Consider $R: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ given by $R(a,b) = (b,2.5b-a).$ Let $p_0 = (x_0, y_0)$ be arbitrary and $p_{i+1} = R(p_i).$ Most starting points $p_0$ give a divergent path. One can show that the set of $p_0$ which give a convergent path are $(2t, t),$ and these all converge to $(0,0).$
Next consider $S(a,b) = (b, 2.5b-a-a^2).$ Let $p_0 = (x_0, y_0)$ be arbitrary and $p_{i+1} = R(p_i).$ Let $T$ be the set of points $p_0$ in $\mathbb{R}^2$ or maybe $\mathbb{C}^2$ which give converge to $(0,0).$ (Note that $(0.5, 0.5)$ is another fixed point, but not what I'm interested in. oops, edited)
Edit: The comment about possibly being a fractal seems correct. Instead, let $U$ be the set of points $p_0$ such that the path converges to $(0,0)$ and also never goes outside the set $[-0.1,0.1] \times [-0.1,0.1].$
Intuitively, $U$ should be a curve or part of a curve which is locally given by $(2t, t)$ near $(0,0).$ Furthermore, one can assume that $U$ is locally given by $(t, P(t))$ where $P$ is a Taylor series, and then one can actually solve for the coefficients recursively, although with no obvious closed form. My question is what is $U$? Could it be part of an algebraic curve or solution set of a reasonable equation?
Note: I am actually more interested in the following more complicated recursion: $T(a,b,c,d) = (c,d,e,f),$ where $c = (fa)/(f+a+e), d = (ad+b-bc-af)/(ad+b+1+cf-c-bc-af-f)$ and the fixed point $(1-\sqrt{1/2}, 2-\sqrt{2}, 1-\sqrt{1/2}, 2-\sqrt{2}).$ I do not even know the asymptotic form near this fixed point.