Krdytkyu

Let $f(x) = 1 - |1 - 2x|$, $a_1 = a$, $a_{n+1} = f(a_n)$. Prove there exists $a in [0, 1]$ such that for every $x in [0, 1]$ there exists a subsequence of $a_n$ convergent to $x$. I've tried to analyze the graph of this function, but couldn't spot anything useful.

Note that $f(x)=2min{x,1-x}$. If the binary expansion of $a$ is
$$a=0.b_1ldots b_n0x_1x_2ldots $$
(where not all $x_i$ are $=1$),
we conclude that the binary expansion of $a_{n+2}$ is $$a_{n+2}=0.x_1x_2ldots$$
So in order to find a subsequence converging to $x$, we only need to make sure that longer and longer prefixes of the binary expansion of $x$ occur after a $0$ in the expansion of $a$ (where we use the expansion $0.1111ldots$ for $x=1$). In order to achieve this for all possible $xin[0,1]$, we just need to make sure that all possible bit patterns occur after a $0$. So let
$$ a=0.0color{red}00color{red}10color{red}{00}0color{red}{01}0color{red}{10}0color{red}{11}0color{red}{000}0color{red}{001}0color{red}{010}0color{red}{011}0color{red}{100}0color{red}{101}0color{red}{110}0color{red}{111}0color{red}{0000}0color{red}{0001}ldots$$