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I know that the set $\{x \in \mathbb R : \lim_{n\to \infty} \sin (n!\pi x)=0 \}$ is a proper uncountable subgroup of $(\mathbb R,+)$ . How to show that it a $F_{\sigma \delta}$ set i.e. that it is a countable union of countable intersections of open sets ?

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  • $\begingroup$ Caution : a countable union of countable intersections of open sets is a $G_{\delta \sigma}$, and a $F_{\sigma \delta}$ is a countable intersection of countable unions of closed set $\endgroup$ – charmd Jan 17 '17 at 18:20
  • $\begingroup$ Do you have a link to the result you know? $\endgroup$ – Colas Dec 2 '18 at 11:14
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We denote $E = \{ x \in \mathbb{R},\ \lim \limits_{n \to +\infty} \sin (n! \pi x) = 0\}$, and $d(t,A) = \inf \{|t-a|,\ a \in A \}$ stands for the distance to a set. First we remind a concavity inequality : $$\forall t \in [-\frac{\pi}{2},\frac{\pi}{2}],\ \frac{2}{\pi}|t| \le |\sin(t)|\le |t|$$

Proof : $\sin$ is odd, so we must prove it only on $[0,\frac{\pi}{2}]$. Denote $\varphi : t \in [0,\frac{\pi}{2}] \mapsto \sin(t) - \frac{2}{\pi}t$. Then $\varphi(0)=\varphi(\frac{\pi}{2})=0$, and $\varphi'(t)=\cos(t)-\frac{2}{\pi}$, so $\varphi$ is increasing and then decreasing on $[0,\frac{\pi}{2}]$, so $\varphi \ge 0$ and thus $\forall t \in [0,\frac{\pi}{2}], \sin(t) \ge \frac{2}{\pi}t$. For the other inequality, we use that $\sin'=\cos$ so $\sin$ is Lipschitz continuous with Lipschitz constant $1$, so $|\sin(t)-\sin(0)|\le 1 |t-0|$ i.e. $|\sin(t)|\le |t|$.

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Now we will use :

$$\forall y \in \mathbb{R},\ \frac{2}{\pi}d\big(y,\pi \mathbb{Z}\big) \le |\sin(y)| \le d\big(y,\pi \mathbb{Z}\big)$$

Proof : We take $y \in \mathbb{R}$. Then it is easy to prove that there exists $n \in \mathbb{Z}$ s.t. $|y-\pi n| = d\big(y,\pi \mathbb{Z}\big)$. Moreover $d\big(y,\pi \mathbb{Z} \big) \le \frac{\pi}{2}$ so $|y-\pi n| \le \frac{\pi}{2}$. Thus we can apply our first inequality, which yields $\frac{2}{\pi} d\big(y,\pi \mathbb{Z} \big) = \frac{2}{\pi}|y-\pi n| \le |\sin (y-\pi n)| \le |y-\pi n| = d\big(y, \pi \mathbb{Z}\big)$. Moreover, as $n\in \mathbb{Z}$, we have $|\sin(y-\pi n)| = |\sin (y)|$, so finally $\frac{2}{\pi} d\big(y, \pi \mathbb{Z}\big) \le |\sin(y)| \le d\big(y, \pi \mathbb{Z}\big)$.


Now back to your original problem. If $x \in E$, then $|\sin(n!\pi x)| \underset{n\to +\infty}{\longrightarrow} 0$, so using the left part of the inequality, $\frac{2}{\pi}d\big( n! \pi x,\ \pi \mathbb{Z} \big) \underset{n\to +\infty}{\longrightarrow} 0$, i.e. $d\big( n! \pi x,\ \pi \mathbb{Z} \big) \underset{n\to +\infty}{\longrightarrow} 0$. Thus, for all $k \in \mathbb{N}$, $$\exists n_0 \in \mathbb{N}:\ \forall n\ge n_0,\ n!\pi x \in \bigcup \limits_{p \in \mathbb{Z}} \left [\pi p - 2^{-k}, \pi p+2^{-k} \right]$$

Dividing by $n! \pi$, we obtain easily that $$x \in \bigcap \limits_{k \in \mathbb{N}} \bigcup \limits_{n_0 \in \mathbb{N}} \bigcap \limits_{n \ge n_0} \bigcup \limits_{p \in \mathbb{Z}} \left [ \frac{p}{n!}-\frac{2^{-k}}{\pi n!}, \frac{p}{n!}+\frac{2^{-k}}{\pi n!} \right ]$$

We denote, for $n_0 \in \mathbb{N}$, $S_{n_0} = \bigcap \limits_{n \ge n_0} \bigcup \limits_{p \in \mathbb{Z}} \left [ \frac{p}{n!}-\frac{2^{-k}}{\pi n!}, \frac{p}{n!}+\frac{2^{-k}}{\pi n!} \right ]$. It is not hard to prove that $\bigcup \limits_{p \in \mathbb{Z}} \left [ \frac{p}{n!}-\frac{2^{-k}}{\pi n!}, \frac{p}{n!}+\frac{2^{-k}}{\pi n!} \right ]$ is closed (the intersection with any compact is the union of finitely many closed sets, so a closed set too...), and thus, because intersections of closed sets are closed, for all $n_0 \in \mathbb{N}$, $S_{n_0}$ is closed.

Now if you take $x \in \bigcap \limits_{k \in \mathbb{N}} \bigcup \limits_{n_0 \in \mathbb{N}} S_{n_0}$, then for every $k \in \mathbb{N}$, there exists $n_0 \in \mathbb{N}$ such that for all $n \ge n_0$, $d \big( n! \pi x, \pi \mathbb{Z} \big) \le 2^{-k}$. Thus $d\big(n! \pi x, \pi \mathbb{Z} \big) \underset{n \to +\infty}{\longrightarrow} 0$, so using the left part of the inequality, we have $\sin (n! \pi x)\underset{n \to +\infty}{\longrightarrow} 0$, that is to say $x \in E$. Hence :

$$ E = \bigcap \limits_{k \in \mathbb{N}} \bigcup \limits_{n_0 \in \mathbb{N}} S_{n_0} = \bigcap \limits_{k \in \mathbb{N}} \bigcup \limits_{n_0 \in \mathbb{N}} \bigcap \limits_{n \ge n_0} \bigcup \limits_{p \in \mathbb{Z}} \left [ \frac{p}{n!}-\frac{2^{-k}}{\pi n!}, \frac{p}{n!}+\frac{2^{-k}}{\pi n!} \right ]$$

As pointed out above, the $S_{n_0}$ are closed, so $E$ is a $F_{\sigma \delta}$ set, that is to say a countable intersection of countable unions of closed sets.

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