the geometric description of limsup $A_n$ and liminf $A_n$ suppose $A_n=\{(x,y):|x|\leq1 ,|y|\leq1\}$ is square where is rotated to size $2\pi n\theta$.
What is the geometric description of limsup $A_n$ and liminf $A_n$
I: if $\theta$ be Rational
II: if $\theta$ be Irrational
 A: I will consider only the case II.

Lemma. Let $r > 0$ and $U$ be a nonempty open subset of $C_r := \{ (x, y) : x^2+y^2 = r^2\}$, and $R_{\varphi}$ denote the rotation by angle $\varphi$. Then we claim that for any irrational $\theta$,
$$\limsup_{n\to\infty} R_{2\pi n\theta} U = C_r.$$
Proof. In the proof, we use the polar coordinate.
Pick any $p \in U$ and write $p = (r, \varphi_0)$. Then there exists $0 < \delta \ll 1$ such that
$$V_p = \{ (r, \varphi) : \left|\varphi - \varphi_0\right| < \delta \}$$
is in $U$. Now recall that $\left< n \theta \right> = n\theta - \lfloor n\theta \rfloor$, the sequence of the fractional part of $n\theta$, is dense in $[0, 1)$. Thus for any $\varphi$ satisfying $\varphi-\varphi_0 \in [0, 2\pi)$, we can find infinitely many positive integers $n$ such that
$$\left|2\pi\left<n \theta \right> + \varphi_0 - \varphi \right| = \left|2\pi\left<n \theta \right> - (\varphi-\varphi_0) \right| < \delta. $$
This implies that $(r, \varphi) \in R_{2\pi n\theta} U$ is satisfied for infinitely many $n$, hence $(r, \varphi) \in \limsup R_{2\pi n\theta} U$. Therefore we have $C_r = \limsup_n R_{2\pi n\theta} U$.

Since $A_n = R_{2\pi n\theta} A_0$ and
$$\limsup_{n\to\infty}A_n = \bigcup_{r \geq 0} \left(\limsup_{n\to\infty}A_n\right) \cap C_r = \bigcup_{r \geq 0} \limsup_{n\to\infty}(A_n \cap C_r), $$
we are going to apply the Lemma to the following observation
$$ A_0 \cap C_r = \begin{cases}
\text{has nonempty interior in } C_r & r < \sqrt{2} \\
\{(\pm1, \pm1)\} & r = \sqrt{2} \\
\varnothing & r > \sqrt{2}.
\end{cases}$$
Thus applying the lemma to each nonempty open subset $\mathrm{int}(A_0) \cap C_r$ of $C_r$, for $r < \sqrt{2}$, we find that $B_{\sqrt{2}} \subset \limsup_n A_n$. Now, since each element of $\bigcup_{n} \{R_{2\pi n\theta}(\pm1, \pm1)\}$ occurs exactly once in the sequence $R_{2\pi n\theta}(\pm1,\pm1)$ by irrationality of $\theta$, we have $\limsup_n \{R_{2\pi n\theta}(\pm1, \pm1)\} = \varnothing$. Therefore we have
$$ \limsup_{n\to\infty} A_n = B_{\sqrt{2}}.$$
Applying similar argument to $\Bbb{R}^2\setminus A_n$, we have
$$ \liminf_{n\to\infty} A_n = \Bbb{R}^2\setminus \left(\limsup_{n\to\infty} \Bbb{R}^2\setminus A_n\right) = \mathrm{cl}(B_1).$$
