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I will rewrite my question using Matt F. suggestion.

Consider the logical structure $L = (\mathbb{R}, +, *, 1, 0, =)$ and a function $f:\mathbb{R}\to\mathbb{R}$. Consider the map $Q:2^\mathbb{R}→2^\mathbb{R}$ by $$Q(S)=\{a:∃x,y∈S \: a \in [1, 3]∧a=f(x)∧ x = y^2\}$$ which has a definition in $L$ and uses $f$ only once. For a given logical structure with one function $f$, can we characterize the maps from $2^\mathbb{R}→2^\mathbb{R}$ which have similar first-order definitions in $L$ and use $f$ linearly (in the sence that there are no iterations $f(\dots f(\ldots))$)?


Motivation: I wanted to give logically correct definition of all constructions of fractals such as Apollonian gasket and Sierpinski triangle. In the setting of Apollonian gasket it is natural to replace $\mathbb{R}$ with the set $\mathcal{S}$ of circles in $\mathbb{R}^2$ and $f$ with the map that inputs three pairwase tangent circles and outputs two tangent circles to a given three ones. So the map $$Q(S) = \Big\{a: \exists b, x, y, z\in S\: \{a, b\} = f(x, y, z)∧ a\notin S\Big\}$$ realize Apollonian fractal as $Q(S_0)\sqcup Q(Q(S_0))\sqcup\ldots$ where $S_0$ denote three pairwise tangent circles.

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    $\begingroup$ I might ask this as: Consider $\mathbb{R}$ in the language $L$ with one function $f$, and a family of relations like $$ \big\{Sq(x,y):=x=y^2,\ \ In(x):=x\in[1,3]\big\} $$ Consider the map $\mathcal{Q}: 2^\mathbb{R}\to 2^\mathbb{R}$ by $$\mathcal{Q}(S) = \{a: \exists x,y\in S\ In(a) \wedge a=f(x) \wedge Sq(x,y)\}$$ which has a first-order definition in $L$ and uses $f$ only once. For a given language with one function $f$ and a family of relations, can we characterize the maps from $2^\mathbb{R}\to 2^\mathbb{R}$ which have similar first-order definitions in $L$ and use $f$ only once? $\endgroup$
    – user44143
    Commented Sep 9, 2021 at 12:31

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