Elementary proof for non-existence of a pointwise convergent subsequence of $\{\sin (nx)\}$ My teacher showed this proof using the dominated convergence theorem or Fourier analysis, but I wonder if there is an elementary proof of this problem. My teacher said it is difficult to solve this in an elementary way, but do you know how?
 A: Here is an elementary proof inspired by this proof.
Fix the subsequence $1\leq n_1<n_2<\ldots$. I will inductively construct sequences $a_1\leq a_2\leq\ldots$ and $b_1\geq b_2\geq\ldots$ such that:


*

*$a_i<b_i$

*$\sin(n_k[a_i,b_i])\subseteq(-1)^i[1/2,1]$ for some $n_k\geq i$.


Let $a_1=(-\pi/2)/n_1$ and $b_1=(-\pi/6)/n_1$. Suppose $[a_i,b_i]$ have been constructed. Pick $n_k$ larger than $i$ and larger than $3\pi/(b_i-a_i)$. Let $m=\lfloor n_kb_i/(2\pi)-1/4\rfloor$. Note that
$$
  2\pi m+\pi/2\leq n_kb_i,
$$
$$
  2\pi m-\pi/2\geq n_kb_i-3\pi\geq n_ka_i.
$$
Thus we can take
$$
  a_{i+1}=\left(2\pi m+\pi/6\right)/n_k,\,
  b_{i+1}=\left(2\pi m+\pi/2\right)/n_k
$$
if $i+1$ is even, and
$$
  a_{i+1}=\left(2\pi m-\pi/2\right)/n_k,\,
  b_{i+1}=\left(2\pi m-\pi/6\right)/n_k
$$
if $i+1$ is odd. This concludes the inductive construction.
Now $a_i$ is a bounded monotonic sequence, so it converges to some $a$. For each $i$ we have $a_i\leq a\leq b_i$, so
$$
  \sin(n_ka)\in(-1)^i[1/2,1]
$$
for some $n_k\geq i$. This shows $\sin(n_ka)$ cannot converge as $k\to\infty$.
A: Here is a proof using measure theory (so unfortunately not elementary).
Fix the subsequence $n_1<n_2<\ldots$ and let $X$ be the set of $x\in\mathbb R$ for which $\sin(n_kx)$ converges. For any $\epsilon\in(0,1)$ let
$$\begin{eqnarray*}
  Y_k^\epsilon&=&\{x\in[0,4\pi]\mid|\sin(n_kx)-\sin(n_{k+1}x)|<2\epsilon^2\},\\
  X_k^\epsilon&=&\bigcap_{l\geq k}Y_k^\epsilon.
\end{eqnarray*}$$
Note that each $Y_k^\epsilon$ is open and therefore measurable, so $X_k^\epsilon$ is also measurable. Moreover
$$
  |\sin(n_kx)-\sin(n_{k+1}x)|=2|\cos(ax)\sin(bx)|
$$
where $a=\frac{n_k+n_{k+1}}2$, $b=\frac{n_k-n_{k+1}}2$. Hence $Y_k^\epsilon\subseteq A_k^\epsilon\cup B_k^\epsilon$ where
$$\begin{eqnarray*}
  A_k^\epsilon&=&\{x\in[0,4\pi]\mid|\cos(ax)|<\epsilon\},\\
  B_k^\epsilon&=&\{x\in[0,4\pi]\mid|\sin(bx)|<\epsilon\}.
\end{eqnarray*}$$
We have
$$\begin{eqnarray*}
  \mu(A_k^\epsilon)&=&\frac1a\mu(\{u\in[0,4a\pi]\mid|\cos(u)|<\epsilon\})\\
    &=&2\mu(\{u\in[0,2\pi]\mid|\cos(u)|<\epsilon\})\\
    &=&2\mu\left((\pi/2-\delta,\pi/2+\delta)\cup(3\pi/2-\delta,3\pi/2+\delta)\right)\\
    &=&8\delta
\end{eqnarray*}$$
where $\delta=\sin^{-1}(\epsilon)$. A similar calculation applies to $B_k^\epsilon$. Hence
$$
  \mu(X_k^\epsilon)\leq\mu(Y_k^\epsilon)\leq16\sin^{-1}(\epsilon).
$$
Note that $X_1^\epsilon\subseteq X_2^\epsilon\subseteq\ldots$, so
$$
  \mu\left(\bigcup_kX_k^\epsilon\right)\leq16\sin^{-1}(\epsilon).
$$
Finally
$$
  X\cap[0,4\pi]\subseteq\bigcup_kX_k^\epsilon.
$$
Since $\epsilon\in(0,1)$ was arbitrary and $X$ is $2\pi$-periodic, $X$ cannot contain any set of positive measure. (I'd like to say that $\mu(X)=0$, but it may not be measurable). In particular $X$ cannot equal $\mathbb R$.

I tried to find a proof which shows that $X$ is countable, but this is not always true. For example, let $n_k=k!$. Let $A$ be the (uncountable) set of sequences with values in $\{0,1\}$, and define $f:A\rightarrow\mathbb R$ by
$$
  f(a)=2\pi\sum_{k=1}^\infty\frac{(-1)^{a_k}}{k!}.
$$
Note that for $k\geq1$ we have
$$
  \left|\sum_{l>k}\frac{(-1)^{a_l}}{l!}\right|\leq
    \frac2{k+1}\sum_{l>k}\frac{1}{l!2^{l-k}}=\frac2{(k+1)!}\leq\frac1{k!}.
$$
This implies $f$ is injective, so $f(A)$ is uncountable. Moreover
$$
  |\sin(n_kf(a))|=\left|\sin\left(2\pi k!\sum_{l>k}\frac{(-1)^{a_l}}{l!}\right)\right|
    \leq\frac{4\pi}{k+1}
$$
so $\sin(n_kf(a))\rightarrow0$ as $k\rightarrow\infty$. Thus $f(A)\subseteq X$.
