Doubly infinite sequence is a closed set Let $a_i$ be a doubly infinite sequence($i=...-1,0,1,...$) of points in [0,1] such that for all $i$ either $a_i\equiv a_{i-1}+\sqrt2$ (mod 1) or $a_i\equiv a_{i-1}+\sqrt3$ (mod 1). Can the set $A=\{a_i|i\in \mathbb{Z}\}$ be closed?
 A: First note that we need only consider an ordinary sequence $\langle a_k:k\in\Bbb N\rangle$: if we can make $\{a_k:k\in\Bbb N\}$ closed, we can make a closed doubly infinite sequence by letting $$a_{-n}=a_0-(a_n-a_0)=2a_0-a_n$$ for $n\in\Bbb Z^+$.
Let $\langle a_k:k\in\Bbb N\rangle$ be as specified in the problem statement, let $A=\{a_k:k\in\Bbb N\}$, and suppose that $A$ is closed in $[0,1]$. Set $x_0=a_0$ and $y_0=0$. Given $\langle x_k,y_k\rangle$, let
$$\langle x_{k+1},y_{k+1}\rangle=\begin{cases}
\left\langle (x_k+\sqrt2)\bmod 1,y_k\right\rangle,&\text{if }a_{k+1}=(a_k+\sqrt2)\bmod1\\
\left\langle x_k,(y_k+\sqrt3)\bmod 1\right\rangle,&\text{if }a_{k+1}=(a_k+\sqrt3)\bmod1\;,
\end{cases}$$
so that $a_k=(x_k+y_k)\bmod 1$ for each $k\in\Bbb N$. Let $S=\{\langle x_k,y_k\rangle:k\in\Bbb N\}$; it’s easy to check that $S$ is closed in $[0,1]\times[0,1]$.
If $\langle y_k:k\in\Bbb N\rangle$ is eventually constant, then $A$ is dense in $[0,1]$ and therefore not closed, so assume that $\langle y_k:k\in\Bbb N\rangle$ is not eventually constant; it must then be dense in $[0,1]$.
Let $y\in[0,1]$ be arbitrary; there is a subsequence $\langle y_{n_k}:k\in\Bbb N\rangle$ of $\langle y_n:n\in\Bbb N\rangle$ converging to $y$. The sequence $\langle x_{n_k}:k\in\Bbb N\rangle$ is bounded, so it has a subsequence converging to some $x\in[0,1]$, and to avoid getting too deep in subscripts we may as well assume that $\langle x_{n_k}:k\in\Bbb N\rangle$ itself converges to $x$. Then $\big\langle\langle x_{n_k},y_{n_k}\rangle:k\in\Bbb N\big\rangle$ is a sequence in the closed set $S$ converging to $\langle x,y\rangle$, so $\langle x,y\rangle\in S$, and hence $y=y_m$ for some $m\in\Bbb N$. Thus, $[0,1]\subseteq\{y_m:m\in\Bbb N\}$, which is absurd. It follows that $A$ cannot be closed.
