Generating Sets that are "Closed" under a linear equation

I have the two linear equations $$ p_{1,2} = kx\pm\frac{k-1}{k} $$

with $k\in[1,\infty)$. I now want to generate a finite set $G\subseteq[-1,1]$ with the following property.$\forall g\in G,\forall j\in\{1,2\}$, either $p_j(g)\in G$ or $|p_j(g)|>1$.

In plain words this just means that I want a finite set such that if I apply my mapping to any point in my set, I either get back another point in the set, or I exit the region the set contains.

My question is, is there a way to generate these finite sets for any value of $k$, such that the largest difference between two consecutive items in the set becomes arbitrarily small?

Solution 1:

Not A Full Answer
One thing immediately noticeable, is that your sets are invariable under union.
If $G,Q$ satisfies your condition, then so does $G\cup Q$ satisfies it.
Now, if we remove your second condition, we get a simpler problem, we'll later take the intersection with $[-1,1]$ and check if it's finite.
Now we are looking for sets which are invariant under $p_{1,2}$.
An even simpler question we can ask is what sets are invariant under just one of $p_{1,2}$.
The union of two such sets is clearly also such a set, the simplest type of these set's are what happends if we start at some $x$ and recursively apply $p_1,p_2$.
Assuming you know some basic linear algebra it is easy to prove $p_{1}^n(x)=\frac{k^n(kx+1)-1}{k},p_{2}^n(x)=\frac{k^n(kx-1)-1}{k}$.
So for any set(of reals) $S$, we have $G-\{\frac{k^n(kx\pm 1)-1}{k}|n\in\mathbb N,x\in S\}$ satisfying the condition of the respective $p_1,p_2$.