$. But that is not difficult :  Let $r_n = (-1/n)^n$. We see that $r_n < 0$ for $n$ odd, and $r_n > 0$ for $n$ even, but $r_n \to 0$ as $n \to \infty$. From this , $A_n$ are (open) circles with radius $1$ and center that is converging to $(0,0)$, but the center alternates between the right and the left of the $x$-axis. This information is very useful.

We first run up a triangle inequality corollary : we have $\|x - (r_n,0)\| \geq \|x\| - |r_n|$, very trivially. Using this we can proceed to the trivial cases.

Consider first any $x$ with $\|x\| > 1$. We will show that $x \notin A_n$ for all large enough $n$, but this is clear : if $|r_n| < \|x\|-1$ then $x \notin A_n$ (exercise). Since this holds for large enough $n$ we are done! Consequently, no such $x$ can be in $\limsup A_n$, or therefore in $\liminf A_n$.

Now, we will show that if $\|x\| =1$ then $x \in A_n$ for infinitely many $n$,but $x \notin A_n$ for infinitely many $A_n$ (barring two very special points). Indeed, WLOG let $x$ has positive first coordinate. Then, it is easy to see that if $r_n > 0$  and $|r_n|$ is smaller than the first coordinate of $x$, we have $\|x-(r_n,0)\| < \|x\| = 1$, and if $r_n < 0$ we have $\|x - (r_n,0)\| > \|x\| = 1$ (To see this, write down $x = (x_1,x_2)$ with $x_1>0$ and see what happens). Therefore, $x$ doesn't belong in $\liminf A_n$, but does in $\limsup A_n$.
Of course, if $x$ has negative first coordinate things are similar. But look at $x = (0,1)$ : it doesn't belong in any of the $A_n$! Similarly $(0,-1)$. These two special points don't belong in either the $\limsup$ or $\liminf$ of the sets.

Now, let $\|x\| < 1$. Then there exists $\delta > 1$ such that for all $\|y\|< \delta$ we have $\|x -y\| < 1$. Taking $|r_n| < \delta$, for large enough $n$ it is seen that $\|x - (r_n,0)\| < 1$ so $x \in A_n$. Since this happens for all $n$ after some large enough $N$, we conclude that $x \in \liminf A_n$ and so in $\limsup A_n$.
The characterizations are :

*

*$\limsup A_n = \overline{B(0,1)} \setminus \{(0,1),(0,-1)\}$.


*$\liminf A_n = B(0,1)$.
