# 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

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Could you clarify what you mean by "rotated to size $2\pi n\theta$? – Pedro Tamaroff Aug 21 '12 at 1:34
@PeterTamaroff "rotated by the angle $2\pi n\theta$", I suppose. – user31373 Aug 21 '12 at 1:48
@PeterTamaroff There is dependence on $n$: namely, $A_n$ is the set that you get when you rotate the square $\max(|x|,|y|)\le 1$ by $2\pi \pi \theta$. – user31373 Aug 21 '12 at 1:52
@PeterTamaroff This is more or less standard: $\limsup A_n:=\bigcap_{m=1}^\infty \bigcup_{n=m}^\infty A_n$; $\liminf A_n:=\bigcup_{m=1}^\infty \bigcap_{n=m}^\infty A_n$. If you think in terms of characteristic functions $\chi_{A_n}$, this is just the usual $\limsup$ and $\liminf$ of functions. – user31373 Aug 21 '12 at 2:07

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

$$\liminf_{n\to\infty} A_n = \Bbb{R}^2\setminus \left(\limsup_{n\to\infty} \Bbb{R}^2\setminus A_n\right) = \mathrm{cl}(B_1).$$

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Nice argument. Just for completeness, I'll.add that the rational case will have two polygons: union and intersection of all rotations by multiples of $2\pi\theta$. – user31373 Aug 23 '12 at 3:16