Let $B(x)=\frac{3x+1}{4}$ and $C(x)=\frac{3x-1}{4}$. Then $(-1,1)$ is a stable interval for both $B$ and $C$. Let $S\subseteq (-1,1)$ be nonempty and stable by both $B$ and $C$. Note that for any initial value $x_0$, the iterates $B^n(x_0)$ and $C^n(x_0)$ converge to $1$ and $-1$ respectively, so $1$ and $-1$ are accumulation points for $S$.
My question: Is $0$ an accumulation point for $S$ ?
What I tried : With the help of a computer, I found that if we put
$$ \begin{array}{lcl} f(x)&=&CBBCBCBCCBBCCBCBBCBB \\ & & CCBBCBCCBBCBCCBBCCBB \\ & & CCBCBBCCBBCCBCBBBCCC \\ & & BBBCBBCCB(x) \\ r&=&\frac{15571040882549573754344675288597}{348449142892655818255415064238273319835661} \approx 4 \times 10^{-11} \end{array} $$
then $f$ has the property that $0<f(x)<x$ whenever $x\gt r$, because $x-f(x)=\bigg(1-\big(\frac{3}{4}\big)^{69}\bigg)(x-r)$. In particular $S$ always contains an element in $(0,r]$. I see no clear pattern in the above long string of $C$'s and $B$'s though.
Proof: Every point in $(-1,1)$ is mapped to by either $B$ or $C$ i.e. $B(-1,1)\cup C(-1,1)=(-1,1)$. Hence $BB(-1,1)\cup BC(-1,1)\cup CB(-1,1) \cup CC(-1,1)=(-1,1)$ and likewise for the length $n$ strings of $B$s and $C$s. Therefore for any $n$ there exists a string of $B$s and $C$s of length $n$ whose image contains $0$. By choosing sufficiently large $n$ we can force this image to be within an arbitrarily small interval.
If $S$ is non-empty then there exists a string of $B$s and $C$s which maps any $x_0\in S$ to a point in $(-\varepsilon,\varepsilon)$. Therefore zero is an accumulation point of $S$.