A metric geometry problem which calculates the limitation of human eyes

Question

This is the update version of this question A functional inequality which calculates the limitation of human eyes

Let an Euclidean space $M$ (or a path connected metric space) be partitioned into finite path-connected parts $M_1,...,M_k$, we call space $M$ with a partition is a figure $F$. Consider this test: Give a figure $F$, we ask Bob to find a unique point $X$ on $M$ just by his bare eye. Bob knows all informations about figure $F$ and the point $X$. Of course Bob almost can't choose the point $X$ exactly, but may choose a point $X'$ which is close to $X$. Lets $E^F_X$ be error set of $X$, the set of all posible points that Bob may choose when find $X$.

For example, give a triangle $ABC$ and ask Bob find its centroid $G$, you can think as decompose $M$ into $3$ point $A,B,C$, $3$ open segment $AB,BC,CA$ and $2$ region inside and outside $ABC$

We want to find $E^F_X$ for all $X\in M$ and to do it mathematically, I've made some axioms:

1. $X\in E^F_X$ and $E^F_X$ is path-connected closed set. Of course Bob may choose $X$ exactly and the set $E^F_X$ should be nice
   
2. If $X\in M_i$ then $E^F_X\subset M_i$. It's a obvious property of $X$ that Bob can't be obey.
   
3. If a figure $F'$ come from a partition finer than the partition of $F$ then $E^{F'}_X\subset E^F_X$. Bob can choose more exactly with more information. (a partition $P'$ is finer than $P$ if every elements of $P'$ is a subset of some elements of $P$).
   
4. Let $f$ be a similarity transformation projective transformation and two figure $F,F'$ such that $f(F)=F'$, then $f(E^F_X)=E^{F'}_{f(X)}$. Symmetry and Weber-Fechner Laws. $f(F)=F'$ mean $f$ sends each element $M_i$ of $F$ partition to an element $M'_i$ of $F'$ partition
   
5. If $X\in E^F_Y$ then $Y\in E^F_X$. We call $X,Y$ indistinguishable.
   
6. There exist $a>1$ such that $\frac{1}{a}<\frac{d(X,M_i)}{d(X',M_i)}<a$ for all $X'\in E^F_X,X\notin M_i$. We have good estimation for all scale.
   
7. There no algorithm that can choose $X$ more exactly. It's quite hard to state formally, so I give an example: Assume Bob find centroid $G$ of triangle $ABC$, instead of finding $G$ by his bare eye, consider this simple algorithm: Bob can find midpoint $M$ of $AB$ then choosing $G$ such that $CG=\frac{2}{3}CM$, of course the point $M$ or $G$ that Bob choose is usually not exactly the midpoint or centroid. Let $E'^F_G$ be a set that contain all point $G'$ that Bob can choose this way. Then $E^F_G\in E'^F_G$. This axiom means human brain is quite perfect
   
   Those axioms are quite complicative, and axiom 7 is weird and hard to state formally. I see that axiom 3,4 can be think as axiom about morphism between figure and property between their error sets, but i can't find a general axiom, but if we can then that axiom may even include some remain axioms, especially axiom 7.
   

Can we state those axioms in more simple and formal way?

We have a trivial solution $E^F_X=\{X\}$, so we want to know is there any non-trivial solution and moreover, the structure of all solution. But it seems very hard to answer that question here so I want to know:

Any method or theory that is related and could be used to solve this problem?

To solve the general case, we should consider the simpliest non-trivial case of this problem: Give a line segment $AB$ and a real number $x\in [0,1]$, find $C$ such that $\frac{AC}{AB}=x$. From axiom 1, $E^F_x=E^F_C$ is a closed segment $[f^-(x),f^+(x)]$. I have writen a lot about the pair of functions $f^-,f^+$ in above post, so I just rewrite the functional inequality here (which I fix a bit):

Find all pair of function $f^-,f^+:[0,1]\rightarrow[0,1]$ such that:
(1) $f^-(0)=f^+(0)=0,f^-(1)=f^+(1)=1$.
(2) $f^-(x)\leq x\leq f^+(x)$.
(3) $f^-(x)+f^+(1-x)=1$.
(4) $f^-(x)f^-(y)\leq f^-(xy)\leq f^-(x)f^+(y)$.
(5) $f^+(x)f^-(y)\leq f^+(xy)\leq f^+(x)f^+(y)$.
(6) $f^-(f^+(x))=f^+(f^-(x))=x$.
(7) $\frac{x}{f^-(x)},\frac{f^+(x)}{x}$ is bound for $x\neq 0$.
for all $x,y\in [0,1].$
(The equation (1) come from axiom 2 and equation (6),(7) ones come from axiom 5,6)

Is there any non-trivial solution of that functional inequality?

Update: I've found a possible approach to the third question. Consider this problem: Give a circle $C$ with radius $1$ have a point $A$ on it. Find a point $X$ on $C$ such that $\overset{\frown}{AX}=2\pi x$. We have a functional inequality like above, but the inequation (4),(5) become: $f^-(x)+f^-(y)\leq f^-(x+y)\leq f^-(x)+f^+(y)\leq f^+(x+y)\leq f^+(x)+f^+(y)$.
It may be easier because it's quite linearity. Then consider the same problem but with two point $A,B$ on circle $C$. It's more complicative but still quite linearity. Then take $A,B$ infinitesimal close, $AB$ become a line segment then we have the solution of the above functional inequality.

New update 15-Mar-2023: In Axiom 4, we should change similarity transformation by projective transformation/homography, which mean the problem should be still the same under different point of view. Here we extend Euclidean space to projective space, and choose suitable projective transformation that not bring our objects to infinity. But the axiom work only in Euclide space.

Generate axiom 4 to arbitrary metric space.

We can now immediately solve the third question by using suitable projective transformation that change the problem to the case $x=\frac{1}{2}$, note that the cross-ration is unchange, by some caculation we have $f^-(x)=\frac{\alpha x}{2\alpha x-x-\alpha+1},f^+(x)=\frac{(1-\alpha)x}{(1-2\alpha)x+\alpha}$ with $0\leq\alpha=f^-(\frac{1}{2})\leq\frac{1}{2}$. It can be check that those functions statisfy conditions (1)-(7).

Solve the general problem in one dimension.